Answer:
Explanation:
given,
mass of wheel(M) = 3 Kg
radius(r) = 35 cm
revolution (ω_i)= 800 rev/s
mass (m)= 1.1 Kg
I_{wheel} = Mr²
when mass attached at the edge
I' = Mr² + mr²
using conservation of angular momentum
The minimum value of the coefficient of static friction between the block and the slope is 0.53.
<h3>Minimum coefficient of static friction</h3>
Apply Newton's second law of motion;
F - μFs = 0
μFs = F
where;
- μ is coefficient of static friction
- Fs is frictional force
- F is applied force
μ = F/Fs
μ = F/(mgcosθ)
μ = (250)/(50 x 9.8 x cos15)
μ = 0.53
Thus, the minimum value of the coefficient of static friction between the block and the slope is 0.53.
Learn more about coefficient of friction here: brainly.com/question/20241845
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Hello,
<u>Solution for A:</u>
Force = 3.00N
Mass = 0.50 Kgs
Time = 1.50 Seconds
According to newton's second law of motion;
Force = Mass times Acceleration(a)
3.00 = 0.50 * a
a = 3.00/0.50 = 6.00 m/s^2
We know that acceleration = Velocity / time
So Velocity = time * acceleration = 1.50 * 6 = 9.00 m/s^2
<u>Solution for B:</u>
The net force = 4.00N - 3.00N = 1.00N to the left
Force = 1.00N
Mass = 0.50Kg
Time = 3.00 Seconds
Again; F = MA (Where F is force, M is mass and A is acceleration)
1.00N = 0.5 * A
A = 1/0.5 = 2 m/s^2
Velocity = Acceleration * Time = 2 * 3 = 6 m/s
