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jok3333 [9.3K]
2 years ago
9

A planet has two moons, Moon A and Moon B, that orbit at different distances from the planet's center, as shown. Astronomerscoll

ect data regarding the planet, the two moons, and their obits. The astronomers are able to estimate the planet's radius and
mass.
The masses of the two moons are determined to be 2M for Moon A and M for Moon B. It is observed that the distance
between Moon B and the planet is two times that of the distance between Moon A and the planet. How does force exerted from
the planet on Moon A compare to the force exerted from the planet on Moon B?




READ CAREFULLY PLEASE


A) The gravitational force exerted from the planet on Moon A is two times larger than the gravitational force exerted from the planet on Moon B

B) The gravitational force exerted from the planet on Moon A is eight times larger than the gravitational force exerted from the planet on Moon B

C) The gravitational force exerted from the planet on Moon A is two times smaller than the gravitational force exerted from the planet on Moon B

D) The gravitational force exerted from the planet on Moon A is eight times smaller than the gravitational force exerted from the planet on Moon B
​
Physics
2 answers:
Crazy boy [7]2 years ago
8 0

B) The gravitational force exerted from the planet on Moon A is eight times larger than the gravitational force exerted from the planet on Moon B

Gala2k [10]2 years ago
5 0
C: The gravitational force exerted from the planet on Moon A is two times smaller than the gravitational force exerted from the planet on Moon B.
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A car travels along a clear 10.0 km section of motorway in 6.0 minutes. It then drives through 3.0 km of roadwork in 3.0 minutes
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\\ \bull\tt\dashrightarrow Avg\:Speed=\dfrac{Total\:Displacement}{Total\:Time}

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A baseball, which has a mass of 0.685 kg., is moving with a velocity of 38.0 m/s when it contacts the baseball bat duringwhich t
Evgen [1.6K]

Answers:

a) 65.075 kgm/s

b) 10.526 s

c) 61.82 N

Explanation:

<h3>a) Impulse delivered to the ball</h3>

According to the Impulse-Momentum theorem we have the following:

I=\Delta p=p_{2}-p_{1} (1)

Where:

I is the impulse

\Delta p is the change in momentum

p_{2}=mV_{2} is the final momentum of the ball with mass m=0.685 kg and final velocity (to the right) V_{2}=57 m/s

p_{1}=mV_{1} is the initial momentum of the ball with initial velocity (to the left) V_{1}=-38 m/s

So:

I=\Delta p=mV_{2}-mV_{1} (2)

I=\Delta p=m(V_{2}-V_{1}) (3)

I=\Delta p=0.685 kg (57 m/s-(-38 m/s)) (4)

I=\Delta p=65.075 kg m/s (5)

<h3>b) Time </h3>

This time can be calculated by the following equations, taking into account the ball undergoes a maximum compression of approximately 1.0 cm=0.01 m:

V_{2}=V_{1}+at (6)

V_{2}^{2}=V_{1}^{2}+2ad (7)

Where:

a is the acceleration

d=0.01 m is the length the ball was compressed

t is the time

Finding a from (7):

a=\frac{V_{2}^{2}-V_{1}^{2}}{2d} (8)

a=\frac{(57 m/s)^{2}-(-38 m/s)^{2}}{2(0.01 m)} (9)

a=90.25 m/s^{2} (10)

Substituting (10) in (6):

57 m/s=-38 m/s+(90.25 m/s^{2})t (11)

Finding t:

t=1.052 s (12)

<h3>c) Force applied to the ball by the bat </h3>

According to Newton's second law of motion, the force F is proportional to the variation of momentum  \Delta p in time  \Delta t:

F=\frac{\Delta p}{\Delta t} (13)

F=\frac{65.075 kgm/s}{1.052 s} (14)

Finally:

F=61.82 N

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