(a) [tex]1.78\cdot 10^{-6}N[/tex]
Here we want to find the gravitational force exerted on the particle at a distance of 3.7 m from the center of the sphere. Since the radius of the sphere is 1.9 m, we are outside the sphere, so we can use Newton's law of gravitation:
[tex]F=G\frac{mM}{r^2}[/tex]
where
G is the gravitational constant
m = 8.3 kg is the mass of the particle
[tex]M=4.4\cdot 10^4 kg[/tex] is the mass of the sphere
r = 3.7 m is the distance
Substituting into the formula, we find
[tex]F=(6.67\cdot 10^{-11})\frac{(8.3 kg)(4.4\cdot 10^4 kg)}{(3.7 m)^2}=1.78\cdot 10^{-6} N[/tex]
(b) [tex]1.46\cdot 10^{-6}N[/tex]
Here we want to find the gravitational force exerted on the particle at a distance of 0.41 m from the center of the sphere. Since the radius of the sphere is 1.9 m, this time we are inside the sphere, so the formula for the gravitational force is different:
[tex]F=G\frac{mMr}{R^3}[/tex]
where
G is the gravitational constant
m = 8.3 kg is the mass of the particle
[tex]M=4.4\cdot 10^4 kg[/tex] is the mass of the sphere
r = 0.41 m is the distance from the centre of the sphere
R = 1.9 m is the radius of the sphere
Substituting numbers into the formula, we find
[tex]F=(6.67\cdot 10^{-11})\frac{(8.3 kg)(4.4\cdot 10^4 kg)(0.41 m)}{(1.9 m)^3}=1.46\cdot 10^{-6}N[/tex]
(c) [tex]F=G\frac{mMr}{R^3}[/tex]
The magnitude of the gravitational force on the particle when located inside the sphere can be found starting from Newton's law of gravitation:
[tex]F=G\frac{mM'}{r^2}[/tex] (1)
where the only difference compared to the standard law is that M' is not the total mass of the sphere, but only the amount of mass of the sphere enclosed by the spherical surface of radius r centered in the center of the sphere.
The mass enclosed is
[tex]M'=\rho V' = \rho (\frac{4}{3}\pi r^3)[/tex] (2)
where [tex]\rho[/tex] is the density of the sphere and V' is the enclosed volume. We can rewrite the density of the sphere as ratio between mass of the sphere (M) and volume of the sphere:
[tex]\rho=\frac{M}{V}=\frac{M}{\frac{4}{3}\pi R^3}[/tex] (3)
where R is the radius of the sphere.
Substituting (3) into (2):
[tex]M' = (\frac{M}{\frac{4}{3}\pi R^3}) (\frac{4}{3}\pi r^3)=\frac{Mr^3}{R^3}[/tex]
And substituting the last equation into (1), we find
[tex]F=G\frac{m(\frac{Mr^3}{R^3})}{r^2}=G\frac{mMr}{R^3}[/tex]
which depends linearly on r.
Which path will the car follow when it leaves the table? A B C D
Answer:
C
Explanation:
Objects in free-fall (also known as projectiles) follow a parabolic curve. So the answer is C.
What vertical distance Δy does a free-falling particle travel from the moment it starts to the moment it reaches a speed of 7.9 m/s if it starts from rest? Work out your solution using one of the equations for vertical motion with constant acceleration, specifically, v2f=v2i+2aΔy, where vi and vf are the particle’s initial and final speeds, respectively, and a is the particle’s acceleration.
Answer:
3.2 m
Explanation:
The equation to use to solve this problem is:
[tex]v_f^2 = v_i^2 + 2 a \Delta y[/tex]
where
[tex]v_f[/tex] is the final velocity
[tex]v_i[/tex] is the initial velocity
a is the acceleration
[tex]\Delta y[/tex] is the distance covered
For the particle in free-fall in this problem, we have
[tex]v_i = 0[/tex] (it starts from rest)
[tex]v_f = 7.9 m/s[/tex]
[tex]g=9.8 m/s^2[/tex] (acceleration due to gravity)
By re-arranging the equation, we can find the distance travelled:
[tex]\Delta y = \frac{v_f^2 -v_i^2}{2a}=\frac{(7.9 m/s)^2-0^2}{2(9.8 m/s^2)}=3.2 m[/tex]
Electromagnetic radiation behaves both as particles (called photons) and as waves. Wavelength (λ) and frequency (ν) are related according to the equation c=λ×ν where c is the speed of light (3.00×108 m/s). The energy (E in joules) contained in one quantum of electromagnetic radiation is described by the equation E=h×ν where h is Planck's constant (6.626×10−34 J⋅s). Note that frequency has units of inverse seconds (s−1), which are more commonly expressed as hertz (Hz). A microwave oven operates at 2.20 GHz . What is the wavelength of the radiation produced by this appliance?
Explanation:
We can find the wavelength of the radiation produced by the microwave oven by using the following given equation:
[tex]c=\lambda.\nu[/tex] (1)
Clearing [tex]\lambda[/tex] :
[tex]\lambda=\frac{c}{\nu}[/tex] (2)
Knowing [tex]\nu=2.20 GHz=2.20(10)^{6}Hz=2.20(10)^{6}s^{-1}[/tex]
[tex]\lambda=\frac{3(10)^{8}m/s}{2.20(10)^{6}s^{-1}}[/tex] (3)
[tex]\lambda=136.363m[/tex] This is the wavelength of the radiation produced by the microwave
The wavelength of the radiation produced by a microwave oven operating at 2.20 GHz is approximately 13.6 cm.
Explanation:Calculating the Wavelength of Microwave Oven Radiation
To find the wavelength of the radiation produced by a microwave oven that operates at 2.20 GHz, we use the formula c = λν, where c is the speed of light (3.00 × 10¸ m/s), λ is the wavelength in meters, and ν is the frequency in hertz (Hz).
First, we convert the frequency from gigahertz (GHz) to hertz (Hz) by multiplying it by 10¹:
2.20 GHz × 10¹ = 2.20 × 10¹ Hz.
Next, we rearrange the formula to solve for λ:
λ = c / ν
Now, we plug in the values:
λ = (3.00 × 10¸ m/s) / (2.20 × 10¹ Hz)
Calculating this gives:
λ = (3.00 × 10¸) / (2.20 × 10¹)
λ = 1.36 × 10² m
Therefore, the wavelength of the radiation emitted by the microwave oven is approximately 13.6 cm.
A block of mass 0.254 kg is placed on top of a light, vertical spring of force constant 5 100 N/m and pushed downward so that the spring is compressed by 0.093 m. After the block is released from rest, it travels upward and then leaves the spring. To what maximum height above the point of release does it rise? (Round your answer to two decimal places.)
Answer:
8.86 m
Explanation:
According to the law of conservation of energy, the elastic potential energy initially stored in the spring will be converted into gravitational potential energy of the block when it is at its maximum height:
[tex]\frac{1}{2}kx^2 = mgh[/tex]
where
k = 5100 N/m is the spring constant
x = 0.093 m is the spring compression
m = 0.254 kg is the mass of the block
g = 9.8 m/s^2 is the acceleration due to gravity
h is the maximum height of the block
Solving the equation for h, we find
[tex]h=\frac{kx^2}{2mg}=\frac{(5100 N/m)(0.093 m)^2}{2(0.254 kg)(9.8 m/s^2)}=8.86 m[/tex]
Zero, a hypothetical planet, has a mass of 5.3 x 1023 kg, a radius of 3.3 x 106 m, and no atmosphere. A 10 kg space probe is to be launched vertically from its surface. (a) If the probe is launched with an initial kinetic energy of 5.0 x 107 J, what will be its kinetic energy when it is 4.0 x 106 m from the center of Zero? (b) If the probe is to achieve a maximum distance of 8.0 x 106 m from the center of Zero, with what initial kinetic energy must it be launched from the surface of Zero?
(a) [tex]3.1\cdot 10^7 J[/tex]
The total mechanical energy of the space probe must be constant, so we can write:
[tex]E_i = E_f\\K_i + U_i = K_f + U_f[/tex] (1)
where
[tex]K_i[/tex] is the kinetic energy at the surface, when the probe is launched
[tex]U_i[/tex] is the gravitational potential energy at the surface
[tex]K_f[/tex] is the final kinetic energy of the probe
[tex]U_i[/tex] is the final gravitational potential energy
Here we have
[tex]K_i = 5.0 \cdot 10^7 J[/tex]
at the surface, [tex]R=3.3\cdot 10^6 m[/tex] (radius of the planet), [tex]M=5.3\cdot 10^{23}kg[/tex] (mass of the planet) and m=10 kg (mass of the probe), so the initial gravitational potential energy is
[tex]U_i=-G\frac{mM}{R}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{3.3\cdot 10^6 m}=-1.07\cdot 10^8 J[/tex]
At the final point, the distance of the probe from the centre of Zero is
[tex]r=4.0\cdot 10^6 m[/tex]
so the final potential energy is
[tex]U_f=-G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{4.0\cdot 10^6 m}=-8.8\cdot 10^7 J[/tex]
So now we can use eq.(1) to find the final kinetic energy:
[tex]K_f = K_i + U_i - U_f = 5.0\cdot 10^7 J+(-1.07\cdot 10^8 J)-(-8.8\cdot 10^7 J)=3.1\cdot 10^7 J[/tex]
(b) [tex]6.3\cdot 10^7 J[/tex]
The probe reaches a maximum distance of
[tex]r=8.0\cdot 10^6 m[/tex]
which means that at that point, the kinetic energy is zero: (the probe speed has become zero):
[tex]K_f = 0[/tex]
At that point, the gravitational potential energy is
[tex]U_f=-G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{8.0\cdot 10^6 m}=-4.4\cdot 10^7 J[/tex]
So now we can use eq.(1) to find the initial kinetic energy:
[tex]K_i = K_f + U_f - U_i = 0+(-4.4\cdot 10^7 J)-(-1.07\cdot 10^8 J)=6.3\cdot 10^7 J[/tex]
A ball has a mass of 0.0935 kg just before it strikes the Earth after being dropped from a building 39.8 m tall. What is its de Broglie wavelength? The acceleration of gravity is 9.8 m/s 2 and Planck’s constant is 6.62607 × 10−34 J · s. Answer in units of m.
Answer:
[tex]2.54\cdot 10^{-34}m[/tex]
Explanation:
First of all, we need to find the final velocity of the ball just before it reaches the ground. Since the ball is in free-fall motion, its final velocity is given by
[tex]v^2 = u^2 +2gh[/tex]
where
v is the final velocity
u = 0 is the initial velocity
g = 9.8 m/s^2 is the acceleration due to gravity
h = 39.8 m is the height of the building
Solving for v,
[tex]v=\sqrt{2gh}=\sqrt{2(9.8 m/s^2)(39.8 m)}=27.9 m/s[/tex]
Now we can calculate the ball's momentum:
[tex]p=mv=(0.0935 kg)(27.9 m/s)=2.61 kg m/s[/tex]
And now we can calculate the De Broglie's wavelength of the ball:
[tex]\lambda = \frac{h}{p}=\frac{6.63\cdot 10^{-34} Js}{2.61 kg m/s}=2.54\cdot 10^{-34}m[/tex]
Bob has just finished climbing a sheer cliff above a beach, and wants to figure out how high he climbed. All he has to use, however, is a baseball, a stopwatch, and a friend on the beach below with a long measuring tape. Bob is a pitcher and he knows that the fastest he can throw the ball is about ????0=34.1 m/s.v0=34.1 m/s. Bob starts the stopwatch as he throws the ball (with no way to measure the ball's initial trajectory), and watches carefully. The ball rises and then falls, and after ????1=0.510 st1=0.510 s the ball is once again level with Bob. Bob cannot see well enough to time when the ball hits the ground. Bob's friend then measures that the ball landed ????=126 mx=126 m from the base of the cliff. How high up is Bob, if the ball started exactly 2 m above the edge of the cliff?
Answer:
56.0 m
Explanation:
We know that after 0.510 s, the ball is level with Bob again. We can use this to find the vertical component of the initial velocity.
y = y₀ + v₀ᵧ t + ½ gt²
h+2 = h+2 + v₀ᵧ (0.510) + ½ (-9.8) (0.510)²
v₀ᵧ = 2.50 m/s
Since the magnitude is 34.1 m/s, we can now find the horizontal component:
v₀² = v₀ₓ² + v₀ᵧ²
(34.1)² = v₀ₓ² + (2.50)²
v₀ₓ = 34.0 m/s
And since we know the ball lands 126 m from the base of the cliff, we can find the time it takes to land:
x = x₀ + v₀ₓ t + ½ at²
126 = 0 + (34.0) t + ½ (0) t²
t = 3.71 s
Finally, we can now find the height of the cliff:
y = y₀ + v₀ᵧ t + ½ gt²
0 = h+2 + (2.50) (3.71) + ½ (-9.8) (3.71)²
h = 56.0 m
Bob climbed approximately 4.4 meters.
Explanation:To determine the height that Bob climbed, we can use the kinematic equation for vertical motion:
Y = Yo + Voy*t -1/2gt^2
where:
Y = final height (unknown)
Yo = initial height (2m)
Voy = initial vertical velocity (unknown)
t = time taken to reach the final height (0.510s)
g = acceleration due to gravity (9.8m/s^2)
We need to find the initial vertical velocity. When the ball reaches the final height, it has the same vertical velocity as it did when it was thrown. Using the equation for vertical velocity:
Vy = Voy - gt
where:
Vy = vertical velocity (0 m/s)
Substituting the known values:
0 = Voy - (9.8)(0.510)
Solving for Voy, we get:
Voy ≈ 5.0 m/s
Now we can calculate the final height:
Y = Yo + Voy*t - 1/2gt^2
Y = 2 + (5.0)(0.510) - 1/2(9.8)(0.510)^2
Y ≈ 2 + 2.55 - 0.126
Y ≈ 4.425m
Therefore, Bob climbed approximately 4.4 meters.
Learn more about determining height climbed here:https://brainly.com/question/26203667
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A spring is stretched from x=0 to x=d, where x=0 is the equilibrium position of the spring. It is then compressed from x=0 to x=−d. What can be said about the energy required to stretch or compress the spring? View Available Hint(s) A spring is stretched from to , where is the equilibrium position of the spring. It is then compressed from to . What can be said about the energy required to stretch or compress the spring? More energy is required to stretch the spring than to compress it. The same amount of energy is required to either stretch or compress the spring. Less energy is required to stretch the spring than to compress it.
Answer:
The same amount of energy is required to either stretch or compress the spring.
Explanation:
The amount of energy required to stretch or compress a spring is equal to the elastic potential energy stored by the spring:
[tex]U=\frac{1}{2}k (\Delta x)^2[/tex]
where
k is the spring constant
[tex]\Delta x[/tex] is the stretch/compression of the spring
In the first case, the spring is stretched from x=0 to x=d, so
[tex]\Delta x = d-0=d[/tex]
and the amount of energy required is
[tex]U=\frac{1}{2}k d^2[/tex]
In the second case, the spring is compressed from x=0 to x=-d, so
[tex]\Delta x = -d -0 = -d[/tex]
and the amount of energy required is
[tex]U=\frac{1}{2}k (-d)^2= \frac{1}{2}kd^2[/tex]
so we see that the amount of energy required is the same.
A single resistor is connected across the terminals of a battery. Which one or more of the following changes in voltage and current leaves unchanged the electric power dissipated in the resistor? (A) Doubling the voltage and reducing the current by a factor of two. (B) Doubling the voltage and increasing the resistance by a factor of four. (C) Doubling the current and reducing the resistance by a factor of four.
Answer:
All of the choices are correct
Explanation:
The power dissipated in a single resistor connected to a battery is given by:
[tex]P=VI = I^2 R=\frac{V^2}{R}[/tex]
where
V is the voltage
I is the current
R is the resistance
Let's analyze each case:
A) Doubling the voltage (V'=2V) and reducing the current by a factor of 2 (I'=I/2). The new power dissipated is:
[tex]P'=V'I'=(2V)(\frac{I}{2})=VI=P[/tex] --> the power is unchanged
B) Doubling the voltage (V'=2V) and increasing the resistance by a factor of 4 (R'=4R). The new power dissipated is:
[tex]P'=\frac{V'^2}{R'}=\frac{(2V)^2}{4R}=\frac{V^2}{R}[/tex] --> the power is unchanged
C) Doubling the current (I'=2I) and reducing the resistance by a factor of four (R'=R/4). The new power dissipated is:
[tex]P'=I'^2 R'=(2I)^2(\frac{R}{4})=I^2 R[/tex] --> the power is unchanged
Final answer:
Option (B), which involves doubling the voltage and increasing the resistance by a factor of four, is the correct scenario where the electric power dissipated in the resistor remains unchanged, explained by the formulas P = V^2/R and P = I^2R.
