Determine the speed of sound on a rainy day with the temperature of 18 degrees celsius..
1 answer:
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Answer:
(a) 2.542 cm
(b) 272.7°C
Explanation:
diameter, d = 2.540 cm
T1 = 20°C
α = 11 x 10^-6 /°C
(a) Let d' be the diameter.
T2 = 87°C
Use he formula for the areal expansion
A' = A ( 1 + βΔT)
where, β is the coefficient of areal expansion and ΔT is teh rise in temperature, A' be the area at high temperature and A be the area at low temperature.
β = 2 α = 2 x 11 x 106-6 = 22 x 10^-6 /°C
So,
D'^2 = 2.54^2 ( 1 + 22 x 10^-6 x 67)
D' = 2.542 cm
(b) Let the change in temperature is ΔT.
Use the formula for the volumetric expansion
ΔV = V x γ x ΔT
Where, γ = 3 x α = 3 x 11 x 10^-6 = 33 x 10^-6 /°C
0.9/100 = 33 x 10^-6 x ΔT
ΔT = 272.7°C
Answer:
θ '= 4975 rev
Explanation:
For this exercise let's use the relationship between work and the change in kinetic energy
W = ΔK
the expression for work is
W = - τ Δθ
where the negative sign indicates that the torque is in the opposite direction to rotation
kinetic energy
K = ½ I w²
we substitute
- τ Δθ = 0 - ½ I w²
θ =
if we approach the rotor to a cylinder with an axis of rotation through its center
I = ½ m r²
we substitute
θ = ½ (½ m r²)
How the measurements are in the English system, the weight
W = m g
m = W / g
let's reduce to the english system
w = 3600 rev / min (2pi rad / 1rev) (1 min / 60s) = 376.99 rad / s
r = 9 in (1 ft / 12in) = 0.75 ft
let's calculate
m = 125/32 = 3.91 slug
θ = ¼ 3.91 0.75² 376.99² / 2.5
θ = 3.126 10⁴ rad
let's reduce to revolutions
θ’= 3.126 10⁴ rad (1rev / 2π rad)
θ’= 4974.9 rev
θ '= 4975 rev