Let’s use the *queue dramatic voice* LAW OF THE CONSERVATION OF MOMENTUM!
m1v1i + m2v2i = m1vf + m2vf
m1v1i - m1vf = m2vf - m2v2i
m1(v1i - vf) = m2(vf - v2i)
m2 = [m1(v1i - Vf)] / (vf - v2i)
m2 = [(1)(5 - -1)] / (-1 - -2)
m2 = 6 / 1
m2 = 6 kg
And that’s your final answer! Please press “Thanks”!
The correct answer is B. The atomic weight of the element
Since the spacecraft is two earth radii the surface of the earth, it is three earth radii above the center.
Given: Radius of the earth re = 6.38 x 10⁶ m r = 1.91 x 10⁷ m
Mass of the spacecraft Ms = 1600 Kg
Mass of the earth Me = 5.98 x 10²⁴ Kg
G = 6.67 X 10⁻¹¹ N.m²/Kg²
Formula: F = GMeMs/r²
F = (6.67 X 10⁻¹¹ N.m²/Kg²)(5.98 x 10²⁴ Kg)(1,600 Kg)/(1.91 x 10⁷ m)²
F = 6.38 X 10¹⁷ N/3.65 X 10¹⁴ m²
F = 1,748.5 N
The engine efficiency is 64.73 %
<u>Explanation:</u>
Given data
To find the engine’s efficiency we have the formula,
Energy input- 4565 KJ
Energy output - 2955KJ
Efficiency= energy output/ energy input ×100%
=2955/4565
=0.6473 ×100
η =64.73 %
The engine efficiency is 64.73 %
Answer: set up proportions
Explanation:
