Explanation:
Step1
Absolute pressure is the pressure above zero level of the pressure. Absolute pressure is considering atmospheric pressure in it. Absolute pressure is always positive. There is no negative absolute pressure.
The expression for absolute pressure is given as follows:

Here,
is absolute pressure,
is gauge pressure and
is atmospheric pressure.
Step2
Gauge pressure is the pressure that measure above atmospheric pressure. It is not considering atmospheric pressure. It can be negative called vacuum or negative gauge pressure. Gauge pressure used to simplify the pressure equation for fluid analysis.
Answer:
The energy, that is dissipated in the resistor during this time interval is 153.6 mJ
Explanation:
Given;
number of turns, N = 179
radius of the circular coil, r = 3.95 cm = 0.0395 m
resistance, R = 10.1 Ω
time, t = 0.163 s
magnetic field strength, B = 0.573 T
Induced emf is given as;

where;
ΔФ is change in magnetic flux
ΔФ = BA = B x πr²
ΔФ = 0.573 x π(0.0395)² = 0.002809 T.m²

According to ohm's law;
V = IR
I = V / R
I = 3.0848 / 10.1
I = 0.3054 A
Energy = I²Rt
Energy = (0.3054)² x 10.1 x 0.163
Energy = 0.1536 J
Energy = 153.6 mJ
Therefore, the energy, that is dissipated in the resistor during this time interval is 153.6 mJ
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:
The principal stresses are σp1 = 27 ksi, σp2 = -37 ksi and the shear stress is zero
Explanation:
The expression for the maximum shear stress is given:

Where
σx = stress in vertical plane = 20 ksi
σy = stress in horizontal plane = -30 ksi
τM = 32 ksi
Replacing:

Solving for τxy:
τxy = ±19.98 ksi
The principal stress is:

Where
σp1 = 20 ksi
σp2 = -30 ksi
(equation 1)
equation 2
Solving both equations:
σp1 = 27 ksi
σp2 = -37 ksi
The shear stress on the vertical plane is zero
Answer:
Coat new O-rings (D) with silicone oil or polyalkyleneglycol (PAG) oil, and pull them on the injectors.