The correct answer to the question is : 9375 N.
CALCULATION:
As per the question, the mass of the car m = 1500 Kg.
The diametre of the circular track D = 200 m.
Hence, the radius of the circular path R = 
= 
= 100 m.
The velocity of the truck v = 25 m/s.
When a body moves in a circular path, the body needs a centripetal force which helps the body stick to the orbit. It acts along the radius and towards the centre.
Hence, the force acting on the car is centripetal force.
The magnitude of the centripetal force is calculated as -
Force F = 
= 
= 9375 N. [ANS}
The centripetal force is provided to the car in two ways. It is the friction which provides the necessary centripetal force. Sometimes friction is not sufficient. At that time, the road is banked to some extent which provides the necessary centripetal force.
Answer:
1.F = 256 N
2.a = 8 m/s/s
By looking at the given information, you know force, and an acceleration. Therefore you have enough information to use the first formula.
F = ma
256 N = m * 8 m/s/s
m = 256 N/8 m/s/s
m = 32 Kg
Answer:
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Explanation:
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Answer:
a) F = 680 N, b) W = 215 .4 J
, c) F = 1278.4 N
Explanation:
a) Hooke's law is
F = k x
To find the displacement (x) let's use the elastic energy equation
= ½ k x²
k = 2
/ x²
k = 2 85.0 / 0.250²
k = 2720 N / m
We replace and look for elastic force
F = 2720 0.250
F = 680 N
b) The definition of work is
W = ΔEm
W =
- 
W = ½ k (
² - x₀²)
The final distance
= 0.250 +0.220
= 0.4750 m
We calculate the work
W = ½ 2720 (0.47² - 0.25²)
W = 215 .4 J
We calculate the strength
F = k 
F = 2720 0.470
F = 1278.4 N