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4 years ago

Light energy produces the only voltage in a solar cell. (a)-True(T) (b)- false(F)

Engineering

1 answer:

Burka [1]4 years ago

7 0

Answer:

True

Explanation:

The fundamental steps involved in the operation of a solar cell are as follows:

Light is incident on the solar cell and this light energy is absorbed by the solar cell.
The absorbed energy results in the movement of the electron inside the cell which constitutes the electric current in the cell.
Electric current flows and electricity is produced.
Hence,in solar cell the light energy is responsible for the only voltage in the cell.

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The dc machine of Problem 7.4 is to be operated as a motor supplied by a constant armature terminal voltage of 250 V. If saturat

notka56 [123]

Answer: If saturation effects are ignored, the magnetization curve of Fig. 7.27 becomes a straight line with a constant slope of 150 volts per ampere of field current.

Explanation:

6 0

3 years ago

1. One of these is NOT a type of pneumatic tool. Which one?

Serggg [28]

Answer:Circular

Explanation:

It’s the only thing not list under pneumatic tools‍♂️

5 0

3 years ago

An automobile tire with a volume of 0.8 m3 is inflated to a gage pressure of 200 kPa. Calculate the mass of air in the tire if t

saveliy_v [14]

Answer:

2.83 kg

Explanation:

Given:

Volume, V = 0.8 m³

gage pressure, P = 200 kPa

Absolute pressure = gage pressure + Atmospheric pressure

= 200 + 101 = 301 kPa = 301 × 10³ N/m²

Temperature, T = 23° C = 23 + 273 = 296 K

Now,

From the ideal gas equation

PV = mRT

Where,

m is the mass

R is the ideal gas constant = 287 J/Kg K. (for air)

thus,

301 × 10³ × 0.8 = m × 287 × 296

m = 2.83 kg

3 0

3 years ago

a turbine operating at steady state at 500 kPa, 860 K and exists at 100 kPa. A temperature sensor indicates that the exit air te

noname [10]

Answer:

Given:

P₁ = 500 kPa

T₁ = 860 K

P₂= 100 kPa

T₂ = 460 K

Let's take entropy properties of T1 and T2 from ideal properties of air,

at T = 860K, s(T₁) = 2.79783 kJ/kg.K

at T = 460K, s(T₂) = 2.13407 kJ/kg.K

using entropy balance equation:

$\frac{\sigma _cv}{m} = s(T_2)- s(T_1) - R In [\frac{P_2}{P_1}]$

$\frac{\sigma _cv}{m} = 2.79783 - 2.13407 - 0.287 In [\frac{100}{500}]$

= - 0.2018 kJ/kg. K

In this case the entropy is negative, which means the value of exit temperature is not correct, beacause entropy should always be positive(>0).

8 0

3 years ago

1.The moist unit weights and degrees of saturation of a soil are given: moist unit weight (1) = 16.62 kN/m^3, degree of saturati

alexandr1967 [171]

Answer:

Gs = 2.647

e = 0.7986

Explanation:

We know that moist unit weight of soil is given as

$\gamma_m \ or\ bulk\ density = \frac{(Gs+Se)\times \gamma_w}{(1+e)}$

where, $\gamma_m$ = moist unit weight of the soil

Gs = specific gravity of the soil

S = degree of saturation

e = void ratio

$\gamma_w$ = unit weight of water = 9.81 kN/m3

From data given we know that:

At 50% saturation, $\gamma_m = 16.62 kN/m3$

puttng all value to get Gs value;

$16.62= \frac{(Gs+0.5*e)\timees 9.81}{(1+e)}$

Gs - 1.194*e = 1.694 .........(1)

for saturaion 75%, unit weight = 17.71 KN/m3

$17.71 = \frac{(Gs+0.75*e)\times 9.81}{(1+e)}$

Gs - 1.055*e = 1.805 .........(2)

solving both equations (1) and (2), we obtained;

Gs = 2.647

e = 0.7986

6 0

3 years ago