The answer is c reflection hope this helps :)
Answer:
2.5 m/s east
Explanation:
Let east be the positive direction for velocity.
The change in momentum of the 0.75 kg model car is ...
m1·v2 -m1·v1 = (0.75 kg)(11 m/s) -(0.75 kg)(-9 m/s)
= (0.75 kg)(20 m/s) = 15 kg·m/s
The change in momentum of the 2.0 kg model car is the opposite of this, so the total change in momentum is zero.
m2·v2 -m2·v1 = (2 kg)(v2 m/s) -(2 kg)(10 m/s) = 2(v2 -10) kg·m/s
The required relation is ...
15 kg·m/s = -2(v2 -10) kg·m/s
-7.5 = v2 -10 . . . . divide by -2
2.5 = v2 . . . . . . . add 10
The velocity of the model truck after the collision is 2.5 m/s east.
Answer:
v = 42.92 m/s
Explanation:
Given,
initial speed of the ball, v = 11 m/s
time taken to hit the ground = 5.5 m/s
velocity of the ball just before it hit the ground, v = ?
time taken by the ball to reach the maximum height
using equation of motion
v = u + at
final velocity = 0 m/s
0 = 11 - 9.8 t
t = 1.12 s.
time taken by the ball to reach the water from the maximum height
t' - 5.5 -1.12 = 4.38 s
using equation of motion for the calculation of speed just before it hit the water.
v = u + a t
v = 0 + 9.8 x 4.38
v = 42.92 m/s
Velocity of the ball just before it reaches the water is equal to v = 42.92 m/s
The answer is A because it describes both toy cars
Glad you are coming back tomorrow I am so happy to be with you guys I will
