To find the change in centripetal acceleration, you should first look for the centripetal acceleration at the top of the hill and at the bottom of the hill.
The formula for centripetal acceleration is:
Centripetal Acceleration = v squared divided by r
where:
v = velocity, m/s
r= radium, m
assuming the velocity does not change:
at the top of the hill:
centripetal acceleration = (4.5 m/s^2) divided by 0.25 m
= 81 m/s^2
at the bottom of the hill:
centripetal acceleration = (4.5 m/s^2) divided by 1.25 m
= 16.2 m/s^2
to find the change in centripetal acceleration, take the difference of the two.
change in centripetal acceleration = centripetal acceleration at the top of the hill - centripetal acceleration at the bottom of the hill
= 81 m/s^2 - 16.2 m/s^2
= 64.8 m/s^2 or 65 m/s^2
Answer:
2.5 x 10^{5} J
Explanation:
weight = 5,000 N
coefficient of friction = 0.05
distance = 1000 m
how much work is done by the dogs pulling the sledge
work done = force x coefficient of friction x distance
work done = 5000 x 0.05 x 1000 = 2.5 x 10^{5} J
Answer:
The force of friction acting on block B is approximately 26.7N. Note: this result does not match any value from your multiple choice list. Please see comment at the end of this answer.
Explanation:
The acting force F=75N pushes block A into acceleration to the left. Through a kinetic friction force, block B also accelerates to the left, however, the maximum of the friction force (which is unknown) makes block B accelerate by 0.5 m/s^2 slower than the block A, hence appearing it to accelerate with 0.5 m/s^2 to the right relative to the block A.
To solve this problem, start with setting up the net force equations for both block A and B:

where forces acting to the left are positive and those acting to the right are negative. The friction force F_fr in the first equation is due to A acting on B and in the second equation due to B acting on A. They are opposite in direction but have the same magnitude (Newton's third law). We also know that B accelerates 0.5 slower than A:

Now we can solve the system of 3 equations for a_A, a_B and finally for F_fr:

The force of friction acting on block B is approximately 26.7N.
This answer has been verified by multiple people and is correct for the provided values in your question. I recommend double-checking the text of your question for any typos and letting us know in the comments section.
Answer:
(a) 2.85 m
(b) 16.5 m
(c) 21.7 m
(d) 22.7 m
Explanation:
Given:
v₀ₓ = 19 cos 71° m/s
v₀ᵧ = 19 sin 71° m/s
aₓ = 0 m/s²
aᵧ = -9.8 m/s²
(a) Find Δy when t = 3.5 s.
Δy = v₀ᵧ t + ½ aᵧ t²
Δy = (19 sin 71° m/s) (3.5 s) + ½ (-9.8 m/s²) (3.5 s)²
Δy = 2.85 m
(b) Find Δy when vᵧ = 0 m/s.
vᵧ² = v₀ᵧ² + 2 aᵧ Δy
(0 m/s)² = (19 sin 71° m/s)² + 2 (-9.8 m/s²) Δy
Δy = 16.5 m
(c) Find Δx when t = 3.5 s.
Δx = v₀ₓ t + ½ aₓ t²
Δx = (19 cos 71° m/s) (3.5 s) + ½ (0 m/s²) (3.5 s)²
Δx = 21.7 m
(d) Find Δx when Δy = 0 m.
First, find t when Δy = 0 m.
Δy = v₀ᵧ t + ½ aᵧ t²
(0 m) = (19 sin 71° m/s) t + ½ (-9.8 m/s²) t²
0 = t (18.0 − 4.9 t)
t = 3.67
Next, find Δx when t = 3.67 s.
Δx = v₀ₓ t + ½ aₓ t²
Δx = (19 cos 71° m/s) (3.67 s) + ½ (0 m/s²) (3.67 s)²
Δx = 22.7 m
Answer:
None, both objects will hit ground at the same time.
Explanation:
- Assuming no air resistance present, and that both objects start from rest, we can apply the following kinematic equation for the vertical displacement:

- As the left side in (1) is the same for both objects, the right side will be the same also.
- Since g is constant close to the surface of the Earth, it's also the same for both objects.
- So, the time t must be the same for both objects also.