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olga55 [171]
4 years ago
15

When the distance between two stars decreases by one-third, the force between them

Physics
1 answer:
pashok25 [27]4 years ago
4 0

Answer:

the force will increase by a factor 2.25

Explanation:

The gravitational force between the two stars is given by:

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

where

G is the gravitational constant

m1, m2 are the masses of the two stars

r is the distance between the stars

If the distance is decreased by one-third, it means that the new distance is 2/3 of the previous distance

r'=\frac{2}{3}r

So the new force will be

F'=G\frac{m_1 m_2}{(\frac{2}{3}r)^2}=\frac{9}{4} G\frac{m_1 m_2}{r^2}=2.25 F

So, the force will be 2.25 times the previous value.

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The driver of a car going 86.0 km/h suddenly sees the lights of a barrier 44.0 m ahead. It takes the driver 0.75 s to apply the
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Part a

Answer: NO

We need to calculate the distance traveled once the brakes are applied. Then we would compare the distance traveled and distance of the barrier.

Using the second equation of motion:

s=ut+0.5at^2

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Part b

Answer: 29.6 m/s

The maximum distance that car can travel is s=44.0 m

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The final velocity, v=0

Using the third equation of motion, we can find the maximum initial velocity for car to not hit the barrier:

v^2-u^2=2as

0-u^2=-2 \time 10.0m/s^2 \times 44.0 m\Rightarrow u=\sqrt{880 m^2/s^2}=29.6 m/s

Hence, the maximum speed at which car can travel and not hit the barrier is 29.6 m/s.





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