Anything that has volume or mass

iren2701 [21]

expressing adverse or disapproving comments or judgments.

5 0

2 years ago

What is the approximate value of the gravitational force between a 73 kg astronaut and a 7.1×104 kg spacecraft when they're 89 m

Luden [163]

Answer:

$F = 4.3671 * 10^{-8}\ Newtons$

Explanation:

The gravitational force between two corpses is given by the following equation:

$F = GMm/d^2$

Where F is the force, G is the gravitational constant

( $G = 6.67408*10^{-11}\ m^3kg^{-1}s^{-2}$ ), M and m are the masses of the corpses and d is the distance between them.

So we have that:

$F = 6.67408*10^{-11} * 7.1*10^4 * 73/89^2$

$F = 4.3671 * 10^{-8}\ Newtons$

5 0

2 years ago

A food packet is dropped from a helicopter during a flood-relief operation from a height of 750 meters. Assuming no drag (air fr

Leviafan [203]

1. Velocity at which the packet reaches the ground: 121.2 m/s

The motion of the packet is a uniformly accelerated motion, with constant acceleration $a=g=9.8 m/s^2$ directed downward, initial vertical position $d=750 m$ , and initial vertical velocity $v_0 = 0$ . We can use the following SUVAT equation to find the final velocity of the packet after travelling for d=750 m:

$v_f^2 -v_i^2 =2ad$

substituting, we find

$v_f^2 = 2ad\\v_f = \sqrt{2ad}=\sqrt{2(9.8 m/s^2)(750 m)}=121.2 m/s$

2. height at which the packet has half this velocity: 562.6 m

We need to find the heigth at which the packet has a velocity of

$v_f=\frac{121.2 m/s}{2}=60.6 m/s$

In order to do that, we use again the same SUVAT equation substituting $v_f$ with this value, so that we find the new distance d that the packet travelled from the helicopter to reach this velocity: