Abrick is dropped from the top of a building through the air (friction is present) to the ground below. part a how does the brick's kinetic energy (k) just before striking the ground compare with the gravitational potential energy ugrav at the top of the building? set y=0 at the ground level. how does the brick's kinetic energy () just before striking the ground compare with the gravitational potential energy at the top of the building? set at the ground level. k is equal to ugrav. k is greater than ugrav. k is less than ugrav.

kinetic energy is less than Potential Energy at bottom

Explanation:

Given

Brick is drop from top of tower

suppose brick Potential Energy at top is U=C

Kinetic Energy(K) of brick is =0 at top of building

Just before hitting its Potential Energy is converted to kinetic energy and some is lost due to the friction.

Therefore kinetic Energy is less than the potential energy

at bottom K < U

T = approximately 24 hs.

Explanation:

In order to keep the satellite over a fixed point on the equator, as the earth rotates, the satellite must have the same angular velocity that Earth has, which means that it must have a period equal to the time used by Earth to complete an entire rotation on itself, which is almost exactly 24 Hs.

Mathematically, this can be obtained taking into account that the force that keeps the satellite in orbit is the centripetal force, which is actually the gravitational force exerted by Earth, so we can write the following equality:

Fg = Fc ⇒ G*ms*me / (re +rsat)² = ms*ω²*(re +rsat)

By definition, ω =ΔФ / Δt

For a complete revolution, ΔФ = 2*π, and Δt = T (period of the rotation),

so we can replace ω by (2*π/T), solving then for T:

T= 86,313 sec. (24 hs are exactly 86,400 sec, so the value is actually very close to the theorical one).

Explanation: