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pishuonlain [190]
3 years ago
9

Two small objects each with a net charge of Q (positive) exert a force of magnitude F on each other. We replace one of the objec

ts with another whose net charge is 4Q. The original magnitude of the force on the Q charge was F; what is the magnitude of the force on the Q now?
A. 16F
B. 4F
C. F
D. F/4
Physics
1 answer:
erik [133]3 years ago
5 0

Answer:

option B

Explanation:

we know,

Force between two charge is calculated by

F = \dfrac{kq_1q_2}{r^2}

r is the distance between the two charges

k is Coulomb constant.

give two small object with charge Q

now, Force

F = \dfrac{kQQ}{r^2}......(1)

now,

New force if another charge is equal to 4 Q

F' = \dfrac{kQ(4Q)}{r^2}

F' = 4\dfrac{kQQ}{r^2}

now from equation 1

F' = 4 F

Hence, the correct answer is option B

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An ideal Diesel cycle has a compression ratio of 18 and a cutoff ratio of 1.5. Determine the (1) maximum air temperature and (2)
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Answer:

(1) The maximum air temperature is 1383.002 K

(2) The rate of heat addition is 215.5 kW

Explanation:

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\frac{v_4}{v_3} =\frac{v_4}{v_2} \times \frac{v_2}{v_3}  = \frac{v_1}{v_2} \times \frac{v_2}{v_3} = 18 \times \frac{1}{1.5} = 12

\frac{T_3}{T_4} =(\frac{v_4}{v_3} )^{k-1} = 12^{0.4} = 2.702

Therefore;

T_4 = \frac{1383.002}{2.702} =511.859 \ k

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Heat rejected per kilogram is given by the following relation;

c_v(T_4-T_1)  = 0.718×(511.859 - 290.15) = 159.187 kJ/kg

The efficiency is given by the following relation;

\eta = 1-\frac{\beta ^{k}-1}{\left (\beta -1  \right )r_{v}^{k-1}}

Where:

β = Cut off ratio

Plugging in the values, we get;

\eta = 1-\frac{1.5 ^{1.4}-1}{\left (1.5 -1  \right )18^{1.4-1}}= 0.5191

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Heat supplied = \frac{150}{0.5191}  = 288.978 \ hp

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The rate of heat addition = 215.5 kW.

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f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}

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