Answer:
You will need 450 cells (3 cm each) to meet the voltage/current requirement.
The panel must be 3 cells in one side, by 150 cell in another side. 1350 cm^2 or 0.135 m^2. They must be connected 3 in row in parallel (to add current), then each of the former group must be connected in series to meet the voltage, so it would be 150 rows of connected in series.
The panel can be optimized using a voltage inverter, to convert current to voltage. In this way, less cells can be used achieving the same output specs.
Explanation:
To meet the voltage:
120 [v] required voltage
0.8 [v] voltage of each cell
![\frac{120}{0.8} =150[v]\\](https://tex.z-dn.net/?f=%5Cfrac%7B120%7D%7B0.8%7D%20%3D150%5Bv%5D%5C%5C)
So we need 150 cells in series for the voltage.
To meet the current
1.0 [A] Required current
350[mA]=0.35[A] cell current
1/0.35=3 cell So we need 3 cells in parallel to add the currents and meet the requirement.
See the attached figure
The latent heat of fusion refers to the solid to liquid or liquid to solid states.
Answer: Option C
<u>Explanation:
</u>
It is known that the inter conversion process from the states of solid to liquid is referred as fusion. So, for these conversions, the external energy in the heat form should be supplied to solid.
This external energy should be greater than the latent heat of solid in order to successfully break the bonds to form liquid. So the change in the enthalpy of the reaction while conversion from solids to liquids are termed as latent heats of fusion.
Even the inter-conversion from liquid to solid state will undergo change in enthalpy where the heat will be released and that is termed as latent heats of solidification. It is found that latent heat of solidification is equal in magnitude but opposite in direction with the latent heats of fusion.
Answer:
Burning of canful of petrol will release more energy
Explanation:
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Answer:
Here Strain due to testing is greater than the strain due to yielding that is why computation of load is not possible.
Explanation:
Given that
Yield strength ,Sy= 240 MPa
Tensile strength = 310 MPa
Elastic modulus ,E= 110 GPa
L=380 mm
ΔL = 1.9 mm
Lets find strain:
Case 1 :
Strain due to elongation (testing)
ε = ΔL/L
ε = 1.9/380
ε = 0.005
Case 2 :
Strain due to yielding
ε '=0.0021
Here Strain due to testing is greater than the strain due to yielding that is why computation of load is not possible.
For computation of load strain due to testing should be less than the strain due to yielding.
