Answer:
208.33 W
141.26626 seconds
Explanation:
E = Energy =
t = Time taken = 8 h
m = Mass = 2000 kg
g = Acceleration due to gravity = 9.81 m/s²
h = Height of platform = 1.5 m
Power is obtained when we divide energy by time
The average useful power output of the person is 208.33 W
The energy in the next part would be the potential energy
The time taken would be
The time taken to lift the load is 141.26626 seconds
Answer
given,
L(t) = 10 - 3.5 t
mass of particle = 2 Kg
radius of the circle = 3.1 m
a) torque
τ =
τ =
τ = -3.5 N.m
Particle rotates clockwise as i look down the plane. Hence, its angular velocity is downward.
L decreases the angular acceleration upward. so, net torque is upward.
b) Moment of inertia of the particle
I = m R^2
I = 2 x 3.1²
I = 19.22 kg.m²
L = I ω
ω =
ω =
ω =
A = 0.52 rad/s B = -0.182 rad/s²
Answer:
(a) the final velocity of the cart is 0.857 m/s
(b) the impulse experienced by the package is 21.43 kg.m/s
(c) the fraction of the initial energy lost is 0.71
Explanation:
Given;
mass of the package, m₁ = 10 kg
mass of the cart, m₂ = 25 kg
initial velocity of the package, u₁ = 3 m/s
initial velocity of the cart, u₂ = 0
let the final velocity of the cart = v
(a) Apply the principle of conservation of linear momentum to determine common final velocity for ineleastic collision;
m₁u₁ + m₂u₂ = v(m₁ + m₂)
10 x 3 + 25 x 0 = v(10 + 25)
30 = 35v
v = 30 / 35
v = 0.857 m/s
(b) the impulse experienced by the package;
The impulse = change in momentum of the package
J = ΔP = m₁v - m₁u₁
J = m₁(v - u₁)
J = 10(0.857 - 3)
J = -21.43 kg.m/s
the magnitude of the impulse experienced by the package = 21.43 kg.m/s
(c)
the initial kinetic energy of the package is calculated as;
the final kinetic energy of the package;
the fraction of the initial energy lost;