(a) [tex]2.7\cdot 10^{25} kg[/tex]
The acceleration due to gravity on the surface of the planet is given by
[tex]g=\frac{GM}{R^2}[/tex] (1)
where
G is the gravitational constant
M is the mass of the planet
R is the radius of the planet
Here we know:
[tex]g=22.4 m/s^2[/tex]
[tex]d=1.8\cdot 10^7 m[/tex] is the diameter, so the radius is
[tex]R=\frac{d}{2}=\frac{1.8\cdot 10^7 m}{2}=9\cdot 10^6 m[/tex]
So we can re-arrange eq.(1) to find M, the mass of the planet:
[tex]M=\frac{gR^2}{G}=\frac{(22.4 m/s^2)(9\cdot 10^6 m)^2}{6.67\cdot 10^{-11}}=2.7\cdot 10^{25} kg[/tex]
(b) [tex]4.8\cdot 10^{31}kg[/tex]
The planet is orbiting the star, so the centripetal force is equal to the gravitational attraction between the planet and the star:
[tex]m\frac{v^2}{r}=\frac{GMm}{r^2}[/tex] (1)
where
m is the mass of the planet
M is the mass of the star
v is the orbital speed of the planet
r is the radius of the orbit
The orbital speed is equal to the ratio between the circumference of the orbit and the period, T:
[tex]v=\frac{2\pi r}{T}[/tex]
where
[tex]T=402 days = 3.47\cdot 10^7 s[/tex]
Substituting into (1) and re-arranging the equation
[tex]m\frac{4\pi r^2}{rT^2}=\frac{GMm}{r^2}\\\frac{4\pi r}{T^2}=\frac{GM}{r^2}\\M=\frac{4\pi r^3}{GMT^2}[/tex]
And substituting the numbers, we find the mass of the star:
[tex]M=\frac{4\pi^2 (4.6\cdot 10^{11} m)^3}{(6.67\cdot 10^{-11})(3.47\cdot 10^7 s)^2}=4.8\cdot 10^{31}kg[/tex]
An ideal step-down transformer has a primary coil of 300 turns and a secondary coil of 18 turns. It is plugged into an outlet with 230.0 V (AC) and it draws a current of 0.31 A. Calculate the voltage in the secondary coil. (in V) 1.38×101 V You are correct. Your receipt no. is 154-2147 Help: Receipt Previous Tries Calculate the current in the secondary coil. (in A) 5.17 A You are correct. Your receipt no. is 154-6875 Help: Receipt Previous Tries Calculate the average power dissipated.
1) 13.8 V
We can use the transformer equation:
[tex]\frac{N_p}{N_s}=\frac{V_p}{V_s}[/tex]
where we have
[tex]N_p = 300[/tex] is the number of turns in the primary coil
[tex]N_s=18[/tex] is the number of turns in the secondary coil
[tex]V_p=230.0 V[/tex] is the voltage in the primary coil
[tex]V_s = ?[/tex] is the voltage in the secondary coil
Solving for Vs, we find
[tex]V_s = \frac{N_s}{N_p}V_p=\frac{18}{300}(230.0 V)=13.8 V[/tex]
2) 5.17 A
For an ideal transformer, the power in input is equal to the power in output, so we can write:
[tex]P_{in} = P_{out}\\V_p I_p = V_s I_s[/tex]
where
[tex]V_p=230.0 V[/tex] is the voltage in the primary coil
[tex]V_s = 13.8 V[/tex] is the voltage in the secondary coil
[tex]I_p=0.31 A[/tex] is the current in the primary coil
[tex]I_s = ?[/tex] is the current in the secondary coil
Solving for Is, we find
[tex]I_s = \frac{I_p V_p}{V_s}=\frac{(0.31 A)(230.0 V)}{13.8 V}=5.17 A[/tex]
The Iceberg Kid, infamous for imposing a deep freeze on whoever disagrees with him, discovers that in his advanced teenaged years his hands constantly feel chilled. As he is a cool physics student, the Iceberg Kid quickly calculates the resistance of the resistor needed to warm his hands. Given that 181 J of thermal energy is generated by the resistor during 9.99 s of operation while connected to a 12.0 V battery, find the correct value of this resistance (and avoid a cold, cold fate).
Answer:
[tex]7.96 \Omega[/tex]
Explanation:
First of all, we can calculate the power dissipated by the resistor. We have:
E = 181 J is the energy produced
t = 9.99 s is the time interval
So, the power dissipated is
[tex]P=\frac{E}{t}=\frac{181 J}{9.99 s}=18.1 W[/tex]
But the power dissipated can also be written as
[tex]P=\frac{V^2}{R}[/tex]
where
V = 12.0 V is the potential difference across the resistor
R is the resistance
Solving for R, we find
[tex]R=\frac{V^2}{P}=\frac{(12.0 V)^2}{18.1 W}=7.96 \Omega[/tex]
A rock is tossed straight up from the ground with a speed of 21 m/s . When it returns, it falls into a hole 10 m deep.a.) What is the rock's velocity as it hits the bottom of the hole?b.) How long is the rock in the air, from the instant it is released until it hits the bottom of the hole?
(a) 25.2 m/s
Let's take the initial vertical position of the rock as "zero" (reference height).
According to the law of conservation of energy, the speed of the rock as it reaches again the position "zero" after being thrown upwards is equal to the initial speed of the rock, 21 m/s (in fact, if there is no air resistance, no energy can be lost during the motion; and since the kinetic energy depends only on the speed of the rock:
[tex]K=\frac{1}{2}mv^2[/tex]
and the gravitational potential energy of the rock has not changed, since the rock has returned into its initial position, it means that the speed of the rock should be the same)
This means that we can only analyze the final part of the motion, the one in which the rock falls into the 10 m hole. Since it is a free fall motion, we can find the final speed by using
[tex]v^2 = u^2 + 2gd[/tex]
where
u = 21 m/s is the initial speed of the rock as it enters the hole
g = 9.8 m/s^2 is the acceleration due to gravity
d = 10 m is the depth of the hole
Substituting,
[tex]v=\sqrt{u^2 +2gd}=\sqrt{(21 m/s)^2+2(9.8 m/s^2)(10 m)}=25.2 m/s[/tex]
(b) 4.72 s
The vertical position of the rock at time t is given by
[tex]y(t) = v_y t - \frac{1}{2}gt^2[/tex]
where
[tex]v_y = 21 m/s[/tex] is the initial vertical velocity
Substituting y(t)=-10 m, we can then solve the equation for t to find the time at which the rock reaches the bottom of the hole:
[tex]-10 = 21 t - \frac{1}{2}(9.8)t^2\\10+21 t -4.9t^2 = 0[/tex]
which has two solutions:
t = -0.43 s --> negative, so we discard it
t = 4.72 s --> this is our solution
The rock's velocity as it hits the bottom of the hole is approximately 32.19 m/s. The rock is in the air for approximately 2.14 seconds.
