Answer:
-50.005 KJ
Explanation:
Mass flow rate = 0.147 KJ per kg
mass= 10 kg
Δh= 50 m
Δv= 15 m/s
W= 10×0.147= 1.47 KJ
Δu= -5 kJ/kg
ΔKE + ΔPE+ ΔU= Q-W
0.5×m×(30^2- 15^2)+ mgΔh+mΔu= Q-W
Q= W+ 0.5×m×(30^2- 15^2) +mgΔh+mΔu
= 1.47 +0.5×1/100×(30^2- 15^2)-9.7×50/1000-50
= 1.47 +3.375-4.8450-50
Q=-50.005 KJ
Answer:
Technician A
Explanation:
Galvanic corrosion is not on only one metal alone but caused when two metals are interacting. Thus, Duplicating the original installation method is a better option because re-using a coated bolt doesn't prevent galvanic corrosion because both materials must be coated and not just the bolt and in technician B's case he is coating just the bolt. Thus, technician B's method will not achieve prevention of galvanic corrosion but technician A's method will achieve it.
Answer:
Explanation:
I'm not 100% this is what you want, but here it is:
2
3
13
8
11
A
13
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>
Answer:
401.3 kg/s
Explanation:
The power plant has an efficiency of 36%. This means 64% of the heat form the source (q1) will become waste heat. Of the waste heat, 85% will be taken away by water (qw).
qw = 0.85 * q2
q2 = 0.64 * q1
p = 0.36 * q1
q1 = p /0.36
q2 = 0.64/0.36 * p
qw = 0.85 *0.64/0.36 * p
qw = 0.85 *0.64/0.36 * 600 = 907 MW
In evaporation water becomes vapor absorbing heat without going to the boiling point (similar to how sweating takes heat from the human body)
The latent heat for the vaporization of water is:
SLH = 2.26 MJ/kg
So, to dissipate 907 MW
G = qw * SLH = 907 / 2.26 = 401.3 kg/s
