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Yakvenalex [24]
3 years ago
12

The sum of potential and kinetic energies in the particles of a substance is called

Physics
2 answers:
olganol [36]3 years ago
6 0

Internal energy.

Explanation:

In any substance/object, the particles inside it (atoms/molecules) constantly move in random directions and with random speeds (this motion is called Brownian motion). As a result, the particles have some kinetic energy (which is proportional to the temperature of the substance). Moreover, the particles interact with each other due to the presence of electrostatic intermolecular forces, and as a result, the particles also have some potential energy.

The sum of the kinetic energies and potential energies of the particles in a substance is called internal energy.

Gemiola [76]3 years ago
3 0
Internal energy would be the answer because it says the sum of energies "IN" the particles.
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For each of the following, draw the resulting wave when the two pulses occupy the same space. Unless otherwise noted, each wave
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Answer:

A

Explanation:

because it is are of circle

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3 years ago
Before beginning a long trip on a hot day, a driver inflates anautomobile tire to a gauge pressure of 2.70 atm at 300 K. At the
anygoal [31]

Answer:

The value is the temperature of the air inside the tire T_{2} = 340.54 K

% of the original mass of air in the tire should be released 99.706 %

Explanation:

Initial gauge pressure = 2.7 atm

Absolute pressure at inlet P_{1} = 2.7 + 1 = 3.7 atm

Absolute pressure at outlet P_{2} = 3.2 + 1 = 4.2 atm

Temperature at inlet T_{1} = 300 K

(a) Volume of the system is constant so  pressure is directly proportional to the temperature.

\frac{T_{2} }{T_{1} } = \frac{P_{2} }{P_{1} }

\frac{T_{2} }{300}  = \frac{4.2}{3.7}

T_{2} = 340.54 K

This is the value is the temperature of the air inside the tire

(b). Since volume of the tyre is constant & pressure reaches the original value.

From ideal gas equation P V = m R T

Since P , V & R is constant. So

m T = constant

m_{1}  T_{1} =  m_{2}  T_{2}

\frac{m_{2} }{m_{1}  } = \frac{T_{1} }{T_{2} }

\frac{m_{2} }{m_{1}  } = \frac{300}{354.54}

\frac{m_{2} }{m_{1}  } =0.00294

value of  the original mass of air in the tire should be released is  \frac{m_{2} - m_{1}}{m_{1}}

\frac{0.00294-1}{1}

⇒ -0.99706

% of the original mass of air in the tire should be released 99.706 %.

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