Answer:
You will feel more weight if it is accelerating out of the planet.
You will feel less weight if it is accelerating towards the planet.
Explanation:
The weight that you are observing or feeling is basically due to the change in acceleration of your fall or rising up in the spaceship. When the acceleration is stationary on the surface, you experience your normal weight due to the gravitational acceleration of that planet.
When the spaceship accelerates above or out of the planet you experience acceleration more than the acceleration of gravity hence more weight.
When the spaceship accelerates towards the planet you experience acceleration less than the acceleration of gravity hence less weight.
If the spaceship is free falling at the gravitational acceleration you experience a zero weight
Answer:
(a). The amplitude of the motion is 0.926 m.
(b). The block’s maximum acceleration is 182.31 m/s².
(c). The maximum force the spring exerts on the block is 291.69 N.
Explanation:
Given that,
Mass of block =1.60 kg
Force constant = 315 N/m
Speed = 13.0 m/s
(a). We need to calculate the amplitude of the motion
Using conservation of energy
Put the value into the relation
(b). We need to calculate the block’s maximum acceleration
Using formula of acceleration
Put the value into the formula
(c). We need to calculate the maximum force the spring exerts on the block
Using formula of force
Put the value into the formula
Hence, (a). The amplitude of the motion is 0.926 m.
(b). The block’s maximum acceleration is 182.31 m/s².
(c). The maximum force the spring exerts on the block is 291.69 N.
3.
a)
r = distance of each mass in each hand from center = 0.6 m
m = mass of each mass in each hand = 2 kg
v = linear speed = 1.1 m/s
L = combined angular momentum of the masses = ?
Combined angular momentum of the masses is given as
L = 2 m v r
L = 2 (2) (1.1) (0.6)
L = 2.64 kg m²/s
b)
v' = linear speed when she pulls her arms = ?
r' = distance of each mass from center after she pulls her arms = 0.15 m
Using conservation of momentum , angular momentum remains same, hence
L = 2 m v' r'
2.64 = 2 (2) (0.15) v'
v' = 4.4 m/s
