The magnitude of the induced emf in the coil, calculated using Faraday's Law of Electromagnetic Induction, is 6.18 V when a uniform magnetic field changes linearly over a time of 1.19 seconds.
To find the magnitude of the induced emf in the coil while the magnetic field is changing, we will apply Faraday's Law of Electromagnetic Induction which states that the emf induced in a circuit is proportional to the rate of change of the magnetic flux through the circuit. Given that the coil has 140 turns (N) and each side of the square frame is 35 cm (which we need to convert to meters to maintain SI units, 0.35 m), we can calculate the area (A) of the square frame as:
A = side x side = 0.35 m x 0.35 m = 0.1225 m2
The magnetic field (B) changes from 0 to -0.426 Wb/m2 over a time (t) of 1.19 s. The rate of change of magnetic field (dB/dt) is:
dB/dt = - ( 0 - (-0.426) ) / 1.19 s = 0.358 Wb/m2/s
Now using Faraday’s Law, the induced emf (ε) is given by:
ε = -N x (dΦ/dt) = -N x A x (dB/dt)
Plugging in the values we get:
ε = -140 x 0.1225 m2 x 0.358 Wb/m2/s = -6.18 V
The magnitude of the induced emf is therefore 6.18 V (taking the absolute value as the question asks for magnitude).
A rectangular coil of wire with a dimension of 4 cm x 5 cm and 10 turns is located between the poles of a large magnet that produces a uniform magnetic field of 0.75 T. The surface of the coil which is originally parallel to the field is rotated in 0.10 s, so that its surface is perpendicular to the field. Calculate the average induced emf across the ends of coil as the coil rotates.
Answer:0.15 V
Explanation:
Given
Dimension of coil [tex]4cm\times 5cm[/tex]
Area of coil [tex]A=4\times 5=20\ cm^2[/tex]
Magnetic field [tex]B=0.75\ T[/tex]
Time of rotation [tex]t=0.1\ s[/tex]
No of turns [tex]N=10[/tex]
Initial flux associated with the coil
[tex]\phi_i=N(B\cdot A)[/tex]
[tex]\phi_i=N(BA\cos \theta )[/tex]
where [tex]\theta [/tex]=angle between magnetic field and area vector of coil
[tex]\phi_i=N(BA\cos 90 )[/tex]
Finally when coil is perpendicular to the field
[tex]\phi_f=N(B\cdot A)[/tex]
[tex]\phi_i=N(BA\cos 0 )[/tex]
and induced emf is given by
[tex]e=-\frac{d\phi }{dt}[/tex]
[tex]e=-\frac{\phi_1-\phi_2}{t-0}[/tex]
[tex]e=-\frac{(0-10\times 0.75\times 20\times 10^{-4})}{0.1}[/tex]
[tex]e=0.15\ V[/tex]
The specific heat of a liquid x is 2.09 cal/g°c. A sample amount of grams of this liquid at 101 k is heated to 225 k. the liquid absorbs 5.23 kcals. what is the sample of liquid in grams? (round off decimal in the answer to nearest tenths)
Answer: 20 grams
Explanation:
The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.
[tex]Q=m\times c\times \Delta T[/tex]
Q = Heat absorbed by liquid = 5.23 kcal = 5230 cal (1kcal=1000cal)
C = heat capacity of liquid = [tex]2.09cal/g^0C[/tex]
Initial temperature of the liquid = [tex]T_i[/tex] = 101 K
Final temperature of the liquid = [tex]T_f[/tex] = 225 K
Change in temperature ,[tex]\Delta T=T_f-T_i=(225-101)K=124K[/tex]
Putting in the values, we get:
[tex]5230=m\times 2.09cal/g^0C\times 124K[/tex]
[tex]m=20g[/tex]
Thus the sample of liquid in grams is 20
Mercury is added to a cylindrical container to a depth d and then the rest of the cylinder is filled with water.
If the cylinder is 0.4 m tall and the absolute (or total) pressure at the bottom is 1.1 atmospheres, determine the depth of the mercury. (Assume the density of mercury to be 1.36 104 kg/m^3, and the ambient atmospheric pressure to be 1.013e5 Pa)
Answer:
0.05m
Explanation:
Density of water = ρ(w) = 1000 kg/ m³ ;
Density of Mercury = ρ(m) = 13628.95 kg/ m³
Total pressure at bottom of cylinder=1.1atm
Therefore, pressure due to water and mercury =1.1-1 =atm
0.1atm=10130pa
The pressure at the bottom is given by,
ρ(w) x g[0.4 - d] + ρ(m) x g x d = 10130
1000 x 9.8[0.4 - d] + 13628.95 x 9.8 d = 10130
3924 - 9810d + 133416d= 10130
123606d= 6206
d= 6202/123606
d= 0.05m
Depth of mercury alone = d = 0.05m
We have that for the Question " determine the depth of the mercury."
It can be said that
The depth of the mercury = [tex]4.333*10^{-2}[/tex]
From the question we are told
the cylinder is 0.4 m tall and the absolute (or total) pressure at the bottom is 1.1 atmospheres, determine the depth of the mercury. (Assume the density of mercury to be 1.36 104 kg/m^3, and the ambient atmospheric pressure to be 1.013e5 Pa)
Therefore,
Absolute pressure at the bottom of the container =
[tex]P = 1.1 atm = 1.1 * (1.013*105) Pa\\\\= 1.1143 * 10^5 Pa[/tex]
Where,
Height of the cylinder = H = [tex]0.4 m[/tex]
Height of the water in the cylinder = [tex]H_1[/tex]
Height of the mercury in the cylinder = [tex]H_2[/tex]
Therefore,[tex]H = H_1 + H_2\\\\H_1 = H - H_2\\\\P = P_{atm} + \rho_1gH_1 + \rho_2gH_2\\\\P = P_{atm} + \rho_1g(H - H_2) + \rho_2gH_2\\\\1.1143*10^5 = 1.013*10^5 + (1000)(9.81)(0.4 - H_2) + (1.36*10^4)(9.81)H_2\\\\1.013*10^4 = 3924 - 9810H_2 + 133416H_2\\\\143226H_2 = 6206\\\\H_2 = 4.333*10^{-2} m[/tex]
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In your own words, describe how dog breeds today came from wolves. In other words, describe selective breeding.
Pleaseeeeeeeeeeeeeeeeeeeeee HELPPPPPPPP!!!FASTTT!!!
Answer:
Dogs were probably domesticated by accident, when wolves began trailing ancient hunter-gatherers to snack on their garbage. Docile wolves may have been slipped extra food scraps, the theory goes, so they survived better, and passed on their genes. Eventually, these friendly wolves evolved into dogs
Selective breeding, also known as artificial selection, is a process used by humans to develop new organisms with desirable characteristics. Breeders select two parents that have beneficial phenotypic traits to reproduce, yielding offspring with those desired traits.
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A 16ft seesaw is pivoted in the center. At what distance from the center would a 200lb person sit to balance a 120lb person on the opposite end?
Answer:
9.6 ft
Explanation:
Distance is inversely proportional to weight
distance = k / (weight), where
k is a constant
or you could say,
distance * weight = k
In this scenario,
120 * 16 = 200 * distance
On rearranging, making, distance the subject of formula, we have
Distance = 120 * 16 / 200
Distance = 1920 / 200
Distance = 9.6 ft
So the 200 pounds person should sit 9.6 feet away from the centre to balance the see saw
Answer:
4.8 ft
Explanation:
torque = wt × distance
t1 = 120lb x 8 ft =960
t2 = 200lb x X ft
set them equal to each other.
120(8) = 200x
960 = 200x
x = 4.8 ft
On earth, what is a child’s mass if the force of gravity on the child’s body is 100 N
Answer: 10.2 kg if g = 9.8, 10 if g = 10.
Explanation:
Weight or the "force of gravity" on a person is simply defined by the equation: F = ma. In this case, the acceleration is g, which is 9.8 but can be rounded up to 10. Based on this, we have:
F = mg
100 = m*9.8
m = 10.2(or 10 if we set g to 10).
The child's mass on the earth if the force of gravity on the child’s body is 100 N will be equal to 10.2 kg.
What is gravity?The fundamental force of attraction operating on all matter is recognized as gravity, also spelled gravity, in mechanics.
It has no impact on identifying the interior properties of common matter because it is the weakest force known to exist in nature.
The formation and growth of planets, galaxies, and the universe are all under the influence of this long-range, cosmic force, which further determines the trajectories of objects throughout the universe and the entire universe.
As per the given information in the question,
Weight, w = 100 N
Use the formula,
W = m × g
100 N = m × 9.8
m = 100/9.8
m = 10.2 kg
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A gas station stores its gasoline in a tank under the ground. The tank is a cylinder lying horizontally on its side. (In other words, the tank is not standing vertically on one of its flat ends.) If the radius of the cylinder is 1.5 meters, its length is 5 meters, and its top is 5 meters under the ground, find the total amount of work needed to pump the gasoline out of the tank. (The density of gasoline is 673 kilograms per cubic meter; use g=9.8 m/s2.)
Answer:
Work needed = 1515.15 KJ
Explanation:
The center of mass of a cylinder lying horizontally on its side would lie on the axis of the cylinder at the center of length l.
Depth of center of mass from ground level;Δh = (r + 5) metres
Now, work done to pump the gasoline out of the tank is equal to the gain in potential energy by gasoline on lifting it from center of mass to the ground level.
Thus;
W = ΔU = mgΔh
We know that mass(m) = volume(V) x density(ρ)
So,
W = (ρV)gΔh
Volume(V) = πr²L
Thus;
W = (ρ(πr²L)) * g(r + 5)
We are given;
Density; ρ = 673 kg/m³
Length; L = 5 m
Radius; r = 1.5 m
Acceleration due to gravity;g = 9.8 m/s²
Thus;
W = (673(π•1.5²•5)) * 9.8(1.5 + 5)
W = 1515154.4 J = 1515.15 KJ
A wire lying along a y axis form y=0 to y=0.25 m carries a current of 2.0 mA in the negative direction of the axis. the wire fully lies in a magnetic field given by (0.3y)i + (0.4y)j Tesla. The magnetic force on the wire is *
Answer:
The force is "19 µN".
Explanation:
The lane's j-component is meaningless, as the current is flowing in the -j line.
Therefore the power is now in the direction of + z (out of the page if x and y are in the page plane) and has the magnitude.
[tex]\ formula: \\\\\ Forec (F) = mA \\\\ \ F \ = 2.0mA \times \int {0.3} \ y \ dy \rightarrow \ from\ 0 \ to \ 0.25 \\\\\ F \ = \ 2.0mA \times 0.15 * 0.25^{2} m\cdot T \\\\ F = 19 \µN[/tex]
What is the acceleration of a 5kg mass pushed by a 10N force?
