The value of the coefficient of kinetic friction between the wagon and inclined surface is 0.78.
<h3>
Coefficient of the kinetic friction</h3>
The value of coefficient of kinetic friction is calculated as follows;
F - Ff = ma
F - μmgcosθ = ma
where;
- F is applied force
- μ is coefficient of kinetic friction
- m is mass of the wagon
- a is acceleration of the wagon
182 - μ(20 x 9.8 x cos30) = 20(2.5)
182 - 169.74μ = 50
182 - 50 = 169.74μ
132 = 169.74μ
μ = 132/169.74
μ = 0.78
Thus, the value of the coefficient of kinetic friction between the wagon and inclined surface is 0.78.
Learn more about coefficient of friction here: brainly.com/question/20241845
Answer:
F = 39.2 N
Explanation:
Since, the object is in uniform motion. Therefore, the frictional force on object will be:
Frictional Force = μk N = μk mg
where,
μk = coefficient of kinetic friction = 0.2
m = mass of crate = 10 kg
g = 9.8 m/s²
Therefore,
Frictional Force = (0.2)(10 kg)(9.8 m/s²)
Frictional Force = 19.6 N
The horizontal component of force must be equal to this frictional force to continue the uniform motion:
F Sin 30° = 19.6 N
F = 19.6 N/Sin 30°
<u>F = 39.2 N</u>
Answer:
The average force ≅ 519.44 N.
Explanation:
Impulse = change in momentum of a body
i.e Ft = m(v - u)
where F is the force, t is the time, m is the mass of the body, v is the final velocity and u is the initial velocity.
m = 55.0 g (0.055 Kg), t = 0.00360 s, v = 34.0 m/s, since the ball was initially at rest; u = 0 m/s
So that,
F x 0.00360 = 0.055(34 - 0)
F x 0.00360 = 0.055 x 34
= 1.87
F = 
= 519.4444
The average force exerted on the ball by the club is approximately 519.44 N.
Answer:
Thats her fault.........................b
Explanation: