(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
Since momentum is a vector quantity, take any direction as positive and other as negative. Answer won't change.
Answer:
the electric field strength of this charge is two times the strength of the other charge
Explanation:
Using the relationship between electric field and the charge, which is inversely proportionality. Let the the magnitude of the first charge be Q and the respective electric field be E. It implies that;
E1/E2 = Q2/Q1
E2 = E1 x Q1/Q2
= E x Q/ (Q/2)
= 2E