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poizon [28]
3 years ago
8

5 ways in which friction can be useful

Physics
1 answer:
saul85 [17]3 years ago
6 0

Friction can be useful when you trying to stop a car, also it can be useful in energy losses, increase wear and tear and produces heat

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What is the relationship between thermal energy, temperature, pressure?
Ira Lisetskai [31]

Answer: When volume is constant, pressure is directly proportional to temperature. When temperature is constant, pressure is inversely proportional to volume. When pressure is constant, volume is directly proportional to temperature.

Explanation:

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Which of the following situations would cause the greatest decrease in the motion of molecules in a system
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At 0 Kelvin everything stops moving ( even electrons ), so you could try to decrease the system's temperature
0 Kelvin = -273 Celsius
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Two closed containers look the same but when is packed with LED and the other with a few feathers how could you determine which
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You would want to push both containers, first one and then the other,
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Suppose you have an unknown clear substance immersed in water, and you wish to identify it by finding its index of refraction. Y
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Answer: 1.65

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3 0
3 years ago
An object of mass 30KG is falling in air and experiences a force due to air resistance of 50 newtons determine the net force act
ryzh [129]

Answer:

Assume that g = 10\; \rm N \cdot kg^{-1}. The net force on this object will be 250\; \rm N (downwards.) The acceleration of this object will be approximately 8.3\; \rm m \cdot s^{-2} (also downwards.)

Explanation:

<h3>Net force</h3>

The object is falling towards the ground because of gravity. The size of the gravitational force on this object depends on its mass and the strength of the gravitational field at its location.

Near the surface of the earth, the gravitational field strength is approximately 10\; \rm N \cdot kg^{-1}. In other words, approximately 10\; \rm N of gravitational force acts on each kilogram of mass near the surface of the earth.

The mass of this object is given as m = 30\; \rm kg. Therefore, the size of the gravitational force on it will be:

W = m \cdot g \approx 30 \; \rm kg \cdot 10\; N \cdot kg^{-1} = 300\; \rm N.

Near the surface of the earth, gravitational forces point towards the ground. On the other hand, the direction of air resistance on this object will be opposite to its direction of motion. Since this objects is moving towards the ground, the air resistance on it will be directed in the opposite direction. That's exactly the opposite of the direction of the gravitational force on this object. The net force on this object will be:

300\; \rm N - 50\; \rm N =250\; \rm N.

<h3>Acceleration</h3>

Let a denote the acceleration on this object. Apply Newton's Second Law of motion:

\begin{aligned} a &= \frac{F(\text{net force})}{m} \approx \frac{250\; \rm N}{30\; \rm kg} \approx 8.3\; \rm m \cdot s^{-2}\end{aligned}.

Note that the acceleration of this object and the net force on it should be in the same direction.

8 0
3 years ago
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