Answer:
Absolute pressure=70.72 KPa
Explanation:
Given that Vacuum gauge pressure= 30 KPa
Barometer reading =755 mm Hg
We know that barometer always reads atmospheric pressure at given situation.So atmospheric pressure is equal to 755 mm Hg.
We know that P= ρ g h
Density of 
So P=13600 x 9.81 x 0.755
P=100.72 KPa
We know that
Absolute pressure=atmospheric pressure + gauge pressure
But here given that 30 KPa is a Vacuum pressure ,so we will take it as negative.
Absolute pressure=atmospheric pressure + gauge pressure
Absolute pressure=100.72 - 30 KPa
So
Absolute pressure=70.72 KPa
Answer:
a)
The crack and connecting rod is used in the design of car.This mechanism is known as slider -crank mechanism.
Components:
1.Inlet tube
2. Wheel
3. Exhaust
4. Engine
5.Air tank
6.Pressure gauge
7.Stand
8. Gate valve
b)
The efficiency of air engine is less as compare to efficiency of electric engine and this is not ecofriendly because it produce green house gases.These gases affect the environment.
c)
it can run around 722 km when it is full charge.
Answer:
I = 0.625 A
Explanation:
Given that,
Power of the light bulb, P = 75 W
Voltage of the circuit, V = 120 V
We need to find the current flowing through it. We know that, Power is given by :
I is the electric current
So, the current is 0.625 A.
Answer:
(a) E = 0 N/C
(b) E = 0 N/C
(c) E = 7.78 x10^5 N/C
Explanation:
We are given a hollow sphere with following parameters:
Q = total charge on its surface = 23.6 μC = 23.6 x 10^-6 C
R = radius of sphere = 26.1 cm = 0.261 m
Permittivity of free space = ε0 = 8.85419 X 10−12 C²/Nm²
The formula for the electric field intensity is:
E = (1/4πεo)(Q/r²)
where, r = the distance from center of sphere where the intensity is to be found.
(a)
At the center of the sphere r = 0. Also, there is no charge inside the sphere to produce an electric field. Thus the electric field at center is zero.
<u>E = 0 N/C</u>
(b)
Since, the distance R/2 from center lies inside the sphere. Therefore, the intensity at that point will be zero, due to absence of charge inside the sphere (q = 0 C).
<u>E = 0 N/C</u>
(c)
Since, the distance of 52.2 cm is outside the circle. So, now we use the formula to calculate the Electric Field:
E = (1/4πεo)[(23.6 x 10^-6 C)/(0.522m)²]
<u>E = 7.78 x10^5 N/C</u>
