To have a weight of 2.21N., the ball's mass is (2.21/9.8) = .226kg.
<span>a) d = 1/2 (vt), = 1/2 (18 x .17), = 1.53m. </span>
<span>b) Acceleration of the ball = (v/t), = 18/.17, = 105.88m/sec^2. </span>
<span>f = (ma), = .226 x 105.88, = 23.92N. </span>
(A) P(v) = 0.135v
(B) P(h) = 0.234v
<u>Explanation:</u>
Given-
Mass of the ball, m = 0.27kg
Force, F = 125N
angle of projection, θ = 30°
Let v be the velocity of the ball.
A) vertical component of the momentum of the volleyball
We know,
P(vertical) = mvsinθ
P(V) = 0.27 X v X sin 30°
P(V) = 0.27 X v X 0.5
P(V) = 0.135v
B) horizontal component of the momentum of the volleyball
We know,
P(Horizontal) = mvcosθ
P(h) = 0.27 X v X cos 30°
P(h) = 0.27 X v X 0.866
P(h) = 0.234v
Answer:A 2.2kg block of ice slides across a rough floor. Its initial velocity is 2.5m/s and its final velocity is 0.50m/s. How much of the ice block melted as a result of the work done by friction? (Latent Heat of water is 3.3*10^5J/kg)
Explanation:
