A swing that is 9.2m long has a child in it who is swinging. what is its period in seconds
1 answer:
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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:
The answers to the question are
(i) The rate of heat loss per-unit-length (W/m) from the pipe is 131.62 W
(ii) The temperature of the outer surface of the insulation is 49.89 °C
Explanation:
To solve the question, we note that the heat transferred is given by
Where
= Temperature at the inside of the pipe = 300 °C
= Temperature at the outside of the pipe = 20 °C
r₁ =internal radius of pipe = 4.0 cm
r₂ = Outer radius of pipe = 4.5 cm
r₃ = Outer radius of the insulation = r₂ + 2.5 = 7.0 cm
= 15 W/m·K
= 0.038 W/m·K
= 75 W/m²·K
= 10 W/m²·K
Plugging in the values in the above equation where for a unit length L = 1 m, we have
Q = 131.32 W
From which we have, for the film of air at the pipe outer boundary layer
Where
for the air film on the pipe outer surface is given by
where A =area of the outside of the pipe
= = 0.227 K/W
Therefore
131.32 W = which gives
= 49.89 °C
Heat transferred by radiation = q' = ε×σ×(T₁⁴ - T₂⁴)
Where ε = 0.9, σ, = 5.67×10⁻⁸W/m²·(K⁴)
T₁ = Surface temperature of the pipe = 49.89 °C and
T₂ = Temperature of the surrounding = 20.00 °C
Plugging in the values gives, q' = 0.307 W per m²
Total heat lost per unit length = 131.32 + 0.307 =131.62 W