T = 4.25 ms = 4 x 10⁻³ s, the time for rebound
v₁ = 25.5 m/s, the impacting velocty
v₂ = -19.5 m/s, the rebounding velocity (n the opposite directon)
The change in velocity is
v₂ - v₁ = - (25.5+19.5) = -45 m/s
The acceleration is
a = (-45 m/s)/(4 s) = -11.25 m/s²
The negative sign indicates that the final velocity is opposiye to the impact velocty.
Answer: The magnitude of the acceleration is 11.25 m/s²
Let north as positive
Fnet=10n-5n
=5n north
The centripetal force acting on the space shuttle as it orbits Earth is equal to the shuttles momentum
Answer:
a. by moving the book without acceleration and keeping the height of the book constant
Explanation:
FOR CONSTANT KINETIC ENERGY:
The kinetic energy of a body depends upon its speed according to its formula:
ΔK.E = (1/2)mΔv²
So, for Δv = 0 m/s
ΔK.E = 0 J
So, for keeping kinetic energy constant, the books must be moved at constant speed without acceleration.
FOR CONSTANT POTENTIAL ENERGY:
The potential energy of a body depends upon its height according to its formula:
ΔP.E = mgΔh
So, for Δh = 0 m/s
ΔP.E = 0 J
So, for keeping potential energy constant, the books must be moved at constant height.
So, the correct option is:
<u>a. by moving the book without acceleration and keeping the height of the book constant</u>