Explanation:
The question addresses how voltage, current, and resistance affect electric power dissipation in a resistor. Electric power (P) dissipated by a resistor can be calculated using the formula P = V2/R or P = I2R, where V is the voltage across the resistor, I is the current through the resistor, and R is the resistance. To keep the power dissipated unchanged, we need to ensure that any changes in voltage, current, or resistance occur in a way that does not change the final calculation of P.
Option (A) suggests doubling the voltage and reducing the current by a factor of two. This will not keep power dissipated the same due to the direct relationship in the formula P = I2R.
Option (B), doubling the voltage and increasing the resistance by a factor of four, will keep the power dissipated in the resistor unchanged because the power can also be expressed as P = V2/R, and thus doubling V while quadrupling R does not change the P.
Option (C) involves doubling the current and reducing the resistance by a factor of four. According to P = I2R, this change will not keep the power dissipated unchanged as the increase in current will significantly raise power dissipation due to its squared relationship in the formula.
Therefore, Option (B) correctly describes a method to change voltage and resistance in such a way that the power dissipated in the resistor remains unchanged.
A power lifter performs a dead lift, raising a barbell with a mass of 305 kg to a height of 0.42 m above the ground, giving the barbell 1256.66 J of potential energy. The power lifter then releases the barbell, letting it drop towards the ground. Determine the magnitude of the vertical velocity of the barbell when it reaches a height of 0.21 m (on the way down) using a mechanical energy approach. Then calculate the velocity right as it reaches the ground using a mechanical energy approach. Confirm your answer for the velocity of the barbell right before it hits the ground by also calculating this velocity using a projectile motion approach.
Answer:
Explanation:
Before it hits the ground:
The initial potential energy = the final potential energy + the kinetic energy
mgH = mgh + 1/2 mv²
gH = gh + 1/2 v²
v = √(2g (H - h))
v = √(2 * 9.81 m/s² * (0.42 m - 0.21 m))
v ≈ 2.0 m/s
When it hits the ground:
Initial potential energy = final kinetic energy
mgH = 1/2 mv²
v = √(2gH)
v = √(2 * 9.81 m/s² * 0.42 m)
v ≈ 2.9 m/s
Using a kinematic equation to check our answer:
v² = v₀² + 2a(x - x₀)
v² = (0 m/s)² + 2(9.8 m/s²)(0.42 m)
v ≈ 2.9 m/s
Wonder Woman and Superman fly to an altitude of 1690 km , carrying between them a chest full of jewels that they intend to put into orbit around Earth. They want to make this tempting treasure inaccessible to their evil enemies who are trying to gain possession of it, yet keep it available for themselves for future use when they retire and settle down. But perhaps the time to retire is now! They accidentally drop the chest, which leaves their weary hands at rest, and discover that they are no longer capable of catching it as it falls into the Pacific Ocean. At what speed does the chest impact the surface of the water? Ignore air resistance, although in the real world it would make a world of difference. The radius and mass of Earth are 6370 km and 5.98×1024 kg , respectively.
Answer:
5120 m/s
Explanation:
The acceleration due to gravity is:
g = MG / r²
where M is the mass of the earth, G is the universal constant of gravitation, and r is the distance from the earth's center to the object's center.
Here, r = h + R, where h is the height of the chest above the surface and R is the radius of the earth.
g = MG / (h + R)²
Acceleration is the derivative of velocity:
dv/dt = MG / (h + R)²
Using chain rule, we can say:
(dv/dh) (dh/dt) = MG / (h + R)²
(dv/dh) v = MG / (h + R)²
Separate the variables:
v dv = MG / (h + R)² dh
Integrating:
∫₀ᵛ v dv = MG ∫₀ʰ dh / (h + R)²
½ v² |₀ᵛ = -MG / (h + R) |₀ʰ
½ (v² − 0²) = -MG / (h + R) − -MG / (0 + R)
½ v² = -MG / (h + R) + MG / R
½ v² = MGh / (R(h + R))
v² = 2MGh / (R(h + R))
Given:
M = 5.98×10²⁴ kg
R = 6.37×10⁶ m
h = 1.69×10⁶ m
G = 6.67×10⁻¹¹ m³/kg/s²
Plugging in:
v² = 2 (5.98×10²⁴) (6.67×10⁻¹¹) (1.69×10⁶) / ((6.37×10⁶) (1.69×10⁶ + 6.37×10⁶))
v² = 2 (5.98) (6.67) (1.69) / ((6.37) (1.69 + 6.37)) × 10⁷
v ≈ 5120 m/s
Notice that if we had approximated g as a constant 9.8 m/s², we would have gotten an answer of:
v² = v₀² + 2a(x - x₀)
v² = (0 m/s)² + 2 (9.8 m/s²) (1.69×10⁶ m - 0 m)
v ≈ 5760 m/s
So we know that our calculated velocity of 5120 m/s is a reasonable answer.
The Newton's second law and the law of universal gravitation allows to find the result for the speed of the chest when reaching the ocean is:
The velocity is v = 5120 m / s
The law of universal gravitation is stable that the gravitational force between bodies is attractive and is proportional to the mass of the bodies and inversely proportional to the square of the distance.
[tex]F= - G \frac{Mm}{r^2}[/tex]
Where M and m are the mass of the two bodies and r is the distance.
Indicate the height from where the chest falls h= 1690 km = 1,690 10⁶ m, they also give the radius and the mass of the earth.
Newton's second law establishes a relationship between force, mass and the acceleration of bodies.
F = m a
[tex]- G \frac{Mm}{r^2} = m a \\a= - G \frac{M}{r^2}[/tex]
The distance of the body from the center of the planet is
r = R + h
The acceleration is defined as the variation of velocity with time.
[tex]a= \frac{dv}{dt} \\ \frac{dv}{dt} = - G \frac{M}{(h+R)^2}[/tex]
Let's use the chain rule
[tex]\frac{dv}{dh}\ \frac{dh}{dt} = - GM \frac{1}{(h+R)^2 }[/tex]
The velocity is the derivative of the position with respect to the time.
[tex]v=\frac{dh}{dt} \\ v \ \frac{dv}{dh} = - GM \ \frac{1}{(h+R)^2}[/tex]
To solve we use the method of separation of variables and we integrate.
[tex]\int\limits^v_0 {v} \, dv = -GM \int\limits^0_h {\frac{1}{(h+R)^2} } \, dh \\\frac{1}{2} ( v^2 - 0) = -GM (-1) [ \frac{1}{(0+R)} - \frac{1}{(h+R)}] \\\frac{1}{2} v^2 = GM \ \frac{h}{(h+R)R}[/tex]
[tex]v^2 = 2GM \ \frac{h}{(h+R) R }[/tex]
Let's calculate
v² = [tex]2 \ 6.67 \ 10^{-11} \ 5.98 \ 10^{24}} \ \frac{1.690 }{(1.609 + 6.370) 6.370} \ 10^{-6}[/tex]
v = 5120 m/s
In conclusion we use Newton's second law and the universal gravitation's law we can find the result for the speed of the chest when reaching the ocean is;
The velocity is v = 5120 m / s
Learn more about the law of universal gravitation here: brainly.com/question/2347945
can someone help me?!!!!!
Answer:
Explanation:
The first one is false. A vector has a magnitude and a direction; a scalar only has a magnitude. The two cannot be added together.
The second one is false. The magnitude of a vector is found using Pythagorean theorem: c² = a² + b². The only way the magnitude of a vector (c) can be 0 is if both components are 0 (a=0 and b=0).
The third one is true. A vector in Quadrant III will have negative components but can still have a positive magnitude. For example, a vector with magnitude 1 and direction 225° has a positive magnitude and negative components.
