Answer:
The force that pulls the car down is Wₓ = 14052.6 N and the one that pushes the car up is F = 26552.6 N
Explanation:
For this exercise we will use Newton's second law, let's set a reference system where the x axis is parallel to the plane, in the adjoint we can see the forces in the system.
sin 35 = Wₓ / W
cos 35 = W_y / W
Wₓ = W sin 35
W_y = W cos 35
Wₓ = 2500 9.8 sin 35
Wₓ = 14052.6 N
let's write the equations for each axis
and
Y axis
N-W_y = 0
N = W_y
X axis
F -Wₓ = m a
F = Wₓ + m a = mg sin 35 + m a
F = m (a + g sin 35)
let's calculate
F = 2500 (5 + 9.8 sin 35)
F = 26552.6 N
The force that pulls the car down is Wₓ = 14052.6 N and the one that pushes the car up is F = 26552.6 N
The answer to this would be ocean floor
Answer:
Technician A
Explanation:
If Technician B was correct, and the master cylinder is defective - then no braking action would occur.
This is not true however, as some breaking action eventually occurs, meaning it must be out of adjustment.
Answer:
idk srry
Explanation:
i wish I could help you out
The shot putter should get out of the way before the ball returns to the launch position.
Assume that the launch height is the reference height of zero.
u = 11.0 m/s, upward launch velocity.
g = 9.8 m/s², acceleration due to gravity.
The time when the ball is at the reference position (of zero) is given by
ut - (1/2)gt² = 0
11t - 0.5*9.8t² = 0
t(11 - 4.9t) = 0
t = 0 or t = 4.9/11 = 0.45 s
t = 0 corresponds to when the ball is launched.
t = 0.45 corresponds to when the ball returns to the launch position.
Answer: 0.45 s
