Answer:
voltage = -0.01116V
power = -0.0249W
Explanation:
The voltage v(t) across an inductor is given by;
v(t) = L
-----------(i)
Where;
L = inductance of the inductor
i(t) = current through the inductor at a given time
t = time for the flow of current
From the question:
i(t) = 
A
L = 10mH = 10 x 10⁻³H
Substitute these values into equation (i) as follows;
v(t) = 
Solve the differential
v(t) = 
v(t) = -0.05 
At t = 8s
v(t) = v(8) = -0.05 
v(t) = v(8) = -0.05 
v(t) = -0.05 x 0.223
v(t) = -0.01116V
(b) To get the power, we use the following relation:
p(t) = i(t) x v(t)
Power at t = 8
p(8) = i(8) x v(8)
i(8) = i(t = 8) = 
i(8) = 
i(8) = 10 x 0.223
i(8) = 2.23
Therefore,
p(8) = 2.23 x -0.01116
p(8) = -0.0249W
To solve this problem it is necessary to apply the concepts related to temperature stagnation and adiabatic pressure in a system.
The stagnation temperature can be defined as
Where
T = Static temperature
V = Velocity of Fluid
Specific Heat
Re-arrange to find the static temperature we have that
Now the pressure of helium by using the Adiabatic pressure temperature is
Where,
= Stagnation pressure of the fluid
k = Specific heat ratio
Replacing we have that
Therefore the static temperature of air at given conditions is 72.88K and the static pressure is 0.399Mpa
<em>Note: I took the exactly temperature of 400 ° C the equivalent of 673.15K. The approach given in the 600K statement could be inaccurate.</em>
C and A I think cause I don’t really remember this I done before it his to be C and A
Answer:
the pressure reading when connected a pressure gauge is 543.44 kPa
Explanation:
Given data
tank volume (V) = 400 L i.e 0.4 m³
temperature (T) = 25°C i.e. 25°C + 273 = 298 K
air mass (m) = 3 kg
atmospheric pressure = 98 kPa
To find out
pressure reading
Solution
we have find out pressure reading by gauge pressure
i.e. gauge pressure = absolute pressure - atmospheric pressure
first we find absolute pressure (p) by the ideal gas condition
i.e pV = mRT
p = mRT / V
p = ( 3 × 0.287 × 298 ) / 0.4
p = 641.44 kPa
so
gauge pressure = absolute pressure - atmospheric pressure
gauge pressure = 641.44 - 98
gauge pressure = 543.44 kPa