Explanation:a) To find the rock's velocity as it hits the bottom of the hole, we can use the equation of motion: v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the displacement. In this case, the rock's initial velocity is 21 m/s, the acceleration is the acceleration due to gravity (-9.8 m/s^2), and the displacement is the depth of the hole (-10 m). Plugging in these values, we get:
v^2 = (21 m/s)^2 + 2(-9.8 m/s^2)(-10 m)
Simplifying, we find that v^2 = 841 + 196 = 1037. Taking the square root of both sides, we get v = √1037 ≈ 32.19 m/s. So, the rock's velocity as it hits the bottom of the hole is approximately 32.19 m/s.
b) To find the time the rock is in the air, we can use another equation of motion: v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. In this case, the initial velocity is 21 m/s, the acceleration is -9.8 m/s^2, and we want to find the time. Plugging in these values, we get:
0 = 21 m/s - 9.8 m/s^2 * t
Simplifying, we find that t ≈ 2.14 seconds. So, the rock is in the air for approximately 2.14 seconds.
A friend asks you how much pressure is in your car tires. You know that the tire manufacturer recommends 30 psi, but it's been a while since you've checked. You can't find a tire gauge in the car, but you do find the owner's manual and a ruler. Fortunately, you've just finished taking physics, so you tell your friend, "I don't know, but I can figure it out." From the owner's manual you find that the car's mass is 1000 kg . It seems reasonable to assume that each tire supports one-fourth of the weight. With the ruler you find that the tires are 13 cm wide and the flattened segment of the tire in contact with the road is 13 cm long.
Answer:
21 psi
Explanation:
The weight of the car is:
W = mg
W = 1000 kg * 9.8 m/s²
W = 9800 N
Divided by 4 tires, each tire supports:
F = W/4
F = 9800 N / 4
F = 2450 N
Pressure is force divided by area, so:
P = F / A
P = (2450 N) / (0.13 m × 0.13 m)
P ≈ 145,000 Pa
101,325 Pa is the same as 14.7 psi, so:
P ≈ 145,000 Pa × (14.7 psi / 101,325 Pa)
P ≈ 21 psi
The pressure in the car's tires can be calculated using physics principles. By calculating the weight of the car and dividing it by the area of contact each tire has with the ground, the pressure is found to be approximately 21.46 psi.
Explanation:To determine the tire pressure, we'll use the following physics concept: Pressure = Force / Area. The force in this case would be the weight of the car distributed among the four tires and the area would be the contact area of the tire with the ground.
First, we calculate the weight of the car by using the formula Weight = mass x gravity. Given that the mass of the car is 1000 kg and acceleration due to gravity is 10 m/s^2 (approx), the weight would be 10000 N (Newtons). Since each tire supports a quarter of the car's weight, the force each tire experiences is 2500 N.
Then, we calculate the area of contact the tire has with the ground. The flattened section would approximate a rectangle with length and width equal to 13 cm each, so Area = length x width = 0.13 m x 0.13 m = 0.0169 m^2.
Finally, we calculate the pressure by dividing the force by the area. Pressure = Force / Area = 2500 N / 0.0169 m^2 ≈ 148,000 Pa (Pascal).
Please note that 1 Pascal (Pa) = 0.000145 psi, so the pressure in psi would be approximately 21.46 psi, which is a bit lower than recommended by the tire manufacturer.
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Irrigation channels that require regular flow monitoring are often equipped with electromagnetic flowmeters in which the magnetic field is produced by horizontal coils embedded in the bottom of the channel. A particular coil has 100 turns and a diameter of 6.0 m. When it's time for a measurement, a 6.5 A current is turned on. The large diameter of the coil means that the field in the water flowing directly above the center of the coil is approximately equal to the field in the center of the coil. The field is directed downward and the water is flowing east. The water is flowing above the center of the coil at 1.5 m/s .Part A) What is the magnitude of the field at the center of the coil?Part B) What is the direction of the force on a positive ion in the water above the center of the coil?The force points west.The force points north.The force points south.The force points east.Part C) What is the magnitude of the force on an ion with a charge +e?
A) [tex]1.36\cdot 10^{-4}T[/tex]
The magnetic field at the center of a coil of N turns is given by
[tex]B=\frac{\mu_0 N I}{2R}[/tex]
where
I is the current in the coil
N is the number of turns
R is the radius of the coil
Here we have
I = 6.5 A is the current in the coil
N = 100 is the number of turns
[tex]R=\frac{6.0 m}{2}=3.0 m[/tex] is the radius of the coil
Substituting,
[tex]B=\frac{(4\pi \cdot 10^{-7} H/m)(100)(6.5 A)}{2(3.0 m)}=1.36\cdot 10^{-4}T[/tex]
B) The force points north
The direction of the force on a positive ion in water can be found by using the right-hand rule. In fact, we have:
- Index finger: direction of motion of the ion --> towards east
- Middle finger: direction of magnetic field --> downward
- Thumb: direction of the force --> towards north
So, the force points north.
C) [tex]3.26\cdot 10^{-23}N[/tex]
The magnitude of the magnetic force on a charged particle moving perpendicularly to the field is
[tex]F=qvB[/tex]
where
q is the charge of the particle
v is the velocity
B is the magnitude of the magnetic field
In this case, we have
[tex]q=+e=1.6\cdot 10^{-19} C[/tex] is the charge
[tex]v=1.5 m/s[/tex] is the velocity
[tex]B=1.36\cdot 10^{-4}T[/tex] is the magnetic field strength
Substituting,
[tex]F=(1.6\cdot 10^{-19} C)(1.5 m/s)(1.36\cdot 10^{-4}T)=3.26\cdot 10^{-23}N[/tex]
The magnetic field at the center of the coil is 0.34 T. The force on a positive ion in the water flowing above the coil points North. Without the value of the charge, the magnitude of the force on an ion cannot be determined.
Explanation:Part A) The magnitude of the magnetic field at the center of the coil, given that the field strength, B is calculated by the equation B = μI/(2r), where μ is the permeability of free space (4*10^-7 T.m/A), I is the current (6.5A), and r is the radius of the loop, which is the diameter divided by 2 (3m), is approximated as 0.34 T (Tesla).
Part B) The direction of the force on a positive ion in the water above the center of the coil is determined by the Right-Hand Rule-2. Given that the water is flowing east and the field is directed downward, the force on a positive ion would point to the north.
Part C) The magnitude of the force on an ion with a charge +e in this electromagnetic field can be calculated with the equation F = qvB, where q is the charge (+e), v is the velocity of the ion (1.5 m/s), and B is the magnetic field. However, without the actual value of the charge, this can't be determined from the current information.
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A 69 kg driver gets into an empty taptap to start the day's work. The springs compress 2×10−2 m . What is the effective spring constant of the spring system in the taptap? Enter the spring constant numerically in newtons per meter using two significant figures.
Answer:
[tex]3.4\cdot 10^4 N/m[/tex]
Explanation:
The spring system in the taptap obey's Hooke's law, which states that:
[tex]F=kx[/tex]
where
F is the magnitude of the force applied
k is the spring constant
x is the compression/stretching of the spring
In this problem:
- The force applied is the weight of the driver of mass m = 69 kg, so
[tex]F=mg=(69 kg)(9.8 m/s^2)=676.2 N[/tex]
- The compression of the spring is
[tex]x=2\cdot 10^{-2} m=0.02 m[/tex]
So, the spring constant is
[tex]k=\frac{F}{x}=\frac{676.2 N}{0.02 m}=3.4\cdot 10^4 N/m[/tex]
The spring constant of the spring compressed 2x10⁻² m by a 69 kg driver is 3.4x10⁴ N/m.
