Answer:
2.11 m/s
Explanation:
Take north to be positive and south to be negative.
Average velocity = displacement / time
v = (82 m + -44 m) / (14 s + 4 s)
v = 2.11 m/s
The velocity is positive, so it is 2.11 m/s north. The magnitude of the velocity is 2.11 m/s.
Answer:
c. low beams and fog lights
Explanation:
When encountering low visibility from rain or fog, use your low beams and fog lights. High beams will only increase the glare. If you can't see at least five seconds in front of you, don't drive. Pull over and put your hazards on until it clears up.
i just got the answer wrong and the drivers ed gave me this explanation !!
Given:
v = 50.0 m/s, the launch velocity
θ = 36.9°, the launch angle above the horizontal
Assume g = 9.8 m/s² and ignore air resistance.
The vertical component of the launch velocity is
Vy = (50 m/s)*sin(50°) = 30.02 m/s
The time, t, to reach maximum height is given by
(30.02 m/s) - (9.8 m/s²)*(t s) = 0
t = 3.0634 s
The time fo flight is 2*t = 6.1268 s
The horizontal velocity is
u = (50 m/s)cos(36.9°) = 39.9842 m/s
The horizontal distance traveled at time t is given in the table below.
Answer:
t, s x, m
------ --------
0 0
1 39.98
2 79.79
3 112.68
4 159.58
5 199.47
6 239.37
Answer:
Explanation:
Parameters given:
Mass of Puck 1, m = 1 kg
Mass of Puck 2, M = 1 kg
Initial velocity of Puck 1, u = 20 m/s
Initial velocity of Puck 2, U = 0 m/s
Final velocity of Puck 1, v = 5 m/s
Since we are told that momentum is conserved, we apply the principle of conservation of momentum:
Total initial momentum of the system = Total final momentum of the system
mu + MU = mv + MV
(1 * 20) + (1 * 0) = (1 * 5) + (1 * V)
20 = 5 + V
V = 20 - 5 = 15 m/s
Puck 2 moves with a velocity of 15 m/s