1.)Two objects, one of m=20,000 kg, and another of 12,500 kg, are placed at a distance of 5 meters apart. What is the force of g

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3 years ago

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1.)Two objects, one of m=20,000 kg, and another of 12,500 kg, are placed at a distance of 5 meters apart. What is the force of g

ravitation between these two objects?
2.) If the two objects in the previous question are moved so that the distance between them is cut exactly in half to 2.5 m, what is the new value of the force between them?

3.) If a perpendicular force is applied to an object at 0.5 m from the radius, causing it to rotate, what is the torque if the force is 25 N?

4.)A hoop that has a moment of inertia (I) of 0.007875 kg*m^2 and an angular velocity of 6.28 rad/s has what value for angular momentum (L)?

Physics

1 answer:

Delvig [45] · Answer 1 · 2022-07-31T04:49:23+03:00

1) $6.67\cdot 10^{-4} N$

The force of gravitation between the two objects is given by:

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

where

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

m1 = 20,000 kg is the mass of the first object

m2 = 12,500 kg is the mass of the second object

r = 5 m is the distance between the two objects

Substituting the numbers inside the equation, we find

$F=(6.67\cdot 10^{-11})\frac{(20,000 kg)(12,500 kg)}{(5 m)^2}=6.67\cdot 10^{-4} N$

2) $2.7\cdot 10^{-3} N$

From the formula in exercise 1), we see that the force is inversely proportional to the square of the distance:

$F \sim \frac{1}{r^2}$

this means that if we cut in a half the distance without changing the masses, the magnitude of the forces changes by a factor

$F'\sim \frac{1}{(r/2)^2}=4 \frac{1}{r^2}=4F$

So, the gravitational force increases by a factor 4. Therefore, the new force will be

$F' = 4 F=4(6.67\cdot 10^{-4} N)=2.7\cdot 10^{-3} N$

3) 12.5 Nm

The torque is equal to the product between the magnitude of the perpendicular force and the distance between the point of application of the force and the centre of rotation:

$\tau=Fd$