The fourth one is false. Rotating a vector will change it.
The fifth one is false. A vector sum can only be 0 if the two vectors have equal magnitudes and opposite directions.
An electron moving at right angles to a 0.14 T magnetic field experiences an acceleration of 6.5×1015 m/ s 2 . What is the electron's speed? Express your answer using two significant figures. By how much does its speed change in 1 ns( 10 −9 s) ?
Answer:
[tex]2.64\cdot 10^5 m/s[/tex], the speed does not change
Explanation:
The magnetic force on the electron is equal to the product between its mass and its acceleration:
[tex]qvB = ma[/tex]
where
q is the electron charge
v is the electron speed
B = 0.14 T is the magnetic field
m is the electron's mass
[tex]a=6.5\cdot 10^{15}m/s^2[/tex] is the acceleration (centripetal acceleration)
Solving for v, we find
[tex]v=\frac{ma}{qB}=\frac{(9.11\cdot 10^{-31} kg)(6.5\cdot 10^{15} m/s^2)}{(1.6\cdot 10^{-19} C)(0.14 T)}=2.64\cdot 10^5 m/s[/tex]
The speed of the electron does not change, because the acceleration is a centripetal acceleration, so it acts perpendicular to the direction of motion of the electron; therefore, no work is done on the electron by the magnetic force, and therefore the electron does not gain kinetic energy, which means that its speed does not change.
Which of the following is a reason why microbes were found to be problematic?
They were found everywhere.
They could cause infection.
They killed specific bacteria.
They did not respond to antibiotics.
Answer:
I believe the answer is B.
Explanation:
Microbes could cause infectious diseases like for Ex: Flu and measles or heart diseases.
Hope my answer has helped you! :)
Microbes are found to be problematic for two reasons that they cause infection and sometime these organisms does not respond to antibiotics.
Explanation:
The microbes can cause infections in humans is a problem. Fungi, helminths and protozoans are the major cause of infectious diseases. Resistance to an antibiotic occurs when an ability to defeat drugs created to kill them are developed in a microbe. When a microbe becomes resistant, the antibiotics cannot fight against them and hence they multiply.
Antibiotic resistance is an urgent threat to the health of the public. They become a threat, when a microbe become resistant to antibiotic and it will be very difficult to destroy them. The resistance to antibiotics can cause serious issues like disability or even cause death.
can someone help me?!!!!!
Answer:
56 m/s
Explanation:
The time we are considering is
t = 15 s
The vertical velocity of the projectile is given by
[tex]v_y(t) = v_{0y}-gt[/tex]
where
[tex]v_{0y}=100 m/s[/tex] is the initial vertical velocity
[tex]g=9.8 m/s^2[/tex] is the acceleration due to gravity
Substituting t=15 s, we find the vertical velocity of the projectile at that time:
[tex]v_y = 100 m/s - (9.8 m/s^2)(15 s)=-47 m/s[/tex]
where the negative sign means the direction is now downward.
The horizontal velocity does not change since there are no forces acting along that direction, so it remains constant:
[tex]v_x = 30 m/s[/tex]
So, the magnitude of the velocity at the moment of impact is
[tex]v=\sqrt{v_x^2 +v_y^2}=\sqrt{(30 m/s)^2+(-47 m/s)^2}=55.8 m/s \sim 56 m/s[/tex]
Which one of the following statements concerning waves is false? A transverse wave is one in which the disturbance is parallel to the direction of travel. A wave can have both transverse and longitudinal components. A wave carries energy from one place to another. A wave does not result in the bulk flow of the material of its medium. A wave is a traveling disturbance.
Answer:
A transverse wave is one in which the disturbance is parallel to the direction of travel
Explanation:
There are two types of waves:
- Transverse waves: in transverse waves, the disturbance (oscillation) occurs in a plane perpendicular to the direction of propagation of the wave
- Longitudinal waves: in longitudinal waves, the disturbance (oscillation) occurs parallel to the direction of propagation of the wave
Therefore, we immediately see that the statement:
"A transverse wave is one in which the disturbance is parallel to the direction of travel"
is wrong, because it is actually the opposite: in a transverse wave, the disturbance is perpendicular to the direction of travel.
The false statement is that a transverse wave has a disturbance parallel to its direction of travel. A transverse wave has the disturbance perpendicular to its direction of travel, not parallel. An example of this would be the waves on a stringed instrument, contrasted with sound waves which are longitudinal (disturbance and wave travel in the same direction).
Explanation:The false statement among the given options is: "A transverse wave is one in which the disturbance is parallel to the direction of travel." This is incorrect because in a transverse wave, the disturbance or oscillation is perpendicular to the direction of wave propagation. For example, the movement of a stringed instrument creates transverse waves, where the strings vibrate upwards and downwards while the wave travels horizontally along the string.
On the other hand, in a longitudinal wave, the disturbance is parallel to the direction of wave propagation. An example would be sound waves, where the air particles vibrate back and forth, in the same direction that the wave propagates.
It's also correct that waves can have both transverse and longitudinal components, such as seismic waves from earthquakes. Furthermore, a wave does carry energy from one location to another but does not result in the bulk movement of the material medium.
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Spider-Man and Ned were testing the distance he could shoot his web depending on the angle at which he points his web shooter. He tested this 3 times at each angle.
0°: 18.2m, 16.7m, and 17.9m
45°: 79.6m, 74.3m, and 76.2m
75°: 62.4m, 61.5m, and 62.7m
What is the Independent variable?
Question 1 options:
The distance spider-man’s web shoots
The angle at which spider-man points his web shooter
Testing each angle 3 times
The location where they are testing
What is the dependent variable?
Question 2 options:
The distance spider-man’s web shoots
The angle at which spider-man points his web shooter
Testing each angle 3 times
The location where they are testing
Look at the 1st graph below, which location has the warmest temperatures in July?
Question 3 options: See Picture below
Barrow, Ak
Christchurch, NZ
Quito, Equador
Madison, WI
According to the 1st graph below, which location does not experience seasonal temperature fluctuations?
Question 4 options:
Barrow, AK
Quito, Equador
Christchurch, NZ
Madison, WI
Use the 2nd chart below to answer the following question. If cool stars are red and hot stars are blue, where would a bright (highly luminous) blue star be found on a Hertzsprung-Russell diagram?
Question 5 options:
Upper right
Lower right
Upper left
Lower left
A group of scientists has developed a computer program that will predict the phase that a star will be after a certain number of years. The scientists used data from real stars to teach the computer how the universe works. Now, it can be used to quickly run a simulation of what would likely happen over billions of years in the universe.
A group of students is wondering the rate at which a star progresses through life. They plan two experiments using computer simulations. They ran each simulation from the protostar phase 100 times and recorded the outcomes.
In Experiment 1 they used one size of star (1 solar mass) and compared seven different lengths of time. In Experiment 2 they used one length of time and 3 different sizes of stars (1, 15, and 30 solar masses). The results of both experiments are below in Figure 2
According to the 3rd data chart in which is the best conclusion of the changes occurring from 5-10 billion years?
Question 6 options:
Main sequence stars are becoming red giants.
Protostars are becoming red giants.
The stars are becoming black holes.
The stars are exploding.
Answer:
3. Madison, WI
4. Quito, Equador
6. Main Sequence stars are becoming red giants
Explanation:
Answer:
1) The angle at which spider-man points his web shooter
2) The distance spider-man’s web shoots
3) Madison, WI
4) Quito, Equador
5) Upper left
6) Main sequence stars are becoming red giants.
Explanation:
1) The experimenters are choosing the angle at which to point the web shooter. This means the angle at which they shoot is independent of the other variables.
2) The distance the web shoots is not chosen by the experimenters and depends on the angle at which it is shot.
3) July is the 7th month and the graph for Madison, WI shows temperatures exceeding all other locations at this time.
4) Fluctuations refer to a changing of temperatures over seasons. Since Quito, Equador has a flat graph, this indicates no fluctuations.
5) This graph plots the hottest (blue) stars on the left and the highest luminosity (bright) stars at the top. Thus, bright blue stars are plotted in the top left.
6) At 5 billion years there are 0 giants and 100 main sequence stars. At 10 billions years there are 94 giants and 6 main sequence stars. This shows that the main sequence stars are becoming giants.
