Explanation:
Given that,
Mass of ball, m = 0.425 kg
Initial speed of the ball, u = 12 m/s
Initial speed of a person, u' = 0
Mass of a person, m' = 68 kg
(a) Let V is the combined speed of the person and the ball. Using conservation of momentum as :
(b) If the ball hits the person and bounces off his chest, so afterwards it is moving horizontally at 9.00 m/s in the opposite direction,. Let v' is the speed of the person after the collision. So,
v = -9 m/s
Hence, this is the required solution.
The magnitude of the kinetic friction force, ƒk, on an object is. Where μk is called the kinetic friction coefficient and |FN| is the magnitude of the normal force of the surface on the sliding object. The kinetic friction coefficient is entirely determined by the materials of the sliding surfaces. hope it helps
You skipped over a number in the question, and you didn't tell me what my average speed is. Lucky for you, my average speed has NO EFFECT on the answer to the question.
When you calculate velocity, you only use the straight-line distance between the start-point and the end-point. It doesn't matter what route the thing took to get there, or how much ground it actually covered.
If I travel in a circle and stop at the same point I started from, then the size of the circle doesn't matter, and neither does my speed. The distance between my start-point and my end-point is zero, and my average velocity is zero.
Respuesta:
24m
Explicación:
Según la ecuación de movimiento
v = u + en
Dado
Velocidad final v = 12 m / s
velocidad inicial u = 0 m / s
tiempo t = 4s
Sustituir
12 = 0 + 4a
a = 12/4
a = 3 m / s²
Lo siguiente es obtener la distancia;
S = ut + 1 / 2at²
S = 0 (4) + 1/2 (3) (4) ²
S = 3 * 16/2
S = 48/2
S = 24 m
Por lo tanto, la distancia requerida es de 24 m.
