Explanation:
thermal expansion ∝L = (δL/δT)÷L ----(1)
δL = L∝L + δT ----(2)
we have δL = 12.5x10⁻⁶
length l = 200mm
δT = 115°c - 15°c = 100°c
putting these values into equation 1, we have
δL = 200*12.5X10⁻⁶x100
= 0.25 MM
L₂ = L + δ L
= 200 + 0.25
L₂ = 200.25mm
12.5X10⁻⁶ *115-15 * 20
= 0.025
20 +0.025
D₂ = 20.025
as this rod undergoes free expansion at 115°c, the stress on this rod would be = 0
Answer:
μ=0.329, 2.671 turns.
Explanation:
(a) ln(T2/T1)=μβ β=angle of contact in radians
take T2 as greater tension value and T1 smaller, otherwise the friction would be opposite.
T2=5000 lb and T1=80 lb
we have two full turns which makes total angle of contact=4π radians
μ=ln(T2/T1)/β=(ln(5000/80))/4π
μ=0.329
(b) using the same relation as above we will now compute the angle of contact.
take greater tension as T2 and smaller as T1.
T2=20000 lb T1=80 lb μ=0.329
β=ln(20000/80)/0.329=16.7825 radians
divide the angle of contact by 2π to obtain number of turns.
16.7825/2π =2.671 turns
Harmonic excitation refers to a sinusoidal external force of a certain frequency applied to a system. ... Resonance occurs when the external excitation has the same frequency as the natural frequency of the system. It leads to large displacements and can cause a system to exceed its elastic range and fail structurally.
Answer:
(A) and (D)
Explanation:
1) P2 is less than P1, that is when P1 increases in pressure, the velocity V1 of the water also increases. Therefore, on the other hand, since P2 is directly proportional to V1, P2 and V2 will be less than P1 and V1 respectively.
2) For P2 greater than P1 and V2 also is greater than V1. Since P2 is directly proportional to V2, hence, when P2 increases in pressure, P1 reduces in pressure. Similarly, velocity, V2 also increases and V1 reduces.