A runner is moving at a constant speed of 8.00 m/s around a circular track. If the distance from the runner to the center of the track is 28.2 m, what is the centripetal acceleration of the runner?
1.13 m/s2
0.284 m/s2
3.53 m/s2
2.27 m/s2
Answer: Last option
2.27 m/s2
Explanation:
As the runner is running at a constant speed then the only acceleration present in the movement is the centripetal acceleration.
If we call a_c to the centripetal acceleration then, by definition
[tex]a_c =w^2r = \frac{v^2}{r}[/tex]
in this case we know the speed of the runner
[tex]v =8.00\ m/s[/tex]
The radius "r" will be the distance from the runner to the center of the track
[tex]r = 28.2\ m[/tex]
[tex]a_c = \frac{8^2}{28.2}\ m/s^2[/tex]
[tex]a_c = 2.27\ m/s^2[/tex]
The answer is the last option
Amy is standing still on the ground; Bill is riding his bicycle at 5 m/s eastward; and Carlos is driving his car at 15 m/s westward. How fast does Bill see Carlos moving and in what direction?
Answer:
20 m/s westward
Explanation:
Taking eastward as positive direction, we have:
[tex]v_B = +5 m/s[/tex] is the velocity of Bill with respect to Amy (which is stationary)
[tex]v_C = -15 m/s[/tex] is the velocity of Carlos with respect to Amy
Bill is moving 5 m/s eastward compared to Amy at rest, so the velocity of Bill's reference frame is
[tex]v_B = +5 m/s[/tex]
Therefore, Carlos velocity in Bill's reference frame will be
[tex]v_C' = v_C - v_B = -15 m/s - (+5 m/s) = -20 m/s[/tex]
and the direction will be westward (negative sign).
Consider an electron with charge −e and mass m orbiting in a circle around a hydrogen nucleus (a single proton) with charge +e. In the classical model, the electron orbits around the nucleus, being held in orbit by the electromagnetic interaction between itself and the protons in the nucleus, much like planets orbit around the sun, being held in orbit by their gravitational interaction. When the electron is in a circular orbit, it must meet the condition for circular motion: The magnitude of the net force toward the center, Fc, is equal to mv 2/r. Given these two pieces of information, deduce the velocity v of the electron as it orbits around the nucleus.
Final answer:
In the classical model of the hydrogen atom, the velocity of the electron as it orbits around the nucleus can be deduced using the equation for the centripetal force. The magnitude of the net force towards the center, Fc, is equal to mv^2/r, and solving for v gives the velocity of the electron.
Explanation:
In the classical model of the hydrogen atom, the electron orbits the proton in a circular path. The magnitude of the net force towards the center, also known as the centripetal force, is equal to the mass of the electron times its velocity squared divided by the radius of the orbit (Fc = mv^2 / r). In this case, we have the mass of the electron, the radius of the orbit, and we need to find the velocity of the electron as it orbits around the nucleus.
Using the given equation of the centripetal force, we can rearrange it to solve for the velocity (v). So, v = sqrt(Fc * r / m). Plugging in the values for the mass of the electron, the radius of the orbit, and the known value of the centripetal force, we can calculate the velocity.
For example, if the radius of the orbit is 5.28 x 10^-11 m and the mass of the electron is 9.11 x 10^-31 kg, the velocity of the electron would be v = sqrt(Fc * r / m) = sqrt(m * v^2 / r * r / m) = v = sqrt(v^2) = v = 2.18 x 10^6 m/s.
Nail tips exert tremendous pressures when they are hit by hammers because they exert a large force over a small area. What force (in N) must be exerted on a nail with a circular tip of 1.15 mm diameter to create a pressure of 2.63 ✕ 109 N/m2? (This high pressure is possible because the hammer striking the nail is brought to rest in such a short distance.)
Answer:
2780 N
Explanation:
Pressure is defined as the ratio between the force applied and the area of the surface:
[tex]p=\frac{F}{A}[/tex]
Here we know the pressure:
[tex]p=2.63 \cdot 10^9 N/m^2[/tex]
we also know the diameter of the tip, d = 1.15 mm, so we can calculate the radius
[tex]r=\frac{1.15 mm}{2}=0.58 mm = 5.8\cdot 10^{-4} m[/tex]
and so the area
[tex]A=\pi r^2 = \pi (5.8\cdot 10^{-4} m)^2=1.057\cdot 10^{-6} m^2[/tex]
And so we can re-arrange the equation to find the force:
[tex]F=pA=(2.63\cdot 10^9 N/m^2)(1.057\cdot 10^{-6} m^2)=2780 N[/tex]
We can calculate the necessary force to drive the nail home by first calculating the area of the nail tip using its given diameter then using the given pressure and the formula for pressure (P = F/A) to solve for the force.
Explanation:To calculate the force required to drive the nail we need to use the equation for pressure which is pressure = force/area. Pressure (P) is given as 2.63 x 109 N/m2. The area (A) can be calculated using the formula for the area of a circle which is A = πr2. The radius (r) of the nail tip can be calculated from its diameter (1.15 mm divided by 2) making sure to convert the units to meters.
After you derive the area, you will use it to calculate the force (F). Rearranging the formula for pressure to solve for force gives F = P * A. This will give the force necessary to drive the nail home exerting the stated pressure.
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A compact disc (CD) stores music in a coded pattern of tiny pits 10−7m deep. The pits are arranged in a track that spirals outward toward the rim of the disc; the inner and outer radii of this spiral are 25.0 mm and 58.0 mm, respectively. As the disc spins inside a CD player, the track is scanned at a constant linear speed of 1.25 m/s. Part A What is the angular speed of the CD when scanning the innermost part of the track? B What is the angular speed of the CD when scanning the outermost part of the track? C The maximum playing time of a CD is 74.0 min. What would be the length of the track on such a maximum-duration CD if it were stretched out in a straight line? D What is the average angular acceleration of a maximum-duration CD during its 74.0-min playing time? Take the direction of rotation of the disc to be positive.
(a) 50 rad/s
The angular speed of the CD is related to the linear speed by:
[tex]\omega=\frac{v}{r}[/tex]
where
[tex]\omega[/tex] is the angular speed
v is the linear speed
r is the distance from the centre of the CD
When scanning the innermost part of the track, we have
v = 1.25 m/s
r = 25.0 mm = 0.025 m
Therefore, the angular speed is
[tex]\omega=\frac{1.25 m/s}{0.025 m}=50 rad/s[/tex]
(b) 21.6 rad/s
As in part a, the angular speed of the CD is given by
[tex]\omega=\frac{v}{r}[/tex]
When scanning the outermost part of the track, we have
v = 1.25 m/s
r = 58.0 mm = 0.058 m
Therefore, the angular speed is
[tex]\omega=\frac{1.25 m/s}{0.058 m}=21.6 rad/s[/tex]
(c) 5550 m
The maximum playing time of the CD is
[tex]t =74.0 min \cdot 60 s/min = 4,440 s[/tex]
And we know that the linear speed of the track is
v = 1.25 m/s
If the track were stretched out in a straight line, then we would have a uniform motion, therefore the total length of the track would be:
[tex]d=vt=(1.25 m/s)(4,440 s)=5,550 m[/tex]
(d) [tex]-6.4\cdot 10^{-3} rad/s^2[/tex]
The angular acceleration of the CD is given by
[tex]\alpha = \frac{\omega_f - \omega_i}{t}[/tex]
where
[tex]\omega_f = 21.6 rad/s[/tex] is the final angular speed (when the CD is scanned at the outermost part)
[tex]\omega_i = 50.0 rad/s[/tex] is the initial angular speed (when the CD is scanned at the innermost part)
[tex]t=4440 s[/tex] is the time elapsed
Substituting into the equation, we find
[tex]\alpha=\frac{21.6 rad/s-50.0 rad/s}{4440 s}=-6.4\cdot 10^{-3} rad/s^2[/tex]
Answer: (a) 50 rad/s, (b) 21.6 rad/s, (c) 5550 m, (d) [tex]-6.4\cdot 10^{-3} rad/s^2[/tex]
Explanation:A compact disc (CD) stores music in a coded pattern of tiny pits 10−7m deep. The pits are arranged in a track that spirals outward toward the rim of the disc; the inner and outer radii of this spiral are 25.0 mm and 58.0 mm, respectively. As the disc spins inside a CD player, the track is scanned at a constant linear speed of 1.25 m/s
Part
A What is the angular speed of the CD when scanning the innermost part of the track?
B What is the angular speed of the CD when scanning the outermost part of the track?
C The maximum playing time of a CD is 74.0 min. What would be the length of the track on such a maximum-duration CD if it were stretched out in a straight line?
D What is the average angular acceleration of a maximum-duration CD during its 74.0-min playing time?
(a) 50 rad/s
[tex]\omega=\frac{v}{r}[/tex]
where
[tex]\omega[/tex] is the angular speed
v is the linear speed, v = 1.25 m/s
r is the distance from the centre of the CD, r = 25.0 mm = 0.025 m
Therefore, the angular speed
[tex]\omega=\frac{1.25 m/s}{0.025 m}=50 rad/s[/tex]
(b) 21.6 rad/s
The angular speed of the CD is
[tex]\omega=\frac{v}{r}[/tex]
When scanning the outermost part of the track
v = 1.25 m/s
r = 58.0 mm = 0.058 m
Therefore, the angular speed is
[tex]\omega=\frac{1.25 m/s}{0.058 m}=21.6 rad/s[/tex]
(c) 5550 m
[tex]t =74.0 min \cdot 60 s/min = 4,440 s[/tex]
the linear speed of the track is v = 1.25 m/s
the total length of the track would be:
d=vt=(1.25 m/s)(4,440 s)=5,550 m
(d) [tex]-6.4\cdot 10^{-3} rad/s^2[/tex]
The angular acceleration of the CD is given by[tex]\alpha = \frac{\omega_f - \omega_i}{t}[/tex]
where
[tex]\omega_f = 21.6 rad/s[/tex] is the final angular speed (when the CD is scanned at the outermost part)
[tex]\omega_i = 50.0 rad/s[/tex] is the initial angular speed (when the CD is scanned at the innermost part)
[tex]t=4440 s[/tex] is the time elapsed
Substituting into the equation, we find
[tex]\alpha=\frac{21.6 rad/s-50.0 rad/s}{4440 s}=-6.4\cdot 10^{-3} rad/s^2[/tex]
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A projectile is fired up into the air at a speed of 165 m/s at an angle of 75° relative to the horizontal. Ignore air drag. A.) Determine the MAXIMUM HEIGHT (above where it was thrown) the projectile will reach. B.) Determine how LONG the projectile will be in the air. Assume it lands at the same height it was launched. C.)Assuming it lands at the same height it was launched at, determine how FAR the projectile will land from where it was launched. D.) Assuming it lands at the same height it was launched at, determine how FAST the projectile will be going upon impact. E.) Assuming it lands at the same height it was launched at, determine the ANGLE (relative to the horizontal) of impact. °
A) 1296.3 m
The initial velocity of the projectile is
[tex]u = 165 m/s[/tex]
and the angle is
[tex]\theta=75^{\circ}[/tex]
So, we can find the initial vertical velocity of the projectile, which is given by
[tex]u_y = u sin \theta = (165 m/s)sin 75^{\circ}=159.4 m/s[/tex]
The motion of the projectile along the vertical axis is a uniformly accelerated motion, with constant acceleration
[tex]g=-9.8 m/s^2[/tex]
where the negative sign means the direction is towards the ground. The maximum height is reached when the vertical velocity becomes zero: therefore, we can use the following SUVAT equation
[tex]v_y ^2 - u_y^2 = 2gh[/tex]
where
[tex]v_y = 0[/tex] is the final vertical velocity
h is the maximum height
Solving for h, we find
[tex]h=-\frac{-u_y^2}{2g}=\frac{-(159.4 m/s)^2}{2(-9.8 m/s^2)}=1296.3 m[/tex]
B) 32.5 s
In order to determine how long the projectile will be in the air, we need to find the time t at which the projectile reaches the ground.
We can find it by analyzing the vertical motion only. The vertical position at time t is given by
[tex]y(t) = u_y t + \frac{1}{2}gt^2[/tex]
By substituting y(t) = 0, we find the time at which the projectile reaches the ground. We have:
[tex]0= u_y t + \frac{1}{2}gt^2\\0 = t(u_y + \frac{1}{2}gt)[/tex]
which has two solutions:
t = 0 --> beginning of the motion
[tex]u_y + \frac{1}{2}gt=0\\t=-\frac{2u_y}{g}=-\frac{2(159.4 m/s)}{-9.8 m/s^2}=32.5 s[/tex]
C) 1387.8 m
The range of the projectile can be found by analyzing the horizontal motion only.
In fact, the projectile travels along the horizontal direction by uniform motion, with constant horizontal velocity, given by:
[tex]u_x = u cos \theta = (165 m/s) cos 75^{\circ}=42.7 m/s[/tex]
So, the horizontal position at time t is given by
[tex]x(t) = u_x t[/tex]
If we substitute
t = 32.5 s
which is the time at which the projectile reaches the ground, we can find the total horizontal distance covered by the projectile.
So we have:
[tex]x= u_x t = (42.7 m/s)(32.5 s)=1387.8 m[/tex]
D) 165 m/s
The speed of the projectile consists of two independent components:
- The horizontal velocity, which is constant during the motion, and which is equal to
[tex]v_x = u_x = 42.7 m/s[/tex]
- The vertical velocity, which changes during the motion, given by
[tex]v_y = u_y + gt[/tex]
Substituting
[tex]u_y = 159.4 m/s[/tex]
and
[tex]t=32.5 s[/tex]
We find the vertical velocity when the projectile reaches the ground
[tex]v_y = 159.4 m/s + (-9.8 m/s^2)(32.5 s)=-159.4 m/s[/tex]
which is the same as the initial vertical velocity, but with opposite direction.
Now that we have the two components, we can calculate the actual speed of the projectile:
[tex]v=\sqrt{v_x^2+v_y^2}=\sqrt{(42.7 m/s)^2+(-159.4 m/s)^2}=165 m/s[/tex]
and the final speed is exactly equal to the initial speed, since according to the conservation of energy, the projectile has lost no energy during the motion.
E) [tex]-75^{\circ}[/tex]
The angle of impact is given by
[tex]\theta = tan^{-1} (\frac{|v_y|}{v_x})[/tex]
where
[tex]|v_y| = 159.4 m/s[/tex] is the final vertical velocity
[tex]v_x = 42.7 m/s[/tex] is the final horizontal velocity
We have taken the absolute value of [tex]v_y[/tex] since [tex]v_y[/tex] is negative; this means that the resulting angle will be BELOW the horizontal. So we have:
[tex]\theta = tan^{-1} (\frac{159.4 m/s}{42.7 m/s})=75^{\circ}[/tex]
which means [tex]-75.0^{\circ}[/tex], below the horizontal.
The temperature of an object is changed when heat is added to or extracted from it. Determine the final temperature (in °C) of a 65.0 g mass of aluminum initially at 60.0°C if 1,050 J of heat energy is extracted from it. The specific heat of aluminum is 900 J/(kg · °C).
Answer:
[tex]42.1^{\circ}C[/tex]
Explanation:
The heat extracted from the aluminium is given by:
[tex]Q=mC_s (T_f-T_i)[/tex]
where we have
Q = -1050 J is the heat extracted
m = 65.0 g = 0.065 kg is the mass
Cs = 900 J/kgC is the specific heat of aluminum
[tex]T_i = 60.0^{\circ}[/tex] is the initial temperature
By solving for [tex]T_f[/tex], we find the final temperature:
[tex]T_f = \frac{Q}{m C_s}+T_i=\frac{-1050 J}{(0.065 kg)(900 J/kg C)}+60.0^{\circ}C=42.1^{\circ}C[/tex]
To find the final temperature, you use the formula for heat transfer, Q = mcΔT, and specific heat value. Given that 1050J of heat is extracted, which is negative heat transfer (-1050 J), plug in the values to get ΔT and then add this to the initial temperature to get the final temperature.
Explanation:The subject of the question is the concept of heat transfer and specific heat in Physics. The specific heat of a substance is the amount of heat required to raise one gram of the substance by one degree Celsius, which is a property intrinsic to the substance. Different substances require different amounts of heat to change their temperature.
To find the final temperature, you use the formula for heat transfer, Q = mcΔT, where Q is heat transferred, m is mass, c is the specific heat, and ΔT is the change in temperature (final temperature - initial temperature). Here, we're asked to find the final temperature of a 65 g (or 0.065 kg) mass of aluminum initially at 60.0°C when 1050 J of heat energy is extracted.
The sign of Q needs to represent that heat energy is extracted from the aluminum, so Q is -1050 J. Now, you can plug in the values and solve the equation for ΔT first, then add the initial temperature to find the final temperature.
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A circular loop of wire with a diameter of 0.626 m is rotated in a uniform electric field to a position where the electric flux through the loop is a maximum. At this position, the electric flux is 7.50 × 105 N⋅m2/C. Determine the magnitude of the electric field. A) 8.88 × 105 N/C B) 1.07 × 106 N/C C) 2.44 × 106 N/C D) 4.24 × 106 N/C E) 6.00 × 106 N/C
Answer:
C) 2.44 × 106 N/C
Explanation:
The electric flux through a circular loop of wire is given by
[tex]\Phi = EA cos \theta[/tex]
where
E is the electric field
A is the cross-sectional area
[tex]\theta[/tex] is the angle between the direction of the electric field and the normal to A
The flux is maximum when [tex]\theta=0^{\circ}[/tex], so we are in this situation and therefore [tex]cos \theta =1[/tex], so we can write
[tex]\Phi = EA[/tex]
Here we have:
[tex]\Phi = 7.50\cdot 10^5 N/m^2 C[/tex] is the flux
d = 0.626 m is the diameter of the coil, so the radius is
r = 0.313 m
and so the area is
[tex]A=\pi r^2 = \pi (0.313 m)^2=0.308 m^2[/tex]
And so, we can find the magnitude of the electric field:
[tex]E=\frac{\Phi}{A}=\frac{7.50\cdot 10^5 Nm^2/C}{0.308 m^2}=2.44\cdot 10^6 N/C[/tex]
CP Global Positioning System (GPS). The GPS network consists of 24 satellites, each of which makes two orbits around the earth per day. Each satellite transmits a 50.0-W (or even less) sinusoidal electromagnetic signal at two frequencies, one of which is 1575.42 MHz. Assume that a satellite transmits half of its power at each frequency and that the waves travel uni- formly in a downward hemisphere. (a) What average intensity does a GPS receiver on the ground, directly below the satellite, receive? (Hint: First use Newton’s laws to find the altitude of the satellite.) (b) What are the amplitudes of the electric and magnetic fields at the GPS receiver in part (a), and how long does it take the signal to reach the receiver? (c) If the receiver is a square panel 1.50 cm on a side that absorbs all of the beam, what average pres- sure does the signal exert on it? (d) What wavelength must the receiver be tuned to?
(a) [tex]9.66\cdot 10^{-15} W/m^2[/tex]
First of all, we need to find the altitude of the satellite. The gravitational attraction between the Earth and the satellite is equal to the centripetal force that keeps the satellite in circular motion, so
[tex]G\frac{mM}{r^2}=m\omega^2 r[/tex] (1)
where
G is the gravitational constant
m is the satellite's mass
M is the earth's mass
r is the distance of the satellite from the Earth's centre
[tex]\omega[/tex] is the angular frequency of the satellite
The satellite here makes two orbits around the Earth per day, so its frequency is
[tex]\omega = \frac {2 \frac{rev}{day}}{24 \frac{h}{day} \cdot 60 \frac{min}{h}} \cdot \frac{2\pi rad/rev}{s/min}=1.45\cdot 10^{-4} rad/s[/tex]
And by solving eq.(1) for r, we find
[tex]r=\sqrt[3]{\frac{GM}{\omega^2}}=\sqrt[3]{\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}kg)}{(1.45\cdot 10^{-4} rad/s)^2}}=2.67\cdot 10^{7} m[/tex]
The radius of the Earth is
[tex]R=6.37\cdot 10^6 m[/tex]
So the altitude of the satellite is
[tex]h=r-R=2.67\cdot 10^7 m-6.37\cdot 10^6m=2.03\cdot 10^7 m[/tex]
The average intensity received by a GPS receiver on the Earth will be given by
[tex]I=\frac{P}{A}[/tex]
where
P = 50.0 W is the power
A is the area of a hemisphere, which is:
[tex]A=4\pi h^2 = 4 \pi (2.03\cdot 10^7 m)^2=5.18\cdot 10^{15} m^2[/tex]
So the intensity is
[tex]I=\frac{50.0 W}{5.18\cdot 10^{15}m^2}=9.66\cdot 10^{-15} W/m^2[/tex]
(b) [tex]2.70\cdot 10^{-6} V/m, 9.0\cdot 10^{-15}T[/tex], 0.068 s
The relationship between average intensity of an electromagnetic wave and amplitude of the electric field is
[tex]I=\frac{1}{2}c\epsilon_0 E^2[/tex]
where
c is the speed of light
[tex]\epsilon_0[/tex] is the vacuum permittivity
E is the amplitude of the electric field
Solving for E, we find
[tex]E=\sqrt{\frac{2I}{c\epsilon_0}}=\sqrt{\frac{2(9.66\cdot 10^{-15} W/m^2)}{(3\cdot 10^8 m/s)(8.85\cdot 10^{-12}F/m)}}=2.70\cdot 10^{-6} V/m[/tex]
Instead, the amplitude of the magnetic field is given by:
[tex]B=\frac{E}{c}=\frac{2.70\cdot 10^{-6} V/m}{3\cdot 10^8 m/s}=9.0\cdot 10^{-15}T[/tex]
The signal travels at the speed of light, so the time it takes to reach the Earth is the distance covered divided by the speed of light:
[tex]t=\frac{h}{c}=\frac{2.03\cdot 10^7 m}{3\cdot 10^8 m/s}=0.068 s[/tex]
(c) [tex]3.22\cdot 10^{-23}Pa[/tex]
In case of a perfect absorber (as in this case), the radiation pressure exerted by an electromagnetic wave on a surface is given by
[tex]p=\frac{I}{c}[/tex]
where
I is the average intensity
c is the speed of light
In this case, we have
[tex]I=9.66\cdot 10^{-15} W/m^2[/tex]
So the average pressure is
[tex]p=\frac{9.66\cdot 10^{-15} W/m^2}{3\cdot 10^8 m/s}=3.22\cdot 10^{-23}Pa[/tex]
(d) 0.190 m
The wavelength of the receiver must be tuned to the same wavelength as the transmitter (the satellite), which is given by
[tex]\lambda=\frac{c}{f}[/tex]
where
c is the speed of light
f is the frequency of the signal
For the satellite in the problem, the frequency is
[tex]f=1575.42 MHz=1575.42\cdot 10^6 Hz[/tex]
So the wavelength of the signal is:
[tex]\lambda=\frac{3.0\cdot 10^8 m/s}{1575.42 \cdot 10^6 Hz}=0.190 m[/tex]
Ultraviolet (UV) radiation is a part of the electromagnetic spectrum that reaches the Earth from the Sun. It has wavelengths shorter than those of visible light, making it invisible to the naked eye. These wavelengths are classified as UVA, UVB, or UVC, with UVA the longest of the three at 317 nm to 400 nm. Both the U.S. Department of Health and Human Services and the World Health Organization have identified UV as a proven human carcinogen. Many experts believe that, especially for fair-skinned people, UV radiation frequently plays a key role in melanoma, the deadliest form of skin cancer, which kills more than 8000 Americans each year. UVB has a wavelength between 280 nm and 317 nm. Determine the frequency ranges of UVA and UVB. UVA Hz (smaller value) Hz (larger value) UVB Hz (smaller value) Hz (larger value)
Final answer:
To determine the frequency ranges for UVA and UVB, we use the equation f = c/λ. The UVA frequency range is 7.5 × 10^14 Hz to 9.375 × 10^14 Hz, and the UVB frequency range is 9.375 × 10^14 Hz to 1.0345 × 10^15 Hz.
