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Aneli [31]
3 years ago
10

Particles in an object are constantly moving. The particles also attract and repel each other. Which can this attraction and mov

ement be classified as?
specific heat
heat
thermal energy
Physics
1 answer:
Fed [463]3 years ago
6 0
The cause of attraction and movement in terms of repulsion and attraction is caused electromagnetic properties of the particles. The basic rule of magnetism is like attracts unlike while like repels like. This rule is useful in the application of electricity
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The above Free Body Diagram represents the motion of a toy car across a floor from left to right. The weight of the .5 kg car is
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Answer:

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Since the car is moving to the right, the Normal Force is balancing the Weight and the Net Force is 10 N, right.

Explanation:

as the answers says, the only two forces in the y axis are the normal force and the weight, and they balance each other. On the x axis, you have 20N to the right and the friction is a force that opposes the movement, so the 10N are to the left. The net force is 20 - 10 = 10N.

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Why is a camera lens round but the pictures come out square
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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.
Sliva [168]

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)

Where,

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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3 years ago
PT7 what does ";" mean??
IceJOKER [234]

Answer: do

Explanation:

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