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3 years ago

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A space probe is directly between two moons of a planet. If it is twice as far from moon A as it is from moon B, but the net for

ce on the probe is zero, what can be said about the relative masses of the moons? a. Moon A is twice as massive as moon B. b. Moon A has the same mass as moon B. c. Moon A is four times as massive as moon B. d. Moon A is half as massive as moon B.

1 answer:

Dennis_Churaev [7]3 years ago

3 0

Answer:

c. Moon A is four times as massive as moon B

Explanation:

Let's assume the:

mass of the object = $m\,kilogram$

mass of the moon A = $M_A\,kilogram$

mass of the moon B = $M_B\,kilogram$

distance between the center of masses of the object and moon B = $r\,meters$

According to the given condition the object is twice as far from moon A as it is from moon B

∴distance between the center of masses of the object and moon B = $2r\,meters$

<u>As we know, gravitational force of attraction is given by:</u>

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

<em>According to the condition</em>

Force on m due to $M_B=$ Force on m due to $M_A$

$G\frac{m.M_A}{(2r)^2} =G\frac{m.M_B}{(r)^2}$

$\frac{M_A}{4r^2} =\frac{M_B}{r^2}$

$M_A=4M_B$

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3 years ago

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An experiment is done to compare the initial speed of projectiles fired from high-performance catapults. The catapults are place

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Answer:1.084

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Given

mass of Pendulum M=10 kg

mass of bullet m=5.5 gm

velocity of bullet u

After collision let say velocity is v

conserving momentum we get

$mu=(M+m)v$

$v=\frac{m}{M+m}\times u$

Conserving Energy for Pendulum

Kinetic Energy=Potential Energy

$\frac{(M+m)v^2}{2}=(M+m)gh$

here $h=L(1-\cos \theta )$ from diagram

therefore

$v=\sqrt{2gL(1-\cos \theta )}$

initial velocity in terms of v

$u=\frac{M+m}{m}\times \sqrt{2gL(1-\cos \theta )}$

For first case $\theta =6.8^{\circ}$

$u_1=\frac{M+m_1}{m_1}\times \sqrt{2gL(1-\cos 6.8)}$

for second case $\theta =11.4^{\circ}$

$u_2=\frac{M+m_2}{m_2}\times \sqrt{2gL(1-\cos 11.4)}$

Therefore $\frac{u_1}{u_2}=\frac{\frac{M+m_1}{m_1}\times \sqrt{2gL(1-\cos 6.8)}}{\frac{M+m_2}{m_2}\times \sqrt{2gL(1-\cos 11.4)}}$

$\frac{u_1}{u_2}=\frac{1819.181\times 0.0838}{1001\times 0.1404}$

$\frac{u_1}{u_2}=1.084$

i.e. $\frac{v_1}{v_2}=1.084$

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3 years ago

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Answer:

123.30 m

Explanation:

Given

Speed, u = 22 m/s

acceleration, a = 1.40 m/s²

time, t = 7.30 s

From equation of motion,

v = u + at

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v is the final velocity

u is the initial velocity

a is the acceleration

t is time

V = at + U

using equation v - u = at to get line equation for the graph of the motion of the train on the incline plane

$V_{x} = mt + V_{o}$ where m is the slope

Comparing equation (1) and (2)

$V = V_{x}$

a = m

$U = V_{o}$

Since the train slows down with a constant acceleration of magnitude 1.40 m/s² when going up the incline plane. This implies the train is decelerating. Therefore, the train is experiencing negative acceleration.

a = - 1.40 m/s²

Sunstituting a = - 1.40 m/s² and u = 22 m/s

$V_{x} = -1.40t + 22$

$V_{x} = -1.40(7.30) + 22$

$V_{x} = -10.22 + 22$

$V_{x} = 11. 78 m/s$

The speed of the train at 7.30 s is 11.78 m/s.

The distance traveled after 7.30 sec on the incline is the area cover on the incline under the specific interval.

Area of triangle + Area of rectangle

$[\frac{1}{2} * (22 - 11.78) * (7.30)] + [(11.78 - 0) * (7.30)]$

= 37.303 + 85.994

= 123. 297 m

≈ 123. 30 m

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2 years ago

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