Answer:
C
Explanation:
One of the disadvantages of solar cells is that electricity storage systems are not readily available. Excess energy generated by the solar panels are wasted except they are stored by solar batteries for later use. There are various systems for storing electricity from solar cells apart from solar batteries which is the common storage system. An example of another electricity storage system for solar cell is using the water electrolyzer to store solar energy which can be used to later generate hydroelectricity.
Advantages of a solar cell includes Renewable energy, Economy-friendly and environmental-friendly energy and good durability
Answer:
The Poisson's Ratio of the bar is 0.247
Explanation:
The Poisson's ratio is got by using the formula
Lateral strain / longitudinal strain
Lateral strain = elongation / original width (since we are given the change in width as a result of compession)
Lateral strain = 0.15mm / 40 mm =0.00375
Please note that strain is a dimensionless quantity, hence it has no unit.
The Longitudinal strain is the ratio of the elongation to the original length in the longitudinal direction.
Longitudinal strain = 4.1 mm / 270 mm = 0.015185
Hence, the Poisson's ratio of the bar is 0.00375/0.015185 = 0.247
The Poisson's Ratio of the bar is 0.247
Please note also that this quantity also does not have a dimension
Answer:
The percentage ductility is 35.5%.
Explanation:
Ductility is the ability of being deform under applied load. Ductility can measure by percentage elongation and percentage reduction in area. Here, percentage reduction in area method is taken to measure the ductility.
Step1
Given:
Diameter of shaft is 10.2 mm.
Final area of the shaft is 52.7 mm².
Calculation:
Step2
Initial area is calculated as follows:


A = 81.713 mm².
Step3
Percentage ductility is calculated as follows:


D = 35.5%.
Thus, the percentage ductility is 35.5%.
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.