As the Force of friction is equal to

µ

by using the Law of Action Reaction; as the force normal is the conterforce of gravity

µ


µ

Therefore, by looking at the equation we can infer that
The force of friction is directly proportional to the mass, gravitational constant, and the co-efficent of friction
Because the gravitational constant is dependant on gravitation, a planet's mass and radius also affect the force of friction
But DO NOTE:
That the co-effecient of friction is only applicable between two rubbing surfaces and unaffected by gravitational constants.
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The car with the engine that has the most power
Answer:
no mam
Explanation:
the answer to the question is
Answer:
(a): The normal force on the car from the track when the car's speed is v= 7.6 m/s is FN= -6696 N.
(b): The normal force on the car from the track when the car's speed is v= 17 m/s is FN= 8912.7 N.
Explanation:
m= 1080 kg
r= 16m
v1= 7.6 m/s
v2= 17 m/s
g= 9.81 m/s²
v1= w1*r
w1= v1/r
w1= 0.475 rad/s
ac1= w1² * r
ac1= 3.61 m/s²
FN= m * (ac1 - g)
FN= -6696 N (a)
-----------------------------------------------------
v2= w2*r
w2= v2/r
w2= 1.06 rad/s
ac2= w2² * r
ac2= 18.06 m/s²
FN= m * (ac2 - g)
FN= 8912.7 N (b)
Answer:
The speed of the block when it has returned to the bottom of the ramp is 6.56 m/s.
Explanation:
Given;
mass of block, m = 4 kg
coefficient of kinetic friction, μk = 0.25
angle of inclination, θ = 30°
initial speed of the block, u = 5 m/s
From Newton's second law of motion;
F = ma
a = F/m
Net horizontal force;
∑F = mgsinθ + μkmgcosθ

At the top of the ramp, energy is conserved;
Kinetic energy = potential energy
¹/₂mv² = mgh
¹/₂ v² = gh
¹/₂ x 5² = 9.8h
12.5 = 9.8h
h = 12.5/9.8
h = 1.28 m
Height of the ramp is 1.28 m
Now, calculate the speed of the block (in m/s) when it has returned to the bottom of the ramp;
v² = u² + 2ah
v² = 5² + 2 x 7.022 x 1.28
v² = 25 + 17.976
v² = 42.976
v = √42.976
v = 6.56 m/s
Therefore, the speed of the block when it has returned to the bottom of the ramp is 6.56 m/s.