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konstantin123 [22]
2 years ago
8

Which of the following changes would double the force between two charged particles? O A. Decreasing the distance between the pa

rticles by a factor of 2 B. Increasing the distance between the particles by a factor of 2 O C. Doubling the amount of charge on one of the particles O D. Doubling the amount of charge on each particle
Physics
2 answers:
padilas [110]2 years ago
7 0

Answer:

Decreasing the distance between the particles by a factor of 2

Explanation:

To double the force between the two charge particles, the distance between the particles should be reduced by a factor of two.

According to coulombs law "force is directly proportional to the potential between the two charges and inversely proportional to the square of their distances".

       \sf{F  = \dfrac{kq_{1} q_{2} }{r^{2} }}

F is the electric force

k is the coulomb constant

q is the charge

r is the distance or separation

As the separation between the charges is reduced the force increases and vice versa.

jarptica [38.1K]2 years ago
6 0

Let's check

\\ \rm\dashrightarrow F=\dfrac{k}{q_1q_2}{r^2}

\\ \rm\dashrightarrow F\propto Q

\\ \rm\dashrightarrow F\propto \dfrac{1}{r^2}

So

Option A and C can be used

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3 years ago
Say that you are in a large room at temperature TC = 300 K. Someone gives you a pot of hot soup at a temperature of TH = 340 K.
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To solve the problem it is necessary to take into account the concepts related to energy efficiency in the engines, the work done, the heat input in the systems, the exchange and loss of heat in the soupy the radius between the work done the lost heat ( efficiency).

By definition the efficiency of the heat engine is

\epsilon = 1- \frac{T_c}{T_h}

Where,

T_c = Temperature at the room

T_h  =Temperature of the soup

The work done is defined as,

dW = \epsilon(dQ_h)

Where Q_h represents the input heat and at the same time is defined as

dQ_h =c_v (dT_h)

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c_V =Specific Heat

The change at the work would be defined then as

dW = \epsilon(dQ_h)

dW = \epsilon c_v (dT_h)

dW = (1-\frac{T_c}{T_h})c_v (dT_h)

W = \int dW = \int (1-\frac{T_c}{T_h})c_v (dT_h)

W = c_v (T_h-T_c)-c_v T_c ln(\frac{T_h}{T_c})

On the other hand we have that the heat lost by the soup is equal to

dQ_h =c_v (dT_h)

Q_h =c_v (T_h-T_c)

The ratio between both would be,

\frac{W}{Q_h} = \frac{c_v (T_h-T_c)-c_v T_c ln(\frac{T_h}{T_c})}{c_v (T_h-T_c)}

\frac{W}{Q_h} = \frac{1+ln(\frac{T_h}{T_c})}{1-\frac{T_h}{T_c}}

Replacing with our values we have,

\frac{W}{Q_h} = 1+\frac{ln(\frac{340K}{300K})}{1-\frac{340K}{300K}}

\frac{W}{Q_h} = 0.0613

Therefore the fraction of heat lost by the soup that can be turned into useable work by the engine is 0.0613.

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