We can use the Hooke's law equation to find the spring constant:
[tex] F = -kx [/tex] (1)
Where:
F: is the Hooke's force
k: is the spring constant =?
x: is the distance of compression of the spring = -2x10⁻² m. The negative sign is because it is compressing (negative direction).
The force of equation (1) is equal to the weight force (W) of the driver, so:
[tex] W = F [/tex]
[tex] mg = -kx [/tex] (2)
Where:
m: is the mass of the driver = 69 kg
g: is the acceleration due to gravity = 9.81 m/s²
Solving equation (2) for k, we have:
[tex] k = -\frac{mg}{x} = -\frac{69 kg*9.81 m/s^{2}}{-2\cdot 10^{-2} m} = 3.4 \cdot 10^{4} N/m [/tex]
Therefore, the spring constant is 3.4x10⁴ N/m.
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If you do 50 J of work on a 10 N box, what distance did you move the box?
Answer:
5 m
Explanation:
Work is equal to force times distance.
W = Fd
50 Nm = (10 N) d
d = 5 m
You move the box 5 m.
Which of the following will increase the resistance of a wire? Check all that apply. Check all that apply. Decreasing the length of the wire will increase the resistance of the wire. Decreasing the cross-sectional area of the wire will increase the resistance of the wire. Increasing the cross-sectional area of the wire will increase the resistance of the wire. Increasing the resistivity of the material the wire is composed of will increase the resistance of the wire. Increasing the length of the wire will increase the resistance of the wire. Decreasing the resistivity of the material the wire is composed of will increase the resistance of the wire. SubmitRequest Answer Provide Feedback Next g
Answer:
Decreasing the cross-sectional area of the wire will increase the resistance of the wire.
Increasing the resistivity of the material the wire is composed of will increase the resistance of the wire.
Increasing the length of the wire will increase the resistance of the wire.
Explanation:
The resistance of a piece of wire is given by:
[tex]R=\rho \frac{L}{A}[/tex]
where
[tex]\rho[/tex] is the resistivity of the material
L is the length of the wire
A is the cross-sectional area of the wire
We notice that:
- The resistance of the wire is directly proportional to the resistivity of the material and the to the length of the wire
- The resistance of the wire is inversely proportional to the cross-sectional area of the wire
This means the following are correct:
Decreasing the cross-sectional area of the wire will increase the resistance of the wire.
Increasing the resistivity of the material the wire is composed of will increase the resistance of the wire.
Increasing the length of the wire will increase the resistance of the wire.
(a) Find the energy of the ground state (n = 1) and the first two excited states of an electron in a one-dimensional box of length L = 1.0 10-15 m = 1.00 fm (about the diameter of an atomic nucleus). ground state MeV first excited state MeV second excited state GeV Make an energy-level diagram for the system. (Do this on paper. Your instructor may ask you to turn in this work.) (b) Calculate the wavelength of electromagnetic radiation emitted when the electron makes a transition from n = 2 to n = 1. fm (c) Calculate the wavelength of electromagnetic radiation emitted when the electron makes a transition from n = 3 to n = 2. fm (d) Calculate the wavelength of electromagnetic radiation emitted when the electron makes a transition from n = 3 to n = 1. fm
(a) [tex]3.77\cdot 10^5 MeV, 1.51\cdot 10^6 MeV, 3.39\cdot 10^3 GeV[/tex]
The energy levels of an electron in a box are given by
[tex]E_n = \frac{n^2 h^2}{8mL^2}[/tex]
where
n is the energy level
[tex]h=6.63\cdot 10^{-34}Js[/tex] is the Planck constant
[tex]m=9.11\cdot 10^{-31}kg[/tex] is the mass of the electron
[tex]L=1.0\cdot 10^{-15} m[/tex] is the size of the box
Substituting n=1, we find the energy of the ground state:
[tex]E_1 = \frac{1^2 (6.63\cdot 10^{-34}^2}{8(9.11\cdot 10^{-31}(1.0\cdot 10^{-15})^2}=6.03\cdot 10^{-8}J[/tex]
Converting into MeV,
[tex]E_1 = \frac{6.03\cdot 10^{-8} J}{1.6\cdot 10^{-19} J/eV}\cdot 10^{-6} MeV/eV =3.77\cdot 10^5 MeV[/tex]
Substituting n=2, we find the energy of the first excited state:
[tex]E_2 = \frac{2^2 (6.63\cdot 10^{-34}^2}{8(9.11\cdot 10^{-31}(1.0\cdot 10^{-15})^2}=2.41\cdot 10^{-7}J[/tex]
Converting into MeV,
[tex]E_2 = \frac{2.41\cdot 10^{-7} J}{1.6\cdot 10^{-19} J/eV}\cdot 10^{-6} MeV/eV =1.51\cdot 10^6 MeV[/tex]
Substituting n=3, we find the energy of the second excited state:
[tex]E_3 = \frac{3^2 (6.63\cdot 10^{-34}^2}{8(9.11\cdot 10^{-31}(1.0\cdot 10^{-15})^2}=5.43\cdot 10^{-7}J[/tex]
Converting into GeV,
[tex]E_3 = \frac{5.43\cdot 10^{-7} J}{1.6\cdot 10^{-19} J/eV}\cdot 10^{-9} GeV/eV =3.39\cdot 10^3 GeV[/tex]
(b) [tex]1.10 \cdot 10^{-18} m[/tex]
The energy of the emitted radiation is equal to the energy difference between the two levels, so:
[tex]E=E_2 - E_1 = 2.41\cdot 10^{-7}J - 6.03\cdot 10^{-8} J=1.81\cdot 10^{-7} J[/tex]
And the energy of the electromagnetic radiation is
[tex]E=\frac{hc}{\lambda}[/tex]
where c is the speed of light; so, re-arranging the formula, we find the wavelength:
[tex]\lambda=\frac{hc}{E}=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{1.81\cdot 10^{-7}J}=1.10 \cdot 10^{-18} m[/tex]
(c) [tex]6.59 \cdot 10^{-19} m[/tex]
The energy of the emitted radiation is equal to the energy difference between the two levels, so:
[tex]E=E_3 - E_2 = 5.43\cdot 10^{-7} J - 2.41\cdot 10^{-7}J =3.02\cdot 10^{-7} J[/tex]
Using the same formula as before, we find the corresponding wavelength:
[tex]\lambda=\frac{hc}{E}=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{3.02\cdot 10^{-7}J}=6.59 \cdot 10^{-19} m[/tex]
(d) [tex]4.12 \cdot 10^{-19} m[/tex]
The energy of the emitted radiation is equal to the energy difference between the two levels, so:
[tex]E=E_3 - E_1 = 5.43\cdot 10^{-7} J - 6.03\cdot 10^{-8}J =4.83\cdot 10^{-7} J[/tex]
Using the same formula as before, we find:
[tex]\lambda=\frac{hc}{E}=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{4.83\cdot 10^{-7}J}=4.12 \cdot 10^{-19} m[/tex]
A solid sphere of uniform density has a mass of 4.4 × 104 kg and a radius of 1.9 m. What is the magnitude of the gravitational force due to the sphere on a particle of mass 8.3 kg located at a distance of (a) 3.7 m and (b) 0.41 m from the center of the sphere? (c) Write a general expression for the magnitude of the gravitational force on the particle at a distance r ≤ 1.9 m from the center of the sphere.
(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.
How much heat is absorbed by a 45g iron skillet when its temperature rises from 6oC to 21oC?
Explanation:
The heat (thermal energy) absorbed by the iron skillet can be found using the following equation:
[tex]Q=m.C.\Delta T[/tex] (1)
Where:
[tex]Q[/tex] is the heat
[tex]m=45 g[/tex] is the mass of the element (iron in this case)
[tex]C[/tex] is the specific heat capacity of the material. In the case of iron is [tex]C=0.444\frac{J}{g\°C}[/tex]
[tex]\Delta T=T_{f}-T_{i}=21\°C - 6\°C= 15\°C[/tex] is the variation in temperature
Knowing this, lets rewrite (1) with these values:
[tex]Q=(45 g)(0.444\frac{J}{g\°C})(15\°C)[/tex] (2)
Finally:
[tex]Q=299.7 J[/tex]
Answer:
299.7 J
Explanation:
X-rays with an energy of 301 keV undergo Compton scattering with a target. If the scattered X-rays are detected at 77.5^{\circ} ∘ relative to the incident X-rays, what is the wavelength of the scattered photon?
Answer:
[tex]6.03\cdot 10^{-12} m[/tex]
Explanation:
First of all, we need to find the initial wavelength of the photon.
We know that its energy is
[tex]E=301 keV = 4.82\cdot 10^{-14}J[/tex]
So its wavelength is given by:
[tex]\lambda = \frac{hc}{E}=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{4.82\cdot 10^{-14} J}=4.13\cdot 10^{-12}m[/tex]
The formula for the Compton scattering is:
[tex]\lambda' = \lambda +\frac{h}{mc}(1-cos \theta)[/tex]
where
[tex]\lambda[/tex] is the original wavelength
h is the Planck constant
m is the electron mass
c is the speed of light
[tex]\theta=77.5^{\circ}[/tex] is the angle of the scattered photon
Substituting, we find
[tex]\lambda' = 4.13\cdot 10^{-12} m +\frac{6.63\cdot 10^{-34} Js)}{(9.11\cdot 10^{-31}kg)(3\cdot 10^8 m/s)}(1-cos 77.5^{\circ})=6.03\cdot 10^{-12} m[/tex]
A soccer player takes a free kick from a spot that is 17 m from the goal. The ball leaves his foot at an angle of 38 ∘, and it eventually hits the crossbar of the goal, which is 2.4 m from the ground.Question: At what speed did the ball leave his foot?
Answer:
[tex]v=14.5(m/s)[/tex].
Explanation:
This is a projectile motion problem, so for solving it, we will require some equations of uniform motion and free-fall.
First, we need to recognize the data that we have so we can start solving the problem.
For Free -Fall:
[tex]g= -9.8(m/s^{2})[/tex]
[tex]\Delta y=2.4(m)[/tex]
For uniform motion:
[tex]\Delta x=17(m)[/tex]
The time that takes to the ball to travel 17m horizontally and to hit the crossbar at 2.4m of height is the same, so time is a common variable for free-fall and uniform motion.
Using the equations:
[tex]\Delta y=v_{y0}t+0.5gt^{2}[/tex] and
[tex]\Delta x=v_{x}t[/tex]
and noticing that
[tex]v_{y0}=v*sin(38)[/tex] and
[tex]v_{x}=v*cos(38)[/tex],
we obtain
[tex]\Delta y=v*sin(38)t+0.5gt^{2}[/tex] and
[tex]\Delta x=v*cos(38)t[/tex]
wich is a system of two equations and to variables that can be easily solve.
Making
[tex]v=\frac{\Delta x}{t*cos(38)}[/tex]
we get
[tex]\Delta y=(\frac{\Delta x}{t*cos(38)})sin(38)t+0.5gt^{2}[/tex],
[tex]\Delta y=\Delta x*tan(38)+0.5gt^{2}[/tex],
[tex]\Delta y-\Delta x*tan(38)=0.5gt^{2}[/tex],
[tex]t=\sqrt{\frac{\Delta y-\Delta x*tan(38)}{0.5g}}[/tex],
[tex][tex]t=\sqrt{\frac{2.4-17*tan(38)}{0.5*(-9.8)}}[/tex]
so
[tex]t=1.50s[/tex],
now the speed can be easily compute from one of our equations. Using
[tex]v=\frac{\Delta x}{t*cos(38)}[/tex],
[tex]v=\frac{17}{1.5*cos(38)}[/tex],
[tex]v=14.5(m/s)[/tex].
To find the initial speed of the ball, we can analyze the projectile motion of the ball. We can break down the initial velocity of the ball into horizontal and vertical components, and then use kinematic equations to solve for the initial speed. We can calculate the time it takes for the ball to reach the goal crossbar and then use that time and the horizontal distance to find the initial speed.
Explanation:To find the initial speed of the ball, we need to analyze the projectile motion of the ball. Given that the ball passes over the goal, 2.4 m above the ground, we can use the kinematic equations to solve for the initial speed. In this case, we know the horizontal distance (17 m) and the vertical distance (2.4 m above the ground). We can break down the initial velocity of the ball into horizontal and vertical components.
The horizontal component of the initial velocity (Vx) can be found using the equation:
Vx = V * cos(theta)
where V is the initial speed and theta is the angle of the kick. The vertical component of the initial velocity (Vy) can be found using the equation:
Vy = V * sin(theta)
Using these values, we can calculate the time it takes for the ball to reach the goal crossbar, which is the same as the time it takes for the ball to hit the ground. The time can be found using the equation:
t = 2 * Vy / g
where g is the acceleration due to gravity. Finally, we can use the time and the horizontal distance to find the initial speed using the equation:
Vx = distance / t
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A transformer works when an alternating current flows through the primary coil. This produces an alternating magnetic field in the primary coil. By electromagnetic induction this produces
A) a voltage in the primary coil.
B) a voltage in the secondary coil.
C) a direct current in the primary coil.
D) a direct current in the secondary coil.
a voltage in the secondary coil
Answerr
Explanation:
b
A compressed vertical spring stores 40 J of potential energy. The spring has a 0.1-kg stoneresting on it. The spring is released, throwing the stone straight up into the air. (A) How muchkinetic energy will the stone have when it first leaves the spring? (B) How much gravitationalenergy, relative to the spot where the stone was released, will the stone have when it reachesthe peak of its flight? (C) Calculate the height above the release point to which the stone travels.(D) Suggest something we could change about this situation that would cause the stone toreach a height double that calculated in Part (c).
