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2 years ago

Which process increases the potential energy of the particles of a sample

Physics

1 answer:

Wittaler [7]2 years ago

4 0

Answer:

Vaporization.

Explanation:

When we heat or raise the energy of any sample, it evaporates. Evaporation is known as a form of Vaporization. When some sample evaporates, then they basically take in heat and when the sample condenses, it cools down by giving out the heat into the surrounding, atmosphere and the energy goes into the atmosphere.

An example of evaporation is the rain cycle. Where it's cooler, water on earth's surface evaporates and travels up through the atmosphere. The cool temperature causing the vapor to condense and back into water drops. When the cloud becomes saturated, the droplets fall to the ground in the form of rain.

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What is an example of transparent

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An object of mass 2.0 kg is attached to the top of a vertical spring that is anchored to the floor. The unstressed length of the

poizon [28]

Answer:

The value is $A = 0.014 \ m$

Explanation:

From the question we are told that

The mass of the object is $m = 2.0 \ kg$

The unstressed length of the string is $l = 0.08 \ m$

The length of the spring when it is at equilibrium is $l_e = 5.9 \ cm = 0.059 \ m$

The initial speed (maximum speed)of the spring when given a downward blow $v = 0.30 \ m/s$

Generally the maximum speed of the spring is mathematically represented as

$u = A * w$

Here A is maximum height above the floor (i.e the maximum amplitude)

and $w$ is the angular frequency which is mathematically represented as

$w = \sqrt{\frac{k}{m} }$

$u = A * \sqrt{\frac{k}{m} }$

=> $A = u * \sqrt{\frac{m}{k} }$

Gnerally the length of the compression(Here an assumption that the spring was compressed to the ground by the hammer is made) by the hammer is mathematically represented as

$b = l -l_e$

=> $b = 0.08 - 0.05 9$

=> $b = 0.021 \ m$

Generally at equilibrium position the net force acting on the spring is

$k * b - mg = 0$

=> $k * 0.021 - 2 * 9.8 = 0$

=> $k = 933 \ N/m$

$A = 0.30 * \sqrt{\frac{2}{933} }$

=> $A = 0.014 \ m$

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