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dmitriy555 [2]
3 years ago
9

What is the kinetic energy of a 2,000-kilogram ball that is on the ground

Chemistry
1 answer:
Finger [1]3 years ago
6 0
Kinetic energy is associated with the motion of an object.
Since the object is at rest on the ground, its velocity is zero.
Since kinetic energy is directly proportional to the square of velocity, kinetic energy of the ball on the ground is zero.
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The amount of kinetic energy an object has depends on
Alex777 [14]
How much it has to drop and how heavy it is. Hope this is what you're looking for:)
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3 years ago
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If 2,035 cal of heat is added to a 500.0 g sample of water at 35.0°C, what is the final
Lynna [10]

Answer:

39.1 °C

Explanation:

Recall the equation for specific heat:

q=mc \Delta T\\

Where q is the heat, m is the mass, c is the specific heat of the substance (in this case water), and delta T is the change in temperature.

You should know that the specific heat of water is 1 cal/g/C.

Using the information in the question:

2035=500(1)(T-35)\\2035=500T-17500\\500T=19535\\T=39.07

The final temperature is about 39.1 °C.

5 0
3 years ago
Please help!! Balancing Nuclear Equations
sineoko [7]

The missing part of the equation is found to be 4/2He. Option A

<h3>What are nuclear equations?</h3>

The term nuclear equations have to do with the type of equation in which one type of nucleus is transformed into another sometimes by the bombardment or loss of a particle.

Now the full equation ought to be written as 7/3Li + 1/1H -----> 4/2He + 4/2He. This is because the total mass on the left is 8 and the total charge on the left is 4.

Learn more about nuclear equations:brainly.com/question/19752321

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8 0
2 years ago
Calculate the standard entropy of vaporization of ethanol at its boiling point, 352 K. The standard molar enthalpy of vaporizati
Korvikt [17]

Answer : The correct option is, (b) +115 J/mol.K

Explanation :

Formula used :

\Delta S=\frac{\Delta H_{vap}}{T_b}

where,

\Delta S = change in entropy

\Delta H_{vap} = change in enthalpy of vaporization = 40.5 kJ/mol

T_b = boiling point temperature = 352 K

Now put all the given values in the above formula, we get:

\Delta S=\frac{\Delta H_{vap}}{T_b}

\Delta S=\frac{40.5kJ/mol}{352K}

\Delta S=115J/mol.K

Therefore, the standard entropy of vaporization of ethanol at its boiling point is +115 J/mol.K

8 0
3 years ago
Given the different molecular weights, dipole moments, and molecular shapes, why are their molar volumes nearly the same?
Marina CMI [18]

Answer:

option d

Explanation:

Molecular sizes of gaseous molecules are very less. Volume occupied by the all the molecules of the gases are very less or negligible as compared to the container in which it is kept. Therefore, most of the volume occupied by gaseous molecules are negligible.

Volume occupied by the gaseous molecules are actually the volume of the container and its does not depend upon the amount, molecular mass or dipole moment of the gaseous molecules.

Therefore, the correct option is d ‘Because most of the volume occupied by the substance is empty space.’

8 0
3 years ago
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