The question is not complete. We are supposed to find the average value of v_o.
Answer:
v_o,avg = 0.441V
Explanation:
Let t1 and t2 be the start and stop times of the output waveforms. Thus, from the diagram i attached, using similar triangles, we have;
3/(T/4) = 0.7/t1
So, 12/T = 0.7/t1
So, t1 = 0.7T/12
t1 = 0.0583 T
Also, from symmetry of triangles,
t2 = T/2 - t1
So, t2 = T/2 - 0.0583 T
t2 = 0.4417T
Average of voltage output is;
v_o,avg = (1/T) x Area under small triangle
v_o,avg = (1/T) x (3 - 0.7) x (T/4 - t1)
v_o,avg = (1/T) x (2.3) x (T/4 - 0.0583 T)
v_o,avg = (1/T) x 2.3 x 0.1917T
T will cancel out to give;
v_o,avg = 0.441V
Answer:240m
Explanation:
Given rpm increases from 1750 rpm to 3500 rpm
initial head 60 m and flow rate=
Since unit speed remains same
therefore
=
H=240m
Also unit Flow remains same
=
=
Answer:
T=833.8 °C
Explanation:
Given that
m= 2 kg
T₁=200 °C
time ,t= 10 min = 600 s
Work input = 1 KW
Work input = 1 x 600 KJ=600 KJ
Heat input = 0.5 KW
Q= 05 x 600 = 300 KJ
Gas is ideal gas.
We know that for ideal gas internal energy change given as
ΔU= m Cv ΔT
For air Cv= 0.71 KJ/kgK
From first law of thermodynamics
Q = ΔU +W
Heat input taken as positive and work in put taken as negative.
300 KJ = - 600 KJ + ΔU
ΔU = 900 KJ
ΔU= m Cv ΔT
900 KJ = 2 x 0.71 x (T- 200 )
T=833.8 °C
So the final temperature is T=833.8 °C
I believe my house is manufactured
Answer:
The stress in the rod is 39.11 psi.
Explanation:
The stress due to a pulling force is obtained dividing the pulling force by the the area of the cross section of the rod. The respective area for a cylinder is:
Replacing the diameter the area results:
Therefore the the stress results: