Answer:
Explanation:
I is the moment of inertia of the pulley, α is the angular acceleration of the pulley and T is the tension in the rope. Let a is the linear acceleration.
The relation between the linear acceleration and the angular acceleration is
a = R α .... (1)
According to the diagram,
T x R = I x α
T x R = I x a / R from equation (1)
T = I x a / R² .... (2)
mg - T = ma .... (3)
Substitute the value of T from equation (2) in equation (3)
T is the acceleration in the system
Substitute the value of a in equation (2)
This is the tension in the string.
Answer:
6N
Explanation:
Given parameters:
Pressure applied by the woman = 300N/m²
Area = 0.02m²
Unknown:
Force applied = ?
Solution:
Pressure is the force per unit area on a body
Pressure = 
Force = Pressure x area
Force = 300 x 0.02 = 6N
Answer:
a. 21.68 rad/s b. 30.78 m/s c. 897 rev/min² d. 1085 revolutions
Explanation:
a. Its angular speed in radians per second ω = angular speed in rev/min × 2π/60 = 207 rev/min × 2π/60 = 21.68 rad/s
b. The linear speed of a point on the flywheel is gotten from v = rω where r = radius of flywheel = 1.42 m
So, v = rω = 1.42 m × 21.68 rad/s = 30.78 m/s
c. Using α = (ω₁ - ω)/t where α = angular acceleration of flywheel, ω = initial angular speed of wheel in rev/min = 21.68 rad/s = 207 rev/min, ω₁ = final angular speed of wheel in rev/min = 1410 rev/min = 147.65 rad/s, t = time in minutes = 80.5/60 min = 1.342 min
α = (ω₁ - ω)/t
= (1410 - 207)/(80.5/60)
= 60(1410 - 207)/80.5
= 60(1203)80.5
= 896.65 rev/min² ≅ 897 rev/min²
d. Using θ = ωt + 1/2αt²
where θ = number of revolutions of flywheel. Substituting the values of the variables from above, ω = 207 rev/min, α = 896.65 rev/min² and t = 80.5/60 min = 1.342 min
θ = ωt + 1/2αt²
= 207 × 1.342 + 1/2 × 896.65 × 1.342²
= 277.725 + 807.417
= 1085.14 revolutions ≅ 1085 revolutions
Answer:
Writing with a pencil. The pencil pushes on the paper. The paper pushes on the pencil.
Explanation:
Newton's third law.
