Answer:
See explaination
Explanation:
Let's define tuple as an immutable list of Python objects which means it can not be changed in any way once it has been created.
Take a look at the attached file for a further detailed and step by step solution of the given problem.
Answer:
d) 9.55 psi
Explanation:
pressure at the bottom is =ρgh
weight density is ρg=55 lb/ft³
h=25ft
pressure at the bottom is =
=1375psf
1 ft = 12 inch
pressure at bottom =
= 9.55 psi
so, answer will be option (d) which is 9.55 psi
Answer:
The answer is below
Explanation:
1) The synchronous speed of an induction motor is the speed of the magnetic field of the stator. It is given by:
2) The speed of the rotor is the motor speed. The slip is given by:
3) The frequency of the rotor is given as:
4) At standstill, the speed of the motor is 0, therefore the slip is 1.
The frequency of the rotor is given as:
Answer:
Half-wave rectifier converts an AC signal into a DC signal. It's called a half-wave because it only rectify the positive part of an AC signal.
AC Signal = An electrical signal that alternates between positive and negative voltage.
DC Signal = An electrical signal that only has positive voltage.
Rectify = A fancy word for converting something.
Adding a capacitor helps the positive part of the signal stay on longer. This work because the capacitor stores energy kinda like a battery. During the negative part of the AC signal, the energy stored in the capacitor will be drained and used, then the cycle repeats.
The load resistor is just there to prevent a short circuit from happening.
Answer:
(a) 2.39 MPa (b) 3.03 kJ (c) 3.035 kJ
Explanation:
Solution
Recall that:
A 10 gr of air is compressed isentropically
The initial air is at = 27 °C, 110 kPa
After compression air is at = a450 °C
For air, R=287 J/kg.K
cv = 716.5 J/kg.K
y = 1.4
Now,
(a) W efind the pressure on [MPa]
Thus,
T₂/T₁ = (p₂/p₁)^r-1/r
=(450 + 273)/27 + 273) =
=(p₂/110) ^0.4/1.4
p₂ becomes 2390.3 kPa
So, p₂ = 2.39 MPa
(b) For the increase in total internal energy, is given below:
ΔU = mCv (T₂ - T₁)
=(10/100) (716.5) (450 -27)
ΔU =3030 J
ΔU =3.03 kJ
(c) The next step is to find the total work needed in kJ
ΔW = mR ( (T₂ - T₁) / k- 1
(10/100) (287) (450 -27)/1.4 -1
ΔW = 3035 J
Hence, the total work required is = 3.035 kJ