Explanation:
The frequency of electromagnetic radiation, including UV radiation, can be calculated using the formula c = λf, where c is the speed of light (approximately 3 × 10^8 m/s), λ is the wavelength, and f is the frequency. To find the frequency ranges for UVA and UVB, we can rearrange the equation to f = c/λ.
Calculating UVA Frequency
For UVA with a wavelength range of 320-400 nm (or 3.2 × 10^-7 m - 4 × 10^-7 m), we use the formula to calculate the frequency as:
• Lower frequency limit: f = (3 × 10^8 m/s) / (4 × 10^-7 m) = 7.5 × 10^14 Hz
• Upper frequency limit: f = (3 × 10^8 m/s) / (3.2 × 10^-7 m) = 9.375 × 10^14 Hz
Calculating UVB Frequency
For UVB with a wavelength range of 290-320 nm (or 2.9 × 10^-7 m - 3.2 × 10^-7 m), the frequency range is:
• Lower frequency limit: f = (3 × 10^8 m/s) / (3.2 × 10^-7 m) = 9.375 × 10^14 Hz
• Upper frequency limit: f = (3 × 10^8 m/s) / (2.9 × 10^-7 m) = 1.0345 × 10^15 Hz
If a wheel falls from an airplane that is flying horizontally at an altitude of 500 m, how long will it take for the wheel to strike the ground?
10 s
50 s
80 s
100 s
A rifle bullet with mass 8.00 g strikes and embeds itself in a block with mass 0.992 kg that rests on a frictionless, horizontal surface and is attached to a coil spring. The impact compresses the spring 15.0 cm. Calibration of the spring shows that a force of 0.750 N is required to compress the spring 0.250 cma) Find the magnitude of the block's velocity just after impact. b) What was the initial speed of the bullet?
Final Answer:
The magnitude of the block's velocity just after impact is 0.008 times the initial velocity of the bullet. The initial speed of the bullet is 1.25 m/s.
Explanation:
To solve this problem, we can use the principle of conservation of momentum. The mass of the bullet is 8.00 g, or 0.008 kg, and the mass of the block is 0.992 kg. Let's denote the velocity of the block just after impact as V and the initial velocity of the bullet as u. The equation for conservation of momentum is:
m_bullet * u = (m_bullet + m_block) * V
Substituting the given values, we have:
0.008 kg * u = (0.008 kg + 0.992 kg) * V
0.008 u = 1 * V
Now, let's find the magnitude of the block's velocity just after impact:
V = 0.008 u
Given that the impact compresses the spring 15.0 cm, which requires a force of 0.750 N, we can use Hooke's Law to find the spring constant:
Force = spring constant * displacement
0.750 N = k * 0.150 m
k = 5 N/m
Now, to find the initial speed of the bullet, we can use the principle of conservation of mechanical energy. The energy stored in the spring when compressed 0.250 cm is given as 5.0 J. Let's denote the initial speed of the bullet as v_i. The equation for conservation of mechanical energy is:
0.5 * k * (0.250 m)^2 = 0.5 * m_bullet * (v_i)^2
Substituting the given values, we have:
0.5 * 5 N/m * (0.0025 m^2) = 0.5 * 0.008 kg * (v_i)^2
0.00625 J = 0.004 kg * (v_i)^2
(v_i)^2 = 0.00625 J / 0.004 kg
(v_i)^2 = 1.5625 m^2/s^2
v_i = 1.25 m/s
a) The final velocity of the block is approximately 0.0000808 * v_bullet.
b) The initial speed of the bullet is approximately 3.68 m/s.
a) Final Velocity of the Block (v_final):
Given data:
Mass of the bullet (m_bullet): 8.00 g (convert to kg: 0.008 kg)
Mass of the block (m_block): 0.992 kg
Compression of the spring (x): 15.0 cm (convert to meters: 0.15 m)
Initial velocity of the bullet (v_bullet): to be determined
Using the conservation of linear momentum:
m_bullet * v_bullet = (m_bullet + m_block) * v_final
0.008 kg * v_bullet = (0.008 kg + 0.992 kg) * v_final
Solving for v_final:
v_final = (0.008 kg * v_bullet) / (1 kg + 0.992 kg)
v_final ≈ 0.0000808 * v_bullet
b) Initial Speed of the Bullet (v_bullet):
Given data:
Force required to compress the spring (F): 0.750 N
Compression of the spring (x): 0.250 cm (convert to meters: 0.0025 m)
Spring constant (k): to be determined
The work done in compressing the spring is given by W = (1/2) * k * x^2, and this work is equal to the initial kinetic energy (KE_initial) of the bullet:
KE_initial = W = (1/2) * k * x^2
The kinetic energy is also related to the initial speed of the bullet:
KE_initial = (1/2) * m_bullet * v_bullet^2
Setting the two expressions for kinetic energy equal to each other:
(1/2) * k * x^2 = (1/2) * m_bullet * v_bullet^2
Solving for v_bullet:
v_bullet = sqrt((k * x^2) / m_bullet)
Now, using the given information that F = kx, where F is the force required to compress the spring:
k = F / x
Substitute this value of k back into the equation for v_bullet:
v_bullet = sqrt((F * x) / m_bullet)
Substitute the known values:
v_bullet = sqrt((0.750 N * 0.0025 m) / 0.008 kg)
v_bullet ≈ 3.68 m/s
An object oscillates back and forth on the end of a spring. Which of the following statements are true at some time during the course of the motion?a. The object can have zero acceleration and, simultaneously, nonzero velocity.b. The object can have zero velocity and, simultaneously, zero acceleration.c. The object can have nonzero velocity and nonzero acceleration simultaneously.d. The object can have zero velocity and, simultaneously, nonzero acceleration.
Answer:
a. The object can have zero acceleration and, simultaneously, nonzero velocity.
c. The object can have nonzero velocity and nonzero acceleration simultaneously.
d. The object can have zero velocity and, simultaneously, nonzero acceleration.
Explanation:
For an object in simple harmonic motion, the total mechanical energy (sum of elastic potential energy and kinetic energy) is constant:
[tex]E=U+K=\frac{1}{2}kx^2 + \frac{1}{2}mv^2[/tex] (1)
where
k is the spring constant
x is the displacement
m is the mass
v is the speed
We can also notice that the force on the spring is given by Hook's law:
[tex]F=-kx[/tex]
And since according to Newton's law we have F = ma, this can be rewritten as
[tex]ma=-kx\\a=-\frac{k}{m}x[/tex]
which means that the acceleration is proportional to the displacement.
So by looking again at eq.(1), we can now states that:
- when the displacement is zero, x=0, the acceleration is zero, a=0, and the velocity is maximum
- when the velocity is zero, v=0, the acceleration is maximum, which occurs when the displacement is maximum
- in all the other intermediate situations, both velocity and acceleration are non-zero
So the correct answers are
a. The object can have zero acceleration and, simultaneously, nonzero velocity.
c. The object can have nonzero velocity and nonzero acceleration simultaneously.
d. The object can have zero velocity and, simultaneously, nonzero acceleration.
The object oscillating on a spring can have zero acceleration and nonzero velocity, zero velocity and nonzero acceleration, or nonzero velocity and nonzero acceleration at some point during the oscillation. However, it's not possible for it to have both zero velocity and zero acceleration at the same time. These properties are due to the physics of simple harmonic motion.
Explanation:Let's look at the motion of an object oscillating on a spring, a scenario that involves simple harmonic motion:
The object can have zero acceleration and, simultaneously, nonzero velocity. This happens at the equilibrium point (x = 0), where the object's speed is max and acceleration is zero. The object can have zero velocity and, simultaneously, nonzero acceleration. At the maximum displacement (amplitude points x = A or x = -A), the velocity of the object is zero, but the acceleration is nonzero because the spring force, and hence acceleration, is maximum at these points. The object can have nonzero velocity and nonzero acceleration simultaneously. This occurs between the amplitude points and equilibrium where both velocity and acceleration are nonzero. The object cannot have zero velocity and, simultaneously, zero acceleration. It's because when the object has zero velocity at the extreme points, the acceleration is max due to the restoring spring force. At the equilibrium, where the acceleration is zero, the velocity is maximum.
Learn more about Harmonic Motion here:
https://brainly.com/question/32363393
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