Answer:2m/s^2
Explanation:
mass=5kg
Force=10N
Acceleration=force ➗ mass
Acceleration=10 ➗ 5
Acceleration=2m/s^2
Which statement explains the first law of thermodynamics?
Heat is created, but not destroyed or transformed.
Heat is transformed, but not created or destroyed.
Heat is destroyed, but not created or transformed.
Answer:
The right answer among the options is that: Heat is transformed, but not created or destroyed.
Explanation:
The First Law of Thermodynamics states that heat is a form of energy, and that thermodynamic processes are subject to the principle of conservation of energy.
This implies that heat energy cannot be created or destroyed. However, energy can be transferred from one location to another and converted to and from other forms of energy.
Thus, the right answer among the options is that:Heat is transformed, but not created or destroyed.
Answer:
heat is transformed, but not created or destroyed
Explanation:
Traveling with an initial speed of a car accelerates at along a straight road. How long will it take to reach a speed of Also, through what distance does the car travel during this time? (10%) b. At bullet A is fired vertically with an initial (muzzle) velocity of 450 m/s. When bullet B is fired upward with a muzzle velocity of 600 m/s. Determine the time t, after A is fired, as to when bullet B passes bullet A. At what altitude does this occur
Answer:
A) 30 s, 792 m
B) 10.28 s, 4108.2 m = 4.11 km
Explanation:
A) Traveling with an initial speed of 70 km/h, a car accelerates at 6000km/h^2 along a straight road. How long will it take to reach a speed of 120 km/h? Also, through what distance does the car travel during this time?
Using the equations of motion.
v = u + at
v = final velocity = 120 km/h
u = initial velocity = 70 km/h
a = acceleration = 6000 km/h²
t = ?
120 = 70 + 6000t
6000t = 50
t = (50/6000) = 0.0083333333 hours = 30 seconds.
Using the equations of motion further,
v² = u² + 2ax
where x = horizontal distance covered by the car during this time
120² = 70² + 2×6000×x
12000x = 120² - 70² = 9500
x = (9500/12000) = 0.79167 km = 791.67 m = 792 m
B) At t = 0 bullet A is fired vertically with an initial (muzzle) velocity of 450 m/s. When t = 3 s, bullet B is fired upward with a muzzle velocity of 600 m/s. Determine the time t, after A is fired, as to when bullet B passes bullet A. At what altitude does this occur?
Bullet A is fired upwards with velocity 450 m/s
Bullet B is fired upwards with velocity 600 m/s too
Using the equations of motion, we can obtain a relation for when vertical distance covered by the bullets and time since they were fired.
y = ut + ½at²
For the bullet A
u = initial velocity = 450 m/s
a = acceleration due to gravity = -9.8 m/s²
y = 450t - 4.9t² (eqn 1)
For the bullet B, fired 3 seconds later,
u = initial velocity = 600 m/s
a = acceleration due to gravity = -9.8 m/s²
t = T
y = 600T - 4.9T²
At the point where the two bullets pass each other, the vertical heights covered are equal
y = y
450t - 4.9t² = 600T - 4.9T²
But, note that, since T starts reading, 3 seconds after t started reading,
T = (t - 3) s
450t - 4.9t² = 600T - 4.9T²
450t - 4.9t² = 600(t-3) - 4.9(t-3)²
450t - 4.9t² = 600t - 1800 - 4.9(t² - 6t + 9)
450t - 4.9t² = 600t - 1800 - 4.9t² + 29.4t - 44.1
600t - 1800 - 4.9t² + 29.4t - 44.1 - 450t + 4.9t² = 0
179.4t - 1844.1 = 0
t = (1844.1/179.4) = 10.28 s
Putting this t into the expression for either of the two y's, we obtain the altitude at which this occurs.
y = 450t - 4.9t²
= (450×10.28) - (4.9×10.28×10.28)
= 4,108.2 m = 4.11 km
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n order better to map the surface features of the Moon, a 361 kg361 kg imaging satellite is put into circular orbit around the Moon at an altitude of 147 km.147 km. Calculate the satellite's kinetic energy K,K, gravitational potential energy ????,U, and total orbital energy E.E. The radius and mass of the Moon are 1740 km1740 km and 7.36×1022 kg.
Answer:
Explanation:
Mass of satellite
M_s = 361 kg
Distance of satellite from moon
h = 147 km = 147,000m
Radius of the moon is
R_m = 1740 km = 1740,000m
Mass of the moon is
M_m = 7.36 × 10²² kg.
The kinetic energy is equal to the potential energy of the body to the surface of the moon from the conservation of energy.
K.E = P.E = mgh
Gravity on moon is g = 1.62 m/s²
K.E = 361 × 1.62 × 147,000
K.E = 8.597 × 10^7 J.
B. The gravitational potential energy can be calculated using
U = G•M_s × M_m (1/R_s - 1 / R)
R is the total distance from the centre of the moon to the satellite
R = h + R_m = 147 + 1740 = 1887km
R = 1,887,000 m
U = 6.67 × 10^-11 × 361 × 7.36 × 10²² (1/1,740,000 - 1/1,887,000)
U = 6.67 × 10^-11 × 361 × 7.36 × 10²² × 4.48 × 10^-8
U = 7.93 × 10^7 J
Then,
The total energy becomes
E = K.E + U
E= 8.597 × 10^7 + 7.93 × 10^7 J
E = 1.653 × 10^8 J
the gravitational pull of the moon is much less than the gravitational pull of earth, which two statements are true for an object with a mass of 20 kilograms that weighs 44 pounds on earth
Answer:
b
Explanation:
the earths mass is more than the moon .
