Why does a spherometer have three legs
1 answer:
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Answer:
(a) the high of a hill that car can coast up (engine disengaged) if work done by friction is negligible and its initial speed is 110 km/h is 47.6 m
(b) thermal energy was generated by friction is 1.88 x J
(C) the average force of friction if the hill has a slope 2.5º above the horizontal is 373 N
Explanation:
given information:
m = 750 kg
initial velocity, = 110 km/h = 110 x 1000/3600 = 30.6 m/s
initial height, = 22 m
slope, θ = 2.5°
(a) How high a hill can a car coast up (engine disengaged) if work done by friction is negligible and its initial speed is 110 km/h?
according to conservation-energy
EP = EK
mgh =
gh =
h =
= 47.6 m
(b) If, in actuality, a 750-kg car with an initial speed of 110 km/h is observed to coast up a hill to a height 22.0 m above its starting point, how much thermal energy was generated by friction?
thermal energy = mgΔh
= mg (h - )
= 750 x 9.8 x (47.6 - 22)
= 188160 Joule
= 1.88 x J
(c) What is the average force of friction if the hill has a slope 2.5º above the horizontal?
f d = mgΔh
f = mgΔh / d,
where h = d sin θ, d = h/sinθ
therefore
f = (mgΔh) / (h/sinθ)
= 1.88 x /(22/sin 2.5°)
= 373 N
Answer:
3.90 degrees
Explanation:
Let g= 9.81 m/s2. The gravity of the 30kg grocery cart is
W = mg = 30*9.81 = 294.3 N
This gravity is split into 2 components on the ramp, 1 parallel and the other perpendicular to the ramp.
We can calculate the parallel one since it's the one that affects the force required to push up
F = WsinΘ
Since customer would not complain if the force is no more than 20N
F = 20
So the ramp cannot be larger than 3.9 degrees
Answer:
a) -4 N
b) +4 N
Explanation:
Draw a free body diagram for each block.
For the large block, there are 2 forces: 12 N pushing to the right, and F pushing to the left.
For the small block, there is 1 force, F pushing to the right.
There are also weight and normal forces in the vertical direction, but we can ignore those.
Sum of forces on the large block in the x direction:
∑F = ma
12 − F = 4a
Sum of forces on the small block in the x direction:
∑F = ma
F = 2a
2F = 4a
Substitute:
12 − F = 2F
12 = 3F
F = 4
The small block pushes on the large block 4 N to the left (-4 N).
The large block pushes on the small block 4 N to the right (+4 N).
