Answer:
engine B is more efficient.
Explanation:
We know that Carnot cycle is an ideal cycle for all working heat engine.In Carnot cycle there are four processes in which two are constant temperature processes and others two are isentropic process.
We also kn ow that the efficiency of Carnot cycle given as follows

Here temperature should be in Kelvin.
For engine A



For engine B



So from above we can say that engine B is more efficient.
Answer:
The flux (volume of water per unit time) through the hoop will also double.
Explanation:
The flux = volume of water per unit time = flow rate of water through the hoop.
The Flow rate of water through the hoop is proportional to the area of the hoop, and the velocity of the water through the hoop.
This means that
Flow rate = AV
where A is the area of the hoop
V is the velocity of the water through the hoop
This flow rate = volume of water per unit time = Δv/Δt =Q
From all the above statements, we can say
Q = AV
From the equation, if we double the area, and the velocity of the stream of water through the hoop does not change, then, the volume of water per unit time will also double or we can say increases by a factor of 2
Answer:
8.85 Ω
Explanation:
Resistance of a wire is:
R = ρL/A
where ρ is resistivity of the material,
L is the length of the wire,
and A is the cross sectional area.
For a round wire, A = πr² = ¼πd².
For aluminum, ρ is 2.65×10⁻⁸ Ωm, or 8.69×10⁻⁸ Ωft.
Given L = 500 ft and d = 0.03 in = 0.0025 ft:
R = (8.69×10⁻⁸ Ωft) (500 ft) / (¼π (0.0025 ft)²)
R = 8.85 Ω
complete question
A certain amplifier has an open-circuit voltage gain of unity, an input resistance of 1 \mathrm{M} \Omega1MΩ and an output resistance of 100 \Omega100Ω The signal source has an internal voltage of 5 V rms and an internal resistance of 100 \mathrm{k} \Omega.100kΩ. The load resistance is 50 \Omega.50Ω. If the signal source is connected to the amplifier input terminals and the load is connected to the output terminals, find the voltage across the load and the power delivered to the load. Next, consider connecting the load directly across the signal source without the amplifier, and again find the load voltage and power. Compare the results. What do you conclude about the usefulness of a unity-gain amplifier in delivering signal power to a load?
Answer:
3.03 V 0.184 W
2.499 mV 125*10^-9 W
Explanation:
First, apply voltage-divider principle to the input circuit: 1
*5
= 4.545 V
The voltage produced by the voltage-controlled source is:
A_voc*V_i = 4.545 V
We can find voltage across the load, again by using voltage-divider principle:
V_o = A_voc*V_i*(R_o/R_l+R_o)
= 4.545*(100/100+50)
= 3.03 V
Now we can determine delivered power:
P_L = V_o^2/R_L
= 0.184 W
Apply voltage-divider principle to the circuit:
V_o = (R_o/R_o+R_s)*V_s
= 50/50+100*10^3*5
= 2.499 mV
Now we can determine delivered power:
P_l = V_o^2/R_l
= 125*10^-9 W
Delivered power to the load is significantly higher in case when we used amplifier, so a unity gain amplifier can be useful in situation when we want to deliver more power to the load. It is the same case with the voltage, no matter that we used amplifier with voltage open-circuit gain of unity.