Answer:
u = - 38.85 m/s^-1
Explanation:
given data:
acceleration = 2.10*10^4 m/s^2
time = 1.85*10^{-3} s
final velocity = 0 m/s
from equation of motion we have following relation
v = u +at
0 = u + 2.10*10^4 *1.85*10^{-3}
0 = u + (21 *1.85)
0 = u + 38.85
u = - 38.85 m/s^-1
negative sign indicate that the ball bounce in opposite directon
Answer:
the two gliders collide, the mobile glider will transfer a bit of time to the fixed glider, which is why it comes out with a speed that is smaller than that of the bullet glider.
Explanation:
When the two gliders collide, the mobile glider will transfer a bit of time to the fixed glider, which is why it comes out with a speed that is smaller than that of the bullet glider.
Changes can occur that the gliders unite and move with a cosecant speed less than the initial one.
The whole process must be analyzed using conservation of the moment.
p₀ = m v₀
celestines que clash case
p_f = (m + M) v
po = pf
m v₀ = (n + M) v
v =
calculemos
v=
v= 0.09 m/s
elastic shock case
p₀ = m v₀
p_f = m v₁ +M v₂
p₀ = p_f
m v₀ = m v₁ + m v₂
Answer:
15N
Explanation:
According to Newton's Second Law of Motion
F = m*a
mass = m = 5Kg
acceleration = a = 3m/s^2
=> F = 5kg * 3m/s^2
=> F = 15 N
I attached a free body diagram for a better understanding of this problem.
We start making summation of Moments in A,
Then we make a summation of Forces in Y,
At the end we calculate the angle with the sin.