<span>EP (potential energy) = mgy -> (59)(9.8)(-5) = -2,891
EP + EK (kinetic energy) = 0; but rearranging it for EK makes it EK = -EP, such that EK = 2891 when plugged in.
EK = 0.5mv^2, but can also be v = sqrt(2EK/m).
Plugging that in for sqrt((2 * 2891)/59), we get 9.9 m/s^2 with respect to significant figures.</span>
Answer:
i hope the answear is D becuase went over this long time ago when i was like you
Explanation:
Answer:
128.9 N
Explanation:
The force exerted on the golf ball is equal to the rate of change of momentum of the ball, so we can write:

where
F is the force
is the change in momentum
is the time interval
The change in momentum can be written as

where
m = 0.04593 kg is the mass of the ball
u = 0 is the initial velocity of the ball
is the final velocity of the ball
Substituting into the original equation, we find the force exerted on the golf ball:

Answer:
the average speed of the car is 170 mph.
Explanation:
Given;
initial speed, u = 70 mph
time of motion, t₁ = 3 hours
final speed, v = 20 mph
time of motion, t₂ = 2 hours
The average speed of the car is calculated as;

Therefore, the average speed of the car is 170 mph.
Answer:
ΔK = -6 10⁴ J
Explanation:
This is a crash problem, let's start by defining a system formed by the two trucks, so that the forces during the crash have been internal and the moment is preserved
initial instant. Before the crash
p₀ = m v₁ + M 0
final instant. Right after the crash
p_f = (m + M) v
p₀ = p_f
mv₁ = (m + M) v
v =
we substitute
v =
3
v = 1.0 m / s
having the initial and final velocities, let's find the kinetic energy
K₀ = ½ m v₁² + 0
K₀ = ½ 20 10³ 3²
K₀ = 9 10⁴ J
K_f = ½ (m + M) v²
K_f = ½ (20 +40) 10³ 1²
K_f = 3 10⁴ J
the change in energy is
ΔK = K_f - K₀
ΔK = (3 - 9) 10⁴
ΔK = -6 10⁴ J
The negative sign indicates that the energy is ranked in another type of energy