(A) 40 J
Explanation:
The initial potential energy stored in the spring is:
[tex]U=40 J[/tex]
this energy is stored in the spring when the spring is compressed by a certain amount [tex]\Delta x[/tex], such that the elastic potential energy of the spring is
[tex]U=\frac{1}{2}k(\Delta x)^2[/tex]
where k is the spring constant. On the contrary, when it is at rest, the kinetic energy of the stone is zero:
[tex]K=\frac{1}{2}mv^2 = 0[/tex]
because the speed is zero (v=0).
When the spring is released, the spring returns to its equilibrium position, so that
[tex]\Delta x = 0[/tex]
and
[tex]U=0[/tex]
so the elastic potential energy becomes zero: so the total energy must be conserved, this means that all the potential energy has been converted into kinetic energy of the spring, so 40 J.
(B) 40 J
When the stone starts its motion, its kinetic energy is 40 J:
K = 40 J
While its gravitational potential energy is zero:
U = mgh = 0
where m is the mass of the stone, g is the gravitational acceleration, and h=0 is the height when the stone is thrown up.
As the stone goes up, its gravitational potential energy increases, since h in the formula increases; this means that the kinetic energy decreases, since the total energy must be constant.
When the stone reaches its maximum height, its speed becomes zero:
v = 0
This means that
K = 0
And so all the kinetic energy has been converted into gravitational potential energy, therefore
U = 40 J
(C) 40.8 m
At the maximum height of the trajectory of the stone, we have that the gravitational potential energy is
[tex]U=mgh = 40 J[/tex]
where
m = 0.1 kg is the mass of the stone
g = 9.8 m/s^2 is the acceleration due to gravity
h is the maximum height
Solving the formula for h, we find:
[tex]h=\frac{U}{mg}=\frac{40 J}{(0.1 kg)(9.8 m/s^2)}=40.8 m[/tex]
(D) The initial compression of the spring must be increased by a factor [tex]\sqrt{2}[/tex]
Here we want to double the maximum height reached by the stone:
h' = 2h
In order to do that, we must double its gravitational potential energy:
U' = 2U
This means that the initial potential energy stored in the spring must also be doubled, so
U' = 80 J
The elastic potential energy of the spring is
[tex]U' = \frac{1}{2}k(\Delta x)^2[/tex]
We see that the compression of the spring can be rewritten as
[tex]\Delta x = \sqrt{\frac{2U'}{k}}[/tex]
And we see that [tex]\Delta x[/tex] is proportional to the square root of the energy: therefore, if the energy has doubled, the compression must increase by a factor [tex]\sqrt{2}[/tex].
Which of the following explains why doctors practice antiseptic medicine?
A.) to prevent patients from passing their illness to the doctor
B.) to ensure that every patient receives enough antibiotics to kill all bacteria
C.) to ensure that every patient has enough good bacteria to kill harmful bacteria
D.) to prevent doctors from introducing microbes that could infect the patient
Doctors practice antiseptic medicine to D) prevent introducing infection-causing microbes to the patient. Managing bacterial infections requires careful use of antibiotics to avoid killing beneficial bacteria and prevent antibiotic resistance.
The reason why doctors practice antiseptic medicine is to avoid introducing microbes that could infect the patient. This is an essential part of infection control in medical settings. While antibiotics are valuable in treating bacterial infections by killing the bacteria causing the illness, they also impact the good bacteria that protect the body from infection. When deciding on using antimicrobial drugs, doctors must consider the type of bacterial infection and the potential resistance of bacteria to available antibiotics.
Antibiotic resistance is a significant concern in medicine, as it can lead to infections that are harder to treat, requiring longer and more expensive hospital stays. It emerges from the bacteria's adaptation through natural selection, effectively making certain drugs less effective over time.
What was Galileo’s contribution to the study of motion?
Hello! :)
A study of motion is basically like physics.
Hence, using the telescope, Galileo discovered the mountains on the moon, the spots on the sun, and four moons of Jupiter. His discoveries provided the evidence to support the theory that the earth and other planets revolved around the sun.
Hope this helped and I hope I answered in time!
Good luck!
~ Destiny ^_^
A difference in electric potential is commonly known as voltage, and is provided by standard batteries. If you built two identical circuits, the circuit connected to a larger voltage would result in a(n) __________ current.
A) higher
B) lower
C) magnetic
D) unchanged
Answer:
higher
Explanation:
A bigger voltage will result in a faster movement of charge.
wo parallel plates of area 100cm2are given charges of equal magnitudes 8.9 ×10−7C but opposite signs. The electric field within the dielectric material filling the space between the plates is 1.4 ×106V/m. (a) Calculate the dielectric constant of the material. (b) Determine the magnitude of the charge induced on each dielectric surface.
(a) 7.18
The electric field within a parallel plate capacitor with dielectric is given by:
[tex]E=\frac{\sigma}{k \epsilon_0}[/tex] (1)
where
[tex]\sigma[/tex] is the surface charge density
k is the dielectric constant
[tex]\epsilon_0[/tex] is the vacuum permittivity
The area of the plates in this capacitor is
[tex]A=100 cm^2 = 100\cdot 10^{-4} m^2[/tex]
while the charge is
[tex]Q=8.9\cdot 10^{-7}C[/tex]
So the surface charge density is
[tex]\sigma = \frac{Q}{A}=\frac{8.9\cdot 10^{-7} C}{100\cdot 10^{-4} m^2}=8.9\cdot 10^{-5} C/m^2[/tex]
The electric field is
[tex]E=1.4\cdot 10^6 V/m[/tex]
So we can re-arrange eq.(1) to find k:
[tex]k=\frac{\sigma}{E \epsilon_0}=\frac{8.9\cdot 10^{-5} C/m^2}{(1.4\cdot 10^6 V/m)(8.85\cdot 10^{-12} F/m)}=7.18[/tex]
(b) [tex]7.66\cdot 10^{-7}C[/tex]
The surface charge density induced on each dielectric surface is given by
[tex]\sigma' = \sigma (1-\frac{1}{k})[/tex]
where
[tex]\sigma=8.9\cdot 10^{-5} C/m^2[/tex] is the initial charge density
k = 7.18 is the dielectric constant
Substituting,
[tex]\sigma' = (8.9\cdot 10^{-5} C/m^2) (1-\frac{1}{7.18})=7.66\cdot 10^{5} C/m^2[/tex]
And by multiplying by the area, we find the charge induced on each surface:
[tex]Q' = \sigma' A = (7.66\cdot 10^{-5} C/m^2)(100 \cdot 10^{-4}m^2)=7.66\cdot 10^{-7}C[/tex]
Final answer:
The dielectric constant cannot be determined without the electric field value without the dielectric (E0). The induced charge on each dielectric surface can be calculated once the electric field (E) is given, using the permittivity of free space and the area of the plates.
Explanation:
To solve the student's question regarding the dielectric constant and the induced charge on a dielectric material, we need to apply concepts from electrodynamics and the properties of capacitors.
Dielectric Constant (k)
The dielectric constant (k) is given by the ratio of the electric field without the dielectric (E0) to the electric field with the dielectric (E). Here, you provided us with the electric field within the dielectric material (E) which is 1.4 × 106 V/m, but have not given us the electric field without the dielectric (E0). To find k, we would normally use the following equation:
k = E0 / E
Without knowing E0, we cannot calculate the dielectric constant directly.
Induced Charge (Qi)
We can calculate the induced charge on each surface of the dielectric (Qi) using the relation between electric field (E), surface charge density (σ), and the permittivity of free space (ε0).
E = σ / ε0
From that, the induced surface charge density is:
σi = E × ε0
Where the permittivity of free space (ε0) is approximately 8.85 × 10-12 F/m. After calculating σi, we can find Qi by multiplying σi with the plate area (A), which you've given as 100 cm2 or 0.01 m2.
Because of increasing evidence of damage to the ozone layer, chlorofluorocarbon (CFC) production was banned in 1996. However, there are about 100 million auto air conditioners that still use CFC−12 (CF2Cl2). These air conditioners are recharged from stockpiled supplies of CFC−12. If each of the 100 million automobiles contains 1.2 kg of CFC−12 and leaks 20 % of its CFC−12 into the atmosphere per year, how much chlorine, in kg, is added to the atmosphere each year due to auto air conditioners? (Assume two significant figures in your calculations.)
Each automobile adds 0.24 kg of chlorine to the atmosphere each year due to CFC-12 leakage. The total amount of chlorine added each year is 24,000,000 kg.
Explanation:To calculate the amount of chlorine added to the atmosphere each year due to auto air conditioners, we need to consider the amount of CFC-12 leaked by each automobile.
Given that each automobile contains 1.2 kg of CFC-12 and leaks 20% per year, we can calculate the amount of chlorine added per automobile per year:
Chlorine added per automobile per year = 1.2 kg of CFC-12 x 20% = 0.24 kg.
Since there are 100 million automobiles, the total amount of chlorine added to the atmosphere each year due to auto air conditioners is:
Total chlorine added per year = Chlorine added per automobile per year * Number of automobiles = 0.24 kg * 100,000,000 = 24,000,000 kg.
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The range is the horizontal distance from the cannon when the pumpkin hits the ground. This distance is given by the product of the horizontal velocity (which is constant) and the amount of time the pumpkin is in the air (which is determined by the vertical component of the initial velocity, as you just discovered). Set the initial speed to 14 m/s, and fire the pumpkin several times while varying the angle between the cannon and the horizontal. For which angle is the range a maximum (with the initial speed held constant)?
Angle:
0
60
45
90
30
The answer is 45 degrees. I am not doing a field experiment for you that involves a cannon and a day's work, for 5 points.
For 45° angle, the range of the pumpkin is maximum with constant initial velocity.
To find the answer, we need to know about the mathematical expression of range in terms of initial velocity and projectile angle.
What is the time of flight of pumpkin in the air?From Newton's equation of motion, we know the time of flight is 2usinФ/g.Where u is the initial velocity and Ф is the angle of projection.What is the mathematical formulation of range of the pumpkin?Range is given as flight time × horizontal velocity.So, range = (2usinФ/g)×ucosФ= usin2Ф/g
Thus, we can conclude that the range will be maximum when the angle is 45° with constant initial velocity.
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A light source of wavelength λ illuminates a metal with a work function (a.k.a., binding energy) of BE=2.00 eV and ejects electrons with a maximum KE=4.00 eV. A second light source with double the wavelength of the first ejects photoelectrons with what maximum kinetic energy?
Explanation:
The described situation is the photoelectric effect, which consists of the emission of electrons (electric current) that occurs when light falls on a metal surface under certain conditions.
If we consider the light as a stream of photons and each of them has energy, this energy is able to pull an electron out of the crystalline lattice of the metal and communicate, in addition, a kinetic energy.
This is what Einstein proposed:
Light behaves like a stream of particles called photons with an energy [tex]E[/tex]:
[tex]E=h.f[/tex] (1)
So, the energy [tex]E[/tex] of the incident photon must be equal to the sum of the Work function [tex]\Phi[/tex] of the metal and the kinetic energy [tex]K[/tex] of the photoelectron:
[tex]E=\Phi+K[/tex] (2)
Where [tex]\Phi[/tex] is the minimum amount of energy required to induce the photoemission of electrons from the surface of a metal, and its value depends on the metal.
In this case [tex]\Phi=2eV[/tex] and [tex]K_{1}=4eV[/tex]
So, for the first light source of wavelength [tex]\lambda_{1}[/tex], and applying equation (2) we have:
[tex]E_{1}=2eV+4eV[/tex] (3)
[tex]E_{1}=6eV[/tex] (4)
Now, substituting (1) in (4):
[tex]h.f=6eV[/tex] (5)
Where:
[tex]h=4.136(10)^{-15}eV.s[/tex] is the Planck constant
[tex]f[/tex] is the frequency
Now, the frequency has an inverse relation with the wavelength
[tex]\lambda_{1}[/tex]:
[tex]f=\frac{c}{\lambda_{1}}[/tex] (6)
Where [tex]c=3(10)^{8}m/s[/tex] is the speed of light in vacuum
Substituting (6) in (5):
[tex]\frac{hc}{\lambda_{1}}=6eV[/tex] (7)
Then finding [tex]\lambda_{1}[/tex]:
[tex]\lambda_{1}=\frac{hc}{6eV } [/tex] (8)
[tex]\lambda_{1}=\frac{(4.136(10)^{-15} eV.s)(3(10)^{8}m/s)}{6eV}[/tex]
We obtain the wavelength of the first light suorce [tex]\lambda_{1}[/tex]:
[tex]\lambda_{1}=2.06(10)^{-7}m[/tex] (9)
Now, we are told the second light source [tex]\lambda_{2}[/tex] has the double the wavelength of the first:
[tex]\lambda_{2}=2\lambda_{1}=(2)(2.06(10)^{-7}m)[/tex] (10)
Then: [tex]\lambda_{2}=4.12(10)^{-7}m[/tex] (11)
Knowing this value we can find [tex]E_{2}[/tex]:
[tex]E_{2}=\frac{hc}{\lambda_{2}}[/tex] (12)
[tex]E_{2}=\frac{(4.136(10)^{-15} eV.s)(3(10)^{8}m/s)}{4.12(10)^{-7}m}[/tex] (12)
[tex]E_{2}=3.011eV[/tex] (13)
Knowing the value of [tex]E_{2}[/tex] and [tex]\lambda_{2}[/tex], and knowing we are working with the same work function, we can finally find the maximum kinetic energy [tex]K_{2}[/tex] for this wavelength:
[tex]E_{2}=\Phi+K_{2}[/tex] (14)
[tex]K_{2}=E_{2}-\Phi[/tex] (15)
[tex]K_{2}=3.011eV-2eV [/tex]
[tex]K_{2}=1.011 eV[/tex] This is the maximum kinetic energy for the second light source
8. A package was determined to have a mass of 5.7 kilograms. What's the force of gravity acting on the package on earth?
Answer:56 N
Explanation:
Weight is equal to mass times the acceleration of gravity. On earth, that's 9.8 m/s².
W = mg
W = (5.7 kg) (9.8 m/s²)
W = 55.86 N
Since the mass is precise to 2 significant figures, we should round our answer to 2 significant figures: W = 56 N.
The terminals of a battery are connected across two resistors in series. The resistances of the resistors are not the same. Which of the following statements are correct? Choose all that are correct. (a) The resistor with the smaller resistance carries more current than the other resistor. (b) The resistor with the larger resistance carries less current than the other resistor. (c) The current in each resistor is the same. (d) The potential difference across each resistor is the same. (e) The potential difference is greatest across the resistor closest to the positive termina
Answer:
(c) The current in each resistor is the same.
Explanation:
When two resistors are connected in series, we have the following:
- The resistors are connected such that the current passing through the two resistors is the same
- The voltage of the battery is equal to the sum of the voltage drops across each resistor
- the equivalent resistance of the circuit is equal to the sum of the individual resistances:
R = R1 + R2
So, let's analyze each statement:
(a) The resistor with the smaller resistance carries more current than the other resistor. --> FALSE. The current through the two resistors is the same.
(b) The resistor with the larger resistance carries less current than the other resistor. --> FALSE. The current through the two resistors is the same.
(c) The current in each resistor is the same. --> TRUE.
(d) The potential difference across each resistor is the same. --> FALSE: the potential difference across each resistor is given by
V=RI
where I (the current) is the same for both resistors, while R (the resistance) is not, so V is also different for the two resistors.
(e) The potential difference is greatest across the resistor closest to the positive terminal --> FALSE. According to
V=RI
the potential difference depends only on the value of the resistance, so it doesn't matter which resistor is connected to the positive terminal.
In a series circuit, the resistors have the same current flowing through them but can have different potential differences. The resistor with smaller resistance carries more current, while the one with larger resistance carries less current. The potential difference across each resistor is not the same.
Explanation:In a series circuit, the resistors are connected one after the other, so the current passing through each resistor is the same. However, the potential difference (voltage) across each resistor can be different depending on their resistances.
Therefore, the correct statements are:
(a) The resistor with the smaller resistance carries more current than the other resistor. Since the current is the same in both resistors, the one with smaller resistance will have a larger potential difference across it. This means it carries more current.
(b) The resistor with the larger resistance carries less current than the other resistor. Again, since the current is the same in both resistors, the one with larger resistance will have a smaller potential difference across it. This means it carries less current.
(d) The potential difference across each resistor is the same. This statement is incorrect. As mentioned before, the potential difference across each resistor can be different depending on their resistances.
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A pendulum has 532 J of potential energy at the highest point of its swing. How much kinetic energy will it have at the bottom of its swing?
Answer:
532 J
Explanation:
If there are no frictional forces/air resistance involved in the problem, then the mechanical energy of the system is conserved.
This means that:
[tex]E_i = E_f[/tex]
where E_i is the initial mechanical energy and E_f is the final mechanical energy. The mechanical energy is the sum of potential energy U and kinetic energy K:
[tex]U_i + K_i = U_f + K_f[/tex]
At the highest point, the speed of the swing is zero, so
[tex]K_i = 0[/tex]
while at the bottom point, the potential energy is zero (if we take the bottom point of the swing as reference level), so
[tex]U_f =0[/tex]
This means that the previous equation becomes
[tex]U_i = K_f[/tex]
and since
[tex]U_i = 532 J[/tex]
the kinetic energy at the bottom of its swing is
[tex]K_f = 532 J[/tex]
One mole of a substance contains 6.02 × 1023 protons and an equal number of electrons. If the protons could somehow be separated from the electrons and placed in very small, individual containers separated by a million meters, what would be the magnitude of the electrostatic force exerted by one box on the other? A) 8.7 × 103 N B) 9.5 × 104 N C) 2.2 × 105 N D) 8.4 × 107 N E) 1.6 × 108 N
Answer:
[tex]8.4\cdot 10^7 N[/tex]
Explanation:
The electrostatic force between two objects is given by:
[tex]F=k\frac{q_1 q_2}{r^2}[/tex]
where
k is the Coulomb's constant
q1 and q2 are the charges of the two objects
r is the separation between the two objects
In this problem, we have two boxes separated by
[tex]r = 1\cdot 10^6 m[/tex]
The first box contains [tex]6.02\cdot 10^{23}[/tex] protons, so its charge is:
[tex]q_1 = (6.02\cdot 10^{23})(1.6\cdot 10^{-19} C)=9.63\cdot 10^4 C[/tex]
The second box contains [tex]6.02\cdot 10^{23}[/tex] electrons, so its charge is:
[tex]q_2 = (6.02\cdot 10^{23})(-1.6\cdot 10^{-19} C)=-9.63\cdot 10^4 C[/tex]
We are only interested in the magnitude of the force, so we can neglect the negative sign and calculate the electrostatic force as:
[tex]F=(9\cdot 10^9) \frac{(9.63\cdot 10^4 C)(9.63\cdot 10^4 C)}{(1\cdot 10^6 m)^2}=8.4\cdot 10^7 N[/tex]
Sound enters the ear, travels through the auditory canal, and reaches the eardrum. The auditory canal is approximately a tube open at only one end. The other end is closed by the eardrum. A typical length for the auditory canal in an adult is about 2.26 cm. The speed of sound is 348 m/s. What is the fundamental frequency of the canal? (Interestingly, the fundamental frequency is in the frequency range where human hearing is most sensitive.)
The fundamental frequency of the auditory canal, which acts like a tube closed at one end, can be calculated using the speed of sound and the length of the canal. With a typical length of 2.26 cm and a speed of sound of 348 m/s, the fundamental frequency is approximately 3846 Hz.
Explanation:The student is asking about the fundamental frequency of the auditory canal, which is a tube that resonates at a specific frequency when sound waves travel through it and strike the eardrum. To calculate this frequency, we consider that the auditory canal acts like a tube closed at one end with the closed end being the eardrum.
To find the fundamental frequency (f1), we must use the formula for a tube closed at one end:
f1 = v / (4 * L), where v is the speed of sound and L is the length of the auditory canal. Given that the speed of sound (v) is 348 m/s and the typical length (L) of an adult's auditory canal is about 2.26 cm or 0.0226 m, the fundamental frequency can be calculated as follows:
f1 = 348 m/s / (4 * 0.0226 m)
After calculating this expression, we find:
f1 = 348 / (4 * 0.0226)
f1 ≈ 3846 Hz
This result indicates that the fundamental frequency is within the range that the human ear is most sensitive to, specifically in the 2000-5000 Hz range, partly explaining the ear's acute sensitivity to certain sounds.
can someone help me?!!!!!
Answer:
Third Option
[tex]B = -0.5A[/tex]
Explanation:
If we have a vector A = ax + by we know that by definition
cA = cax + cby
Where c is a constant.
In this case we have two vectors
[tex]A = 7.6\^x -9.2\^y\\\\B = -3.8\^x + 4.6\^y[/tex]
You may notice that vector B has an opposite direction to vector A.
You may also notice that | Ax | is the double of | Bx | and | Ay | is double of |By |
That is to say
[tex]3.8 +3.8 = 7.6\\\\4.6 +4.6 = 9.2[/tex]
So the equation that relates to vectors A and B is:
[tex]B = -0.5A[/tex].
You can verify this relationship by performing the operation
[tex]B = -0.5A[/tex]
[tex]-3.8\^x + 4.6\^y = -0.5(7.6\^x -9.2\^y)\\\\\-3.8\^x + 4.6\^y = -3.8\^x + 4.6\^y[/tex]
The answer is:
The third option,
[tex]B=-0.5A[/tex]
Why?To solve the problem, we need to remember the following vector property:
Vector multiplied by a scalar or constant number:
Multiplying a vector by a scalar number results in multiplying each of the components of the vector (x, y and z) by the scalar or constant number (including its sign).
So, we are given the following vectors:
[tex]A=(7.6,-9,2)\\B=(-3.8,4.6)[/tex]
So, writing an equation that involves both given vectors, we have:
[tex]B=-0.5A\\\\(-3.8,4.6)=-0.5((7.6,-9,2))=((-0.5*7.6),(-0.5*-9.2)\\\\(-3.8,4.6)=(-3.8,4.6)[/tex]
Hence, we have that the answer is the third option,
[tex]B=0.5A[/tex]
Have a nice day!
A 22.3-g bullet moving at 1 000 m/s is fired through a one-kg block of wood emerging at a speed of 100 m/s. What is the change in the kinetic energy of the bullet-block system as a result of the collision assuming the block is free to move?
Final answer:
The change in kinetic energy of the bullet-block system can be calculated using the principle of conservation of kinetic energy. The initial kinetic energy of the bullet is given by 0.5 * mass of the bullet * (initial velocity of the bullet)^2, and the final kinetic energy is given by 0.5 * mass of the bullet * (final velocity of the bullet)^2. The change in kinetic energy is the difference between the final and initial kinetic energies.
Explanation:
The change in kinetic energy of the bullet-block system can be calculated using the principle of conservation of kinetic energy. The initial kinetic energy of the bullet is given by 0.5 * mass of the bullet * (initial velocity of the bullet)^2, and the final kinetic energy is given by 0.5 * mass of the bullet * (final velocity of the bullet)^2. The change in kinetic energy is the difference between the final and initial kinetic energies.
Using the values given in the question:
Mass of the bullet = 22.3 g = 0.0223 kg
Initial velocity of the bullet = 1000 m/s
Final velocity of the bullet = 100 m/s
Initial kinetic energy of the bullet = 0.5 * 0.0223 kg * (1000 m/s)^2 = 111.5 J
Final kinetic energy of the bullet = 0.5 * 0.0223 kg * (100 m/s)^2 = 11.15 J
Change in kinetic energy of the bullet-block system = Final kinetic energy - Initial kinetic energy = 11.15 J - 111.5 J = -100.35 J
To practice Problem-Solving Strategy 25.1 Electromagnetic induction. A loop of wire of radius a = 30. mm has an electrical resistance R = 0.032 Ω . The loop is initially inside a uniform magnetic field of magnitude B0 = 1.5 T parallel to the loop's axis. The magnetic field is then reduced slowly at a constant rate, which induces a current I = 0.20 A in the loop. How long does it take for the magnitude of the uniform magnetic field to drop from 1.5 T to zero? As the external magnetic field decreases, an induced current flows in the coil. What is the direction of the induced magnetic field caused by this current?
1. 0.66 s
The current induced in the loop is
I = 0.20 A
while the resistance of the loop is
[tex]R=0.032 \Omega[/tex]
so the emf induced in the loop can be found by using Ohm's law
[tex]\epsilon = RI=(0.032 \Omega)(0.20 A)=0.0064 V[/tex]
We also know that according to Faraday-Newmann-Lenz, the induced emf is
[tex]\epsilon=-\frac{\Delta \Phi}{\Delta t}[/tex] (1)
where
[tex]\Delta \Phi[/tex] is the variation of magnetic flux through the loop
[tex]\Delta t[/tex] is the time interval
We have:
- Initial magnetic field: B = 1.5 T
- Radius of the coil: r = 30 mm = 0.03 m
- So, area of the coil: [tex]A=\pi r^2 = \pi (0.03 m)^2=2.83\cdot 10^{-3} m^2[/tex]
so the initial flux through the coil is
[tex]\Phi_i = B_i A = (1.5 T)(2.83\cdot 10^{-3}m^2)=4.25\cdot 10^{-3} Wb[/tex]
While the final flux through the coil is zero, since the magnetic field at the end is zero, so the change in magnetic flux is
[tex]\Delta \Phi = \Phi_f - \Phi_i = 0-4.25\cdot 10^{-3} Wb=-4.25\cdot 10^{-3} Wb[/tex]
Now re-arranging eq.(1) we find the time interval needed:
[tex]\Delta t = -\frac{\Delta \Phi}{\epsilon}=-\frac{-4.25\cdot 10^{-3} Wb}{0.0064 V}=0.66 s[/tex]
2. Same direction as the external magnetic field
The direction of the induced magnetic field is given by Lenz's law.
In fact, Lenz law states that the induced current in the loop is such that the magnetic field opposes the variation of magnetic flux through the coil.
Here, the magnetic flux through the coil is decreasing, since the external magnetic field is decreasing: this means that the induced magnetic field must be in the same direction as the external magnetic field (in order to restore the flux and to oppose this decrease of flux).
The amount of time it would take for the magnitude of the uniform magnetic field to drop is 0.66 seconds.
Given the following data:
Radius = 30 mm to meters = 0.02 m.
Resistance = 0.032 Ω.
Magnetic field = 1.5 T.
Current = 0.20 A.
How to calculate the time.In order to determine the amount of time it would take for the magnitude of the uniform magnetic field to drop from 1.5 Tesla to zero (0), we would apply Faraday-Newmann-Lenz equation.
For the wire's area:
[tex]A = \pi r^2\\\\A= 3.142 \times 0.03^2\\\\A=2.83 \times 10^{-3}\;m^2[/tex]
For the initial magnetic flux:
[tex]\phi _i = B_i A\\\\\phi _i = 1.5 \times 2.83 \times 10^{-3}\\\\\phi _i = 4.25 \times 10^{-3}\;Wb[/tex]
For the change in magnetic flux:
[tex]\Delta \phi = \phi_f - \phi_i\\\\\Delta \phi =0-4.25\times 10^{-3}\\\\\Delta \phi =-4.25\times 10^{-3}\;Wb[/tex]
From Faraday-Newmann-Lenz equation, we have:
[tex]\epsilon =-\frac{\Delta \phi}{\Delta t} \\\\\Delta t=-\frac{\Delta \phi}{\epsilon} \\\\\Delta t=-(\frac{-4.25\times 10^{-3}}{0.20 \times 0.032})\\\\\Delta t=\frac{4.25\times 10^{-3}}{0.0064}\\\\\Delta t=0.66\;seconds[/tex]
The direction of the induced magnetic field that is caused by this current is in the same direction as the external magnetic field in accordance with Lenz's law.
Read more on magnetic field here: https://brainly.com/question/7802337