A rocket ship is accelerated by the SRB and the main engines for 2.0 minutes and the main engines for 8.5 minutes after the launch. The acceleration of the ship during the first 2.0 minutes is 11 m/s² (D).
A rocket ship has several engines and thrusters. We can divide its initial movement into 2 parts:
- From t = 0 min to t = 2.0 min, the SRB and the main engines act together and the speed goes from 0 m/s (rest) to 1341 m/s.
- From t = 2.0 min to t = 8.5 min, the main engines alone accelerate the ship form 1341 m/s to 7600 m/s.
We want to know the acceleration in the first part (first 2.0 minutes). We need to consider that:
- The speed increases from 0 m/s to 1341 m/s.
- The time elpased is 2.0 min.
- 1 min = 60 s.
The acceleration of the ship during the first 2.0 minutes is:

A rocket ship is accelerated by the SRB and the main engines for 2.0 minutes and the main engines for 8.5 minutes after the launch. The acceleration of the ship during the first 2.0 minutes is 11 m/s² (D).
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Answer:
v₂ = 5.7 m/s
Explanation:
We will apply the law of conservation of momentum here:

where,
Total Initial Momentum = 340 kg.m/s
m₁ = mass of bike
v₁ = final speed of bike = 0 m/s
m₂ = mass of Sheila = 60 kg
v₂ = final speed of Sheila = ?
Therefore,

<u>v₂ = 5.7 m/s </u>
Answer:
C?
Explanation:
My best guess would be C as it's the only answer that gives a reason behind the statement.
Answer:
The velocity of the man is 0.144 m/s
Explanation:
This is a case of conservation of momentum.
The momentum of the moving ball before it was caught must equal the momentum of the man and the ball after he catches the ball.
Mass of ball = 0.65 kg
Mass of the man = 54 kg
Velocity of the ball = 12.1 m/s
Before collision, momentum of the ball = mass x velocity
= 0.65 x 12.1 = 7.865 kg-m/s
After collision the momentum of the man and ball system is
(0.65 + 54)Vf = 54.65Vf
Where Vf is their final common velocity.
Equating the initial and final momentum,
7.865 = 54.65Vf
Vf = 7.865/54.65 = 0.144 m/s