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mafiozo [28]
3 years ago
8

Note that the simulation allows you to also display the force of the smaller moon

Physics
1 answer:
Lelu [443]3 years ago
4 0

the force that the planet exerts on the moon is equal to the force that the moon exerts on the planet

Explanation:

In this problem we are analzying the gravitational force acting between a planet and its moon.

The magnitude of the gravitational attraction between two objects is given by

F=G\frac{m_1 m_2}{r^2}

where :

G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2} is the gravitational constant

m1, m2 are the masses of the two objects

r is the separation between them

In this problem, we are considering a planet and its moon. According to Newton's third law of motion,

"When an object A exerts a force (action force) on an object B, then object B exerts an equal and opposite force (reaction force) on object A"

If we apply this law to this situation, this means that the force that the planet exerts on the moon is equal to the force that the moon exerts on the planet.

Learn more about gravitational force:

brainly.com/question/1724648

brainly.com/question/12785992

#LearnwithBrainly

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A 40-turn coil has a diameter of 11 cm. The coil is placed in a spatially uniform magnetic field of magnitude 0.40 T so that the
seropon [69]

Answer:

(a) emf = 0.507 V

(b) emf = 0.0507 V

(c) emf = 0.00234 V

Explanation:

Given;

number of turns of the coil, N = 40 turns

diameter of the coil, d = 11 cm

radius of the coil, r = 5.5 cm = 0.055 m

magnitude of the magnetic field, B = 0.4 T

The magnitude of the induced emf is calculated as;

emf = - N\frac{d\phi}{dt} \\\\where;\\\\\phi \ is \ magnetic \ flux= BA \\\\A \ is the \ area \ of \ the \ coil = \pi r^2 = \pi (0.055)^2 = 0.0095 \ m^2\\\\emf = - N \frac{dB.A}{dt} = -NA\frac{dB}{dt} \\\\emf = -NA\frac{(B_2 - B_1)}{t} \\\\emf = NA \frac{(B_1 - B_2)}{t} \\\\the \ final \ magnetic \ field \ is \ reduced \ to \ zero;\ B_2 = 0\\\\emf = \frac{NAB_1}{t}

(a) when the time, t = 0.3 s

emf = \frac{NAB_1}{t} = \frac{40\times 0.0095\times 0.4}{0.3} = 0.507 \ V

(b) when the time, t = 3.0 s

emf = \frac{NAB_1}{t} = \frac{40\times 0.0095\times 0.4}{3} = 0.0507 \ V

(c) when the time, t = 65 s

emf = \frac{NAB_1}{t} = \frac{40\times 0.0095\times 0.4}{65} = 0.00234 \ V

7 0
3 years ago
A person strikes a ball with a bat. The temperature of the ball increases by 0.06ᵒC. What accounts for the increase?
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The increase in the average kinetic energy of the ball causes the increase in the temperature of the ball.

Kinetic energy of a particle is directly proportional to its temperature.

A ball initially at rest acquires kinetic energy when an external force is applied to it. As the person strikes the ball with a bat, the ball gains momentum which increases its kinetic energy of the ball.

Temperature on the other hand, is the measure of the average kinetic energy of a particle. Consequently, as the kinetic energy of the ball increases, the temperature of the ball increases as well.

Thus, we can conclude that the increase in the average kinetic energy of the ball causes the increase in the temperature of the ball.

Learn more here: brainly.com/question/18833622

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Answer:

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