Answer:
Wnet, in, = 133.33J
Explanation:
Provided that
Pump heat QH = 1000J
Hot temperature TH= 300K
Cold temperature TL= 260K
Given the heat pump is entirely reversible, the performance coefficient expression is formulated as follows:
According to the first law of thermodynamics,
COP(HP, rev) = 1/(1-TL/TH)
COP(HP, rev) = 1/(1-260/300)
COP(HP, rev) = 1/(1-0.867)
COP(HP, rev) = 1/0.133
COP(HP, rev) = 7.5
The power necessary to operate the heat pump is given by
Wnet, in = QH/COP(HP, rev)
Wnet, in = 1000/7.5
Wnet, in = 133.333J. QED
Thus, the 133.33J represents the initial work input during the heat transfer process.
<padditionally...><pbased on="" the="" first="" law="" rate="" at="" which="" heat="" is="" extracted="" from="" lower="" temperature="" reservoir="" calculated="" as="">
QL=QH-Wnet, in
QL=1000-133.333
QL=866.67J
</pbased></padditionally...>
Answer:
She exerts a force of 40 N.
Explanation:
The fact that the ring remains stationary indicates that the forces are in equilibrium.
Let’s denote Jo's force as x.
The equation to consider is
140 = x + 100
x = 40
Answer:
The outcome of adding 999mm to 100m is 101m.
Explanation:
That's my belief.
Answer:
a, 71.8° C, 51° C
b, 191.8° C
Explanation:
Given the data:
D(i) = 200 mm
D(o) = 400 mm
q' = 24000 W/m³
k(r) = 0.5 W/m.K
k(s) = 4 W/m.K
k(h) = 25 W/m².K
The heat generation formula can be articulated as follows:
q = πr²Lq'
q = π. 0.1². L. 24000
q = 754L W/m
Thermal conduction resistance, R(cond) = 0.0276/L
Thermal conduction resistance, R(conv) = 0.0318/L
Applying the energy balance equation,
Energy In = Energy Out
This equates to q, which is 754L
From the initial analysis, the temperature at the interface between the rod and sleeve is found to be 71.8° C
Additionally, the outer surface temperature records as 51° C
Furthermore, based on the second analysis, the calculated temperature at the center of the rod is determined to be 191.8° C
Complete Question
An aluminum "12 gauge" wire measures a diameter of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The electric field E in the wire varies over time as E(t)=0.0004t2−0.0001t+0.0004 newtons per coulomb, where time is recorded in seconds.
At time 5 seconds, I = 1.2 A.
We need to find the charge Q traveling through a cross-section of the conductor from time 0 to time 5 seconds.
Answer:
The charge is 
Explanation:
The question indicates that
The wire’s diameter is 
The radius of the wire is 
Aluminum's resistivity is 
The electric field variation is described as

The charge is effectively given by the equation

Where A is the area expressed as

Thus,

Therefore

By substituting values
![Q = 120 \int\limits^{t}_{0} { [ 0.0004t^2 - 0.0001t +0.0004] } \, dt](https://tex.z-dn.net/?f=Q%20%3D%20120%20%5Cint%5Climits%5E%7Bt%7D_%7B0%7D%20%7B%20%5B%200.0004t%5E2%20-%200.0001t%20%2B0.0004%5D%20%7D%20%5C%2C%20dt)
![Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | t} \atop {0}} \right.](https://tex.z-dn.net/?f=Q%20%3D%20120%20%5B%20%5Cfrac%7B0.0004t%5E3%20%7D%7B3%7D%20-%20%5Cfrac%7B0.0001%20t%5E2%7D%7B2%7D%20%2B0.0004t%5D%20%7D%20%20%5Cleft%20%7C%20t%7D%20%5Catop%20%7B0%7D%7D%20%5Cright.)
The question states that t = 5 seconds
![Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | 5} \atop {0}} \right.](https://tex.z-dn.net/?f=Q%20%3D%20120%20%5B%20%5Cfrac%7B0.0004t%5E3%20%7D%7B3%7D%20-%20%5Cfrac%7B0.0001%20t%5E2%7D%7B2%7D%20%2B0.0004t%5D%20%7D%20%20%5Cleft%20%7C%205%7D%20%5Catop%20%7B0%7D%7D%20%5Cright.)
![Q = 120 [ \frac{0.0004(5)^3 }{3} - \frac{0.0001 (5)^2}{2} +0.0004(5)] }](https://tex.z-dn.net/?f=Q%20%3D%20120%20%5B%20%5Cfrac%7B0.0004%285%29%5E3%20%7D%7B3%7D%20-%20%5Cfrac%7B0.0001%20%285%29%5E2%7D%7B2%7D%20%2B0.0004%285%29%5D%20%7D)
