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3 years ago

Explain the following defects of a simple electric cell:

Physics

1 answer:

eimsori [14]3 years ago

4 0

Answer:

Explanation:

The two major defects of simple electric cells causes current supplied to be for short time. These defects are: polarization and local action.

a. Polarization: This is a defect caused by an accumulation of hydrogen bubbles at the positive electrode of the cell. It can be prevented by the use of vent, using a hydrogen absorbing material or the use of a depolarizer.

b. Local Action: This is the gradual wearing away of the electrode due to impurities in the zinc plate. It can be controlled by the amalgamation of the zinc plate before it is used.

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If a person lives in china in which two hemispheres does he or she live

Delvig [45]

on the globe , the largest continent on earth , asia lies in between eastern and northern hemisphere. it is also known that country china comes in asian continent. hence we can say that china also lies in between eastern and northern hemisphere.

hence the two hemispheres are northern and eastern.

5 0

4 years ago

Alice's friends Bob and Charlie are having a race to a distant star 10 light years away. Alice is the race official who stays on

hodyreva [135]

Solution :

The distance between the starting point and the end point, $L_0$ = 10 light years

But due to the relativistic motion of Bob and Charlie, the distance will be reduced following the Lorentz contraction. The contracted length will be different since they are moving with different speeds.

For Bob,

Speed of Bob's rocket with respect to Alice, $L_b = 0.7 \ c$

So the distance appeared to Bob due to the length contraction,

$L_b=L_0\sqrt{1-\frac{V_b^2}{c^2}$

$L_b=10\times \sqrt{1-0.49} \ Ly$

$=7.1 \ Ly$

Therefore, the time required to finish the race by Bob is

$t_b = \frac{L_b}{V_b}$

$=\frac{7.1 \ c}{0.7 \ c}$

= 10.143 year

For Charlie,

Speed of Charlie's rocket with respect to Alice, $L_c = 0.866 \ c$

So the distance appeared to Charlie due to the length contraction,

$L_b=L_0\sqrt{1-\frac{V_c^2}{c^2}$

$L_b=10\times \sqrt{1-0.75} \ Ly$

$=5 \ Ly$

The time required to finish the race by Charlie is

$t_b = \frac{L_c}{V_c}$

$=\frac{5 \ c}{0.866 \ c}$

= 5.77 year

5 0

3 years ago

An automobile tire having a temperature of −1.6 ◦C (a cold tire on a cold day) is filled to a gauge pressure of 22 lb/in2 . What

r-ruslan [8.4K]

Answer:

25.8 lb/in²

Explanation:

Gay-Lussac's law tells us that given an ideal gas of a certain mass has a constant volume, the pressure exerted on the sides of its container is directly proportional to its absolute temperature.

$\frac{P_{1} }{T_{1} } = \frac{P_{2} }{T_{2} } \\\\\frac{22}{-1.6+273.15} =\frac{P_{2} }{45+273.15} \\\\P_{2} = \frac{22*318.15}{271.55} = 25.8lb/in^{2}$

4 0

3 years ago

An airplane is cruising at a speed of 250 m/s. If the airplanes engines provide a forward force of 19,540 N, calculate the force

ExtremeBDS [4]

250/300 dived by 500

3 0

3 years ago

An ocean thermal energy conversion system is being proposed for electric power generation. Such a system is based on the standar

defon

Answer:

Explanation:

Dear Student, this question is incomplete, and to attempt this question, we have attached the complete copy of the question in the image below. Please, Kindly refer to it when going through the solution to the question.

To objective is to find the:

(i) required heat exchanger area.

(ii) flow rate to be maintained in the evaporator.

Given that:

water temperature = 300 K

At a reasonable depth, the water is cold and its temperature = 280 K

The power output W = 2 MW

Efficiency $\zeta$ = 3%

where;

$\zeta = \dfrac{W_{out}}{Q_{supplied }}$

$Q_{supplied } = \dfrac{2}{0.03} \ MW$

$Q_{supplied } = 66.66 \ MW$

However, from the evaporator, the heat transfer Q can be determined by using the formula:

Q = UA(L MTD)

where;

$LMTD = \dfrac{\Delta T_1 - \Delta T_2}{In (\dfrac{\Delta T_1}{\Delta T_2} )}$

Also;

$\Delta T_1 = T_{h_{in}}- T_{c_{out}} \\ \\ \Delta T_1 = 300 -290 \\ \\ \Delta T_1 = 10 \ K$

$\Delta T_2 = T_{h_{in}}- T_{c_{out}} \\ \\ \Delta T_2 = 292 -290 \\ \\ \Delta T_2 = 2\ K$

$LMTD = \dfrac{10 -2}{In (\dfrac{10}{2} )}$

$LMTD = \dfrac{8}{In (5)}$

LMTD = 4.97

Thus, the required heat exchanger area A is calculated by using the formula:

$Q_H = UA (LMTD)$

where;

U = overall heat coefficient given as 1200 W/m².K

$66.667 \times 10^6 = 1200 \times A \times 4.97 \\ \\ A= \dfrac{66.667 \times 10^6}{1200 \times 4.97} \\ \\ \mathbf{A = 11178.236 \ m^2}$

The mass flow rate:

$Q_{H} = mC_p(T_{in} -T_{out} ) \\ \\ 66.667 \times 10^6= m \times 4.18 (300 -292) \\ \\ m = \dfrac{ 66.667 \times 10^6}{4.18 \times 8} \\ \\ \mathbf{m = 1993630.383 \ kg/s}$

3 0

3 years ago