Answer:
a) γ =0.055556
b) t = 0.4 MPa
Explanation:
Given:
- The dimensions of rubber block : 18 x 21 x 25
- A load was applied at upper frame P = 420 N
- The rubber deflects dx = 1 mm downwards
Find:
(a) average shear strain in the rubber mounts
(b) average shear stress in the rubber mounts.
Solution:
- For average shear strain we have the definition:
γ = dx / y
Where,
γ: The shear strain.
dx : Deflection along the shear force
y : The length perpendicular to deflection.
- From given data we have dx = 1mm, and the dimension of block perpendicular to deflection is the a dimension. Hence, dx = 0.001 and y =0.018 m:
γ = 0.001 / 0.018 = 0.055556
- The average shear stress along the mating flat surface. We have from definition:
t = F_shear / Area
- Where, F_shear: The shear force on each rubber block is P/2.
Hence,
t = (P/2) / b*c
Plug values in:
t = (420/2) / (0.021*0.025)
t = 0.4 MPa
TIR is possible whenever light is travelling from a more optically dense medium to a less optically dense medium. You didnt provide any scenarios but just look up refractive indexes of the mediums the light is travelling through and if it is travelling from a more dense to a less dense, TIR is possible
Answer:
the gas does no work.
Explanation:
An isochoric process is a process in which the volume of a gas is kept constant.
The work done by a gas during a transformation is given by:
where
p is the gas pressure
is the change in volume of the gas
For an isochoric process, the volume of the gas does not change, so
and so, according to the previous equation, the work done by the gas is zero:
W = 0
Answer:
26.8 %
Explanation:
TH = 100°C = 100 + 273 = 373 K
Tc = 0°C = 273 K
The efficiency of the heat engine is given by
η = 0.268
η = 26.8 %
Thus, the efficiency of the heat engine is 26.8 %.
