The rate of gain for the high reservoir would be 780 kj/s.
A. η = 35%

W = 
W = 420 kj/s
Q2 = Q1-W
= 1200-420
= 780 kJ/S
<h3>What is the workdone by this engine?</h3>
B. W = 420 kj/s
= 420x1000 w
= 4.2x10⁵W
The work done is 4.2x10⁵W
c. 780/308 - 1200/1000
= 2.532 - 1.2
= 1.332kj
The total enthropy gain is 1.332kj
D. Q1 = 1200
T1 = 1000

<h3>Cournot efficiency = W/Q1</h3>
= 1200 - 369.6/1200
= 69.2 percent
change in s is zero for the reversible heat engine.
Read more on enthropy here: brainly.com/question/6364271
Answer:
7.8 Mph
Explanation:
Rate of cycling = 1.1 rev/s
Rear wheel diameter = 26 inches
Diameter of sprocket on pedal = 6 inches
Diameter of sprocket on rear wheel = 4 inches
Circumference of rear wheel = \pi d=26\piπd=26π
Speed would be
\begin{gathered}\text{Rate of cycling}\times \frac{\text{Diameter of sprocket on pedal}}{\text{Diameter of sprocket on rear wheel}}\times{\text{Circumference of rear wheel}}\\ =1.1\times \frac{6}{4}\times 26\pi\\ =134.77432\ inches/s\end{gathered}Rate of cycling×Diameter of sprocket on rear wheelDiameter of sprocket on pedal×Circumference of rear wheel=1.1×46×26π=134.77432 inches/s
Converting to mph
1\ inch/s=\frac{1}{63360}\times 3600\ mph1 inch/s=633601×3600 mph
134.77432\ inches/s=134.77432\times \frac{1}{63360}\times 3600\ mph=7.65763\ mph134.77432 inches/s=134.77432×633601×3600 mph=7.65763 mph
The Speed of the bicycle is 7.8 mph
Answer:
1. Tokyo skytree tower is the tallest tower in the world, measuring 2080 feet. That's almost twice the size of the Eiffel Tower!
2.The process of building the tower began in 2008. The project was completed on 29 February 2012.
3. Pairing form with function, Skytree will serve as a TV and radio broadcast tower.
Answer:
(4.5125 * 10^-3 kg.m^2)ω_A^2
Explanation:
solution:
Moments of inertia:
I = mk^2
Gear A: I_A = (1)(0.030 m)^2 = 0.9*10^-3 kg.m^2
Gear B: I_B = (4)(0.075 m)^2 = 22.5*10^-3 kg.m^2
Gear C: I_C = (9)(0.100 m)^2 = 90*10^-3 kg.m^2
Let r_A be the radius of gear A, r_1 the outer radius of gear B, r_2 the inner radius of gear B, and r_C the radius of gear C.
r_A=50 mm
r_1 =100 mm
r_2 =50 mm
r_C=150 mm
At the contact point between gears A and B,
r_1*ω_b = r_A*ω_A
ω_b = r_A/r_1*ω_A
= 0.5ω_A
At the contact point between gear B and C.
At the contact point between gears A and B,
r_C*ω_C = r_2*ω_B
ω_C = r_2/r_C*ω_B
= 0.1667ω_A
kinetic energy T = 1/2*I_A*ω_A^2+1/2*I_B*ω_B^2+1/2*I_C*ω_C^2
=(4.5125 * 10^-3 kg.m^2)ω_A^2