Answer:
The object's weight would be less on the moon.
The Object's mass would be the same on the moon
Explanation:
Suppose an object of mass m is attached to the end of a spring hanging from the ceiling. The mass is at its equilibrium position y = 0 when the mass hangs at rest. Suppose you push the mass to a position yo units above its equilibrium position and release it. As the mass oscillates up and down (neglecting air friction), the position y of the mass after t seconds is given by the equation below. Use this equation to answer the questions below:
y = yocos (t square root k/m)
a) Find dy/dx, the velocity of the mass. Assumie k and m are constant.
b) How would the velocity be affected if the experiment were repeated with four times the mass on the end of the spring?
c) How would the velocity be affected if the experiment were repeated with a spring that has 4 times the stiffness (if k is increased by a factor of 4)?
d) Assume that y has units of meters, t has units of seconds, m has units of kg and k has units of kg/s2. Show that the units of the velocity in part a) are consistent.
Answer:
Explanation:
y = y₀ cos[tex]\sqrt{\frac{k}{m} }\times t[/tex]
a )
[tex]\frac{dy}{dt}[/tex] = - y₀ x [tex]\sqrt{\frac{k}{m} }[/tex] sin ( [tex]\sqrt{\frac{k}{m} }\times t[/tex] )
b ) If m = 4m
[tex]\frac{dy}{dt}[/tex] = - y [tex]\sqrt{\frac{k}{4m} }[/tex] sin ( [tex]\sqrt{\frac{k}{4m} }\times t[/tex] )
Magnitude of velocity will be decreased .
c )
[tex]\frac{dy}{dt}[/tex] = - y [tex]\sqrt{\frac{4k}{m} }[/tex] sin ( [tex]\sqrt{\frac{4k}{m} }\times t[/tex] )
magnitude of velocity will be increased .
d )
velocity = - y₀ [tex]\sqrt{\frac{k}{m} }[/tex] sin( [tex]\sqrt{\frac{k}{m} }\times t[/tex] )
= L [tex]\sqrt{\frac{ms^{-2}}{m} }[/tex] X 0
= L s⁻¹
= m /s
unit of velocity is consistent .
Final answer:
The velocity of a mass attached to a spring is found by differentiating the position equation with respect to time, revealing how the system's velocity changes with variations in mass and spring stiffness. Increasing the mass decreases the velocity amplitude, while increasing the spring stiffness increases it.
The units of velocity, m/s, are confirmed through dimensional analysis.
Explanation:
To find the velocity of the mass, we need to differentiate the position equation y = yocos (t √ k/m) with respect to time. Using the chain rule for differentiation, the derivative of y with respect to t gives us dy/dt = -y_0(√ k/m)sin(t √ k/m), where dy/dt represents the velocity of the mass.
This equation tells us the velocity at any given moment for a mass m and spring constant k.
b) If the mass is increased by four times, the equation for velocity becomes dy/dt = -y_0(√ k/(4m))sin(t √ k/(4m)). The increase in mass causes the velocity amplitude to decrease, as √(1/4) is in the equation, indicating that velocity decreases in proportion to the square root of the mass increase.
c) Increasing the spring constant k fourfold results in the new velocity equation dy/dt = -y_0(√ (4k)/m)sin(t √ (4k)/m). This shows an increase in the velocity amplitude, as the increase in k results in a velocity proportional to the square root of the increase in k, thus making the system oscillate faster.
d) To confirm the units of velocity are meters per second (m/s), we substitute the units into the derivative of the position equation: [m]*[s√(kg/s2)/kg] simplifies to m/s, thus showing the units of velocity are indeed consistent and correct.
A 54-Ω resistor in a circuit has a voltage difference of 8 V across its leads. Calculate the current through the resistor.
What is the current resistor?
Answer:
The current is 0.148 amps
Explanation:
To find the current, you divide the voltage given by the resistence:
8V ÷ 54 ohms
Which equals 0.1481481481 amps
The current resistor is 0.148 amps.
What is Ohm's law?Ohm's law states that the voltage across a conductor is directly proportional to the current flowing through it, provided all physical conditions and temperatures remain constant.
Ohm's Law formulas V = IR, I = V/R, and R= V/I.
Current through each resistor can be found using Ohm's law I=V/R, where the voltage is constant across each resistor.
The given values are:
V = 8V and R = 54Ω
By using the formula :-
I=V/R
I =8V / 54Ω = 0.148 amps
Therefore,
0.148 amps is the current resistor.
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Calculate the frequency of the 3rd normal mode of a guitar string of length 40.0cm and mass 0.5g when stretched with a tension of 80N.
The frequency of the third normal mode of a guitar string with a length of 40.0 cm, mass of 0.5 g when stretched with a tension of 80N is approximately 948.68 Hz.
To calculate the frequency of the third normal mode (
n = 3) for a guitar string, we can use the formula for the frequency of a string fixed at both ends:
f_n = (n/2L) √(T/μ)
where:
f_n is the frequency of the nth mode,
n is the mode number (which is 3 in this case),
L is the length of the string,
μ is the mass per unit length of the string (linear mass density), and
T is the tension in the string.
