-- As she lands on the air mattress, her momentum is (m v)
Momentum = (60 kg) (5 m/s down) = 300 kg-m/s down
-- As she leaves it after the bounce,
Momentum = (60 kg) (1 m/s up) = 60 kg-m/s up
-- The impulse (change in momentum) is
Change = (60 kg-m/s up) - (300 kg-m/s down)
Magnitude of the change = <em>360 km-m/s </em>
The direction of the change is <em>up /\ </em>.
<span>The weight of the spacecraft keeps changing.
</span>
<span>The mass of the spacecraft remains the same.
These are the correct answers</span>
Answer:
t = 2.2 s
Explanation:
Given that,
A person observes a firework display for A safe distance of 0.750 km.
d = 750 m
The speed of sound in air, v = 340 m/s
We need to find the between the person see and hear a firework explosion. let it is t. So, using the formula of speed.
So, the required time is 2.2 seconds.
Answer:
14 m/s
Explanation:
u = 0, h = 10 m, g = 9.8 m/s^2
Use third equation of motion
v^2 = u^2 + 2 g h
Here, v be the velocity of ball as it just strikes with the ground
v^2 = 0 + 2 x 9.8 x 10
v^2 = 196
v = 14 m/s