Given the length of the string L = 40.0 cm = 0.4 m, the mass m = 0.5 g = 0.0005 kg, and the tension T = 80 N, we first need to calculate the linear mass density:
μ = m/L
In this case,
μ = 0.0005 kg / 0.4 m = 0.00125 kg/m
Now, we use the frequency formula to find f_3:
f_3 = (3/2 0.4 m √(80 N/0.00125 kg/m)
= (3/(0.8 m)) √(64000 N/m)
= 3.75 √(64000 N/m)
= 3.75 √(64000 N/m)
= 3.75 * 252.9822 Hz
= 948.68 Hz
The frequency of the third normal mode of the guitar string is approximately 948.68 Hz.
In part (a), suppose that the box weighs 128 pounds, that the angle of inclination of the plane is θ = 30°, that the coefficient of sliding friction is μ = 3 /4, and that the acceleration due to air resistance is numerically equal to k m = 1 3 . Solve the differential equation in each of the three cases, assuming that the box starts from rest from the highest point 50 ft above ground. (Assume g = 32 ft/s2 and that the downward velocity is positive.)
Answer:
v(t) = 21.3t
v(t) = 5.3t
[tex]v(t) = 48 -48 e ^{ \frac{t}{9}}[/tex]
Explanation:
When no sliding friction and no air resistance occurs:
[tex]m\frac{dv}{dt} = mgsin \theta[/tex]
where;
[tex]\frac{dv}{dt} = gsin \theta , 0 < \theta < \frac{ \pi}{2}[/tex]
Taking m = 3 ; the differential equation is:
[tex]3 \frac{dv}{dt}= 128*\frac{1}{2}[/tex]
[tex]3 \frac{dv}{dt}= 64[/tex]
[tex]\frac{dv}{dt}= 21.3[/tex]
By Integration;
[tex]v(t) = 21.3 t + C[/tex]
since v(0) = 0 ; Then C = 0
v(t) = 21.3t
ii)
When there is sliding friction but no air resistance ;
[tex]m \frac{dv}{dt}= mg sin \theta - \mu mg cos \theta[/tex]
Taking m =3 ; the differential equation is;
[tex]3 \frac{dv}{dt}=128*\frac{1}{2} -\frac{\sqrt{3} }{4}*128*\frac{\sqrt{3} }{4}[/tex]
[tex]\frac{dv}{dt}= 5.3[/tex]
By integration; we have ;
v(t) = 5.3t
iii)
To find the differential equation for the velocity (t) of the box at time (t) with sliding friction and air resistance :
[tex]m \frac{dv}{dt}= mg sin \theta - \mu mg cos \theta - kv[/tex]
The differential equation is :
= [tex]3 \frac{dv}{dt}=128*\frac{1}{2} - \frac{ \sqrt{ 3}}{4}*128 *\frac{ \sqrt{ 3}}{2}-\frac{1}{3}v[/tex]
= [tex]3 \frac{dv}{dt}=16 -\frac{1}{3}v[/tex]
By integration
[tex]v(t) = 48 + Ce ^{\frac{t}{9}[/tex]
Since; V(0) = 0 ; Then C = -48
[tex]v(t) = 48 -48 e ^{ \frac{t}{9}}[/tex]
A particle of positive charge ???? is assumed to have a fixed position at P. A second particle of mass m and negative charge −q moves at constant speed in a circle of radius r1, centered at P. Derive an expression for the work W that must be done by an external agent in the second particle in order to increase the radius of the circle of motion, centered at P, to r2.
Answer:
[tex]W=\frac{1}{2}kq_1q_2[\frac{1}{r_2}-\frac{1}{r_1}][/tex]
Explanation:
To find the work W to put the negative charge in the new orbit you can use the following formula:
[tex]W=\Delta K\\\\K=\frac{1}{2}mv^2[/tex]
That is, the total work is equal to the change in the kinetic energy of the negative charge. Then you calculate the speed of the electron, by using the second Newton Law and the expression for the electrostatic energy:
[tex]F=ma_c\\\\-k\frac{(q_1)(q_2)}{r_1^2}=m\frac{v^2}{r_1}\\\\v^2=k\frac{q_1q_2}{mr_1}[/tex]
r1: radius of the first orbit
m: mass of the negative charge
v: velocity of the charge
k: Coulomb's constant
q1: charge of the fixed particle at point P
q2: charge of the negative charge
Hence, the velocity of the charge in a new orbit with radius r2 is:
[tex]v'^2=k\frac{q_1q_2}{mr_2}[/tex]
Finally the work required to put the charge in the new orbit is:
[tex]W=\Delta K =\frac{1}{2}m[v'^2-v^2]\\\\W=\frac{1}{2}m[k\frac{q_1q_2}{mr_2}-k\frac{q_1q_2}{mr_1}]\\\\W=\frac{1}{2}kq_1q_2[\frac{1}{r_2}-\frac{1}{r_1}][/tex]
ONLINE CALCULATOR .A force of 187 pounds makes an angle of 73 degrees 36 ' with a second force. The resultant of the two forces makes an angle of 29 degrees 1 ' to the first force. Find the magnitudes of the second force and of the resultant.
Answer:
The magnitudes of the second force is [tex]Z = 129.9 N[/tex]
The magnitudes of the resultant force is [tex]R = 256.047 N[/tex]
Explanation:
From the question we are told that
The force is [tex]F = 187 \ lb[/tex]
The angle made with second force [tex]\theta_o = 73 ^o 36' = 73 + \frac{36}{60} = 73.6^o[/tex]
The angle between the resultant force and the first force [tex]\theta _1 = 29 ^o 1 ' = 29 + \frac{1}{60} = 29.0167^o[/tex]
For us to solve problem we are going to assume that
The magnitude of the second force is Z N
The magnitude of the resultant force is R N
According to Sine rule
[tex]\frac{F}{sin (\theta _o - \theta_1 } = \frac{Z}{\theta _1}[/tex]
Substituting values
[tex]\frac{187}{sin(73.3 - 29.01667)} =\frac{Z}{sin (29.01667)}[/tex]
[tex]267.82 =\frac{Z}{0.4851}[/tex]
[tex]Z = 129.9 N[/tex]
According to cosine rule
[tex]R = \sqrt{F ^2 + Z^2 + 2(F) (Z) cos (\theta _o) }[/tex]
Substituting values
[tex]R = \sqrt{187^2 + 129.9 ^2 + 2 (187 ) (129.9) cos (73.6)}[/tex]
[tex]R = 256.047 N[/tex]
Peggy is an astronaut and volunteers for the first manned mission to Alpha Centauri, the nearest star system to the Solar System. Her spacecraft will travel at 80%80% of the speed of light, and the trip there and back will take over 1010 years. Her twin sister Patty is an astronomer and will remain on Earth, studying Alpha Centauri using telescopes. When Peggy returns from her trip, how will their ages compare?
Answer:
If Patty remains on Earth then at the time Peggy will come back from her trip, Peggy will be much younger than her sister Patty because of time-dilation.
Explanation:
Peggy and Patty are sisters. Peggy is an astronaut and Patty is an astronomer.
Peggy goes for mission to Alpha Centauri, the nearest star system to the Solar System at 80% of the speed of light, and will come back after 1010 years.
If Patty remains on Earth then at the time Peggy will come back from her trip, Peggy will be much younger than her sister Patty because of time dilation.
This is due to the fact that time moves slower in Alpha Centauri because of its massive gravitational force which bends space time. Moreover, It is known that Peggy's spacecraft moves at 80% of the speed of light, it will result in velocity time dilation since time moves slow if you travel at a speed near to the speed of light.
Final answer:
Peggy, the astronaut twin traveling at 80% of the speed of light, will experience less time due to time dilation, and upon her return will be younger than her Earth-bound twin sister, Patty.
Explanation:
The question is about the relativistic effects that occur when one twin travels at significant fraction of the speed of light while the other remains on Earth.
According to the theory of relativity, time dilation will cause the traveling twin, Peggy, to age more slowly compared to her twin sister, Patty, who remains on Earth.
If Peggy travels to Alpha Centauri, which is 4.3 light years away, at 80% of the speed of light, and assuming the round trip takes 10 years for the Earth-bound twin, we can calculate that the moving twin will experience less than 10 years of elapsed time due to the effects of time dilation.
This happens because the faster Peggy travels, the more pronounced the effect of time dilation will be. This is a well-known result predicted by Einstein's special theory of relativity and has been confirmed through experiments involving high-speed particles and precise clocks.
Thus, when Peggy returns, she will be younger than her twin sister Patty, who has experienced the full 10 years on Earth.
2. In a series circuit, all resistors have identical currents.
a) What is the relationship between the power and resistance of these resistors?
b) In a parallel circuit, all resistors have identical voltages. What is the relationship
between the power and resistance of these resistors?
Answer:
Explanation:
In series connection of resistors, same current flows in the circuit.
Power dissipated by a resistor is
P = i²R
And since same current flows in them, it implies that the power is directly proportional to Resistance, so the higher the resistance of the resistor the higher the power dissipated in the series connection. Also, the lower the resistance, the lower the power dissipated.
2. In parallel connection, same voltage is applied across the resistor.
So, power dissipated by each resistor is
P = V² / R
So, since the same voltage is applied across parallel connection, then, power dissipated in each resistor is inversely proportional to the resistance.,
So, the higher the resistance, the lower the power and the lower the resistance, the higher the power.
A spring-powered dart gun is unstretched and has a spring constant 16.0 N/m. The spring is compressed by 8.0 cm and a 5.0 gram projectile is placed in the gun. The kinetic energy of the projectile when it is shot from the gun is
Answer:
Explanation:
Given that,
Spring constant = 16N/m
Extension of spring
x = 8cm = 0.08m
Mass
m = 5g =5/1000 = 0.005 kg
The ball will leave with a speed that makes its kinetic energy equal to the potential energy of the compressed spring.
So, Using conservation of energy
Energy in spring is converted to kinectic energy
So, Ux = K.E
Ux = ½ kx²
Then,
Ux = ½ × 16 × 0.08m²
Ux = 0.64 J
Since, K.E = Ux
K.E = 0.64 J
Allie is flying from New York to London. Her plane will most likely fly in the:
stratosphere
thermosphere
troposphere
mesosphere
Answer:
troposphere
Explanation:
because troposphere is the layer of atmosphere closest to the earth . air is thicker at lower altitudes requiring more energy to push themselves to the sky . however , the air is thinner caused flight more fuel efficient .
X rays of wavelength 0.00758 nm are directed in the positive direction of an x axis onto a target containing loosely bound electrons. For Compton scattering from one of those electrons, at an angle of 145°, what are (a) the Compton shift, (b) the corresponding change in photon energy, (c) the kinetic energy of the recoiling electron, and (d) the angle between the positive direction of the x axis and the electron's direction of motion? The electron Compton wavelength is 2.43 × 10-12 m.
A wheel rotating with a constant angular acceleration turns through 22 revolutions during a 5 s time interval. Its angular velocity at the end of this interval is 12 rad/s. What is the angular acceleration of the wheel? Note that the initial angular velocity is not zero. Answer in units of rad/s 2 .
Answer:
0.52rad/s^2
Explanation:
To find the angular acceleration you use the following formula:
[tex]\omega^2=\omega_o^2+2\alpha\theta[/tex] (1)
w: final angular velocity
wo: initial angular velocity
θ: revolutions
α: angular acceleration
you replace the values of the parameters in (1) and calculate α:
[tex]\alpha=\frac{\omega^2-\omega_o^2}{2\theta}[/tex]
you use that θ=22 rev = 22(2π) = 44π
[tex]\alpha=\frac{(12rad/s)^2-(0rad/s)^2}{2(44\pi)}=0.52\frac{rad}{s^2}[/tex]
hence, the angular acceñeration is 0.52rad/s^2
A pumpkin pie in a 9.00 in diameter plate is placed upon a rotating tray. Then, the tray is rotated such that the rim of the pie plate moves through a distance of 233 in. Express the angular distance that the pie plate has moved through in revolutions, radians, and degrees.
Answer:
Explanation:
Given that,
Pie diameter = 9 in
Then, the circumference of the pie is
P = πd = 9π in
Then rim of the pie rotates 233 in,
Then,
1 Revolution of the pie is 9π in,
So, for 233 in, we will have
233 in / 9π in revolution
8.24 revolution
So, the revolution of the pie is 8.24
1 revolution is 2πrad
Then,
8.24 revolution = 8.24 × 2π = 51.78 rad.
And also, 1 revolution is 360°
Then,
8.24 revolution = 8.24 × 360 = 2966.4°
So,
In revolution, θ = 8.24 revolution
In radian = θ = 57.78 rad
In degree θ = 2966.4°
The angular distance should be
In revolution, θ = 8.24 revolution.
In radian = θ = 57.78 rad.
In degree θ = 2966.4°.
Calculation of the angular distance:Since
Pie diameter = 9 in
So, the circumference of the pie should be
P = πd = 9π in
And, rim of the pie rotates 233 in,
So,
1 Revolution of the pie is 9π in,
So, for 233 it should be
= 233 in / 9π in revolution
= 8.24 revolution
Now in the case when
1 revolution is 2πrad
So,
8.24 revolution = 8.24 × 2π = 51.78 rad.
And also, 1 revolution is 360°
So,
8.24 revolution = 8.24 × 360 = 2966.4°
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A large fraction of the ultraviolet (UV) radiation coming from the sun is absorbed by the atmosphere. The main UV absorber in our atmosphere is ozone, \(\rm O_3\). In particular, ozone absorbs radiation with frequencies around 9.38×1014 \({\rm \rm Hz}\) . What is the wavelength \(\texttip{\lambda }{lambda}\) of the radiation absorbed by ozone?
Answer:
The wavelength of the radiation absorbed by ozone is 319.83 nm
Explanation:
Given;
frequency of absorbed ultraviolet (UV) radiation, f = 9.38×10¹⁴ Hz
speed of the absorbed ultraviolet (UV) radiation, equals speed of light, v = 3 x 10⁸ m/s
wavelength of the absorbed ultraviolet (UV) radiation, λ = ?
Apply wave equation for speed, frequency and wavelength;
v = fλ
λ = v / f
λ = (3 x 10⁸) / (9.38×10¹⁴)
λ = 3.1983 x 10⁻⁷ m
λ = 319.83 x 10⁻⁹ m
λ = 319.83 nm
Therefore, the wavelength of the radiation absorbed by ozone is 319.83 nm
A ski lift has a one-way length of 1 km and a vertical rise of 200 m. The chairs are spaced 20 m apart, and each chair can seat three people. The lift is operating a a steady speed of 10 km/h. Neglecting friction and air drag and assuming the average mass of each loaded chair is 250 kg, determine the power required to operate this ski lift. Also estimate the power required to accelerate this ski lift in 5 s to its operating speed when it is first turned on
Answer:
Find attachments for step by step solution.
A power of 68 KW is required to operate the ski lift. A power of 9660.5W is required to accelerate this ski lift.
What is power?The power can be defined as the rate of doing work, it is the work done in unit time. The SI unit of power joules per second (J/s) or Watt (W).
Power is a time-based quantity and the rate at which work is done upon an object. The formula for power can be expressed as mentioned below.
Power = Work/time
P = W/t
Given, the chairs are spaced 20 m apart a length of 1 km = 1000m
Then the number of chairs = 1000/20 = 50
Each chair weighs = 250 kg
Then the weight of M = 50 × 250 = 12500 Kg
Consider, the initial and final heights, h₁ = 0, h₂ = 200 m
The work needed to raise the chairs, W = mgh,
W = 12500 × 9.81 × (200 - 0)
W = 2.54 × 10⁷ J
The rate of work done at a distance of 1 km = 10 km/h,
t = 1/10 = 0.1 hr = 360 s
The power needed to operate this ski lift is, P = W/t
P = 2.54×10⁷ / 360
P = 68125 W = 68 kW
Given, the initial velocity, u = 0 m/s, final velocity, v = 10 km/h = 2.78 m/s
and, t = 5sec
Acceleration during it is first turned on is:
a = (v - u)/t
a = 2.78/ 5
a = 0.556 m/s²
The power required to accelerate this ski lift is:
P = ½ m [(v² - u²)/t]
P = ½ × 12500 × [2.78²/5]
P = 6250 × 1.55
P = 9660.5 W
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A 15.0-kg object and a m^2 =10.0-kg object are joined by a cord that passes over a pulley with a radius of R =10.0 cm and a mass of M = 3.00 kg. The cord has a negligible mass and does not slip on the pulley. The pulley rotates on its axis without friction. The objects are released from rest when they are 3.00m apart and are free to fall. Ignore air resistance. Treat the pulley as a uniform disk, and determine the speeds of the two objects as they pass each other.
Final answer:
The speed of the 15.0-kg object as it passes the 10.0-kg object is 13.3 m/s, and the speed of the 10.0-kg object as it passes the 15.0-kg object is -13.3 m/s.
Explanation:
To determine the speeds of the two objects as they pass each other, we can use the principle of conservation of mechanical energy. When the objects are released from rest, the potential energy of the system is converted into kinetic energy as the objects fall. The sum of the kinetic energies of the two objects will be equal to the initial potential energy of the system.
Using the formula for potential energy (PE=mgh), we can calculate the initial potential energy of the system. The 15.0-kg object will fall a distance of 3.00m, so its potential energy is (15.0 kg)(9.8 m/s^2)(3.00 m) = 441 J. The 10.0-kg object will rise a distance of the same amount, so its potential energy is -441 J (we take the negative sign because the object is moving in the opposite direction).
Now, we can equate the sum of the kinetic energies of the two objects to the initial potential energy of the system. Let v1 be the speed of the 15.0-kg object and v2 be the speed of the 10.0-kg object. The kinetic energy of the 15.0-kg object is (1/2)(15.0 kg)(v1^2) and the kinetic energy of the 10.0-kg object is (1/2)(10.0 kg)(v2^2). Setting the sum of these two kinetic energies equal to 441 J, we can solve for v1 and v2.
441 J = (1/2)(15.0 kg)(v1^2) + (1/2)(10.0 kg)(v2^2)
Simplifying the equation, we have 441 J = (7.5 kg)(v1^2) + (5.0 kg)(v2^2). Since the objects are joined by a cord and the pulley does not slip, the speeds of the two objects will be equal in magnitude but opposite in direction. So we can write v2 = -v1 and substitute into the equation. We can then solve for v1:
441 J = (7.5 kg)(v1^2) + (5.0 kg)(-v1^2)
Simplifying further, 441 J = (2.5 kg)(v1^2)
Solving for v1,
v1^2 = 176.4 m^2/s^2
v1 = 13.3 m/s
Therefore, the speed of the 15.0-kg object as it passes the 10.0-kg object is 13.3 m/s, and the speed of the 10.0-kg object as it passes the 15.0-kg object is -13.3 m/s.
A 2 UC charge q1 and a 2 uC charge q2 are 0.3 m from
the x-axis. A 4 uC charge q3 is 0.4 m from the y-axis.
The distances d13 and d23 are 0.5 m. What is the magnitude and direction
Answer:
Explanation:
The magnitude is 0.5 N
The direction is 0°
Answer:
Magnitude: 0.5 N
Direction: 0 degrees
Explanation:
Going to be honest, I do not understand the magnitude of R, but what I can explain is the direction.
So, following the steps, we find that the magnitude of A (coulomb's constant k times (q1 x q3)/dq13^2) = 1.798 Newtons
The magnitude of B is the same because q1 = q2 in this scenario
Using SohCahToa, we can find the angle of vector A = about 36.9 degrees
since q1 is 0.3 meters away from the x-axis (this is our opposite side) and the distance between q1 and q3 is 0.5 (our hypotenuse because the x-axis and y-axis make a right angle) this means that we can use this for our formula,
the sine of angle A = opposite side / hypotenuse side
Sine of angle A = 0.3m / 0.5m
Angle A = (inverse Sine) of 0.3m/0.5m
Angle A = 36.9 degrees
Using that we can get the x and y components of each vector:
The x component of A will be A cosine angle A which is
Ax = 1.798 N times (cosine(36.9 degrees)) = 1.44 N
The y component of A will be NEGATIVE A sine angle A
The reason why it is negative is because for the A vector, it has a negative slope so its y value is continually decreasing, so it makes sense for it's resulting vector to have the same y decrease.
Ay = (-1.798)(sine(36.9)) = -1.079 N
Now we can do the same with B.
Since B = A IN THIS SCENARIO, the values will be the same,
Bx = 1.798 N times (cosine(36.9 degrees) = 1.44 N
BUT with the y value here, it is actually INCREASING
so:
By = 1.798 N times (sine(36.9 degrees)) = 1.079 N
So the next step is to sum these values.
Rx = Ax + Bx = 1.44 N + 1.44 N = 2.88 N
The x component of vector R is 2.88 N.
Ry = Ay + By = -1.079 N + 1.079 N = 0 N.
The y component of vector R is 0.
Basically, it is traveling straight along the x-axis. This makes sense because the forces of each particle were repelling this one at an EQUAL MAGNITUDE (q1 = q2) So, the direction is 0 degrees because even following the steps, (tan-1 times (Ry over Rx)) is 0/2.88. This will be 0.