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Finger [1]
3 years ago
15

A 2-kg bowling ball is 2.5 meters off the ground on a post when it falls. Just before it reaches the ground, it is traveling 7 m

/s. Assuming that there is no air resistance, which statement is true?
The initial potential energy is less than the final kinetic energy.

The intial potential energy is greater than the final kinetic energy.

The mechanical energy is conserved.

The mechanical energy is not conserved.
Chemistry
2 answers:
notsponge [240]3 years ago
8 0

Answer: The correct answer is option "The mechanical energy is conserved".

Explanation:

Mechanical energy = Kinetic energy + Potential Energy

Mechanical energy of the body is defined a sum of potential energy and kinetic energy possessed by the body in motion. It is the energy associated with position and motion of the body.

The principle of Conservation of Mechanical Energy states that the mechanical energy of the body remains constant as long as the forces acting on the body are conservative forces.

From, this we can say that mechanical energy of the bowling ball with mass of 2 kg falling at speed of 4.4 m/s will remain conserved.

Hence,the correct answer is "The mechanical energy is conserved".

Ksivusya [100]3 years ago
8 0

The answer to your question would be The mechanical energy is conserved. I just took the test and it was right.

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Answer:

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Explanation:

The computation of the molecular mass of the unknown gas is shown below:

As we know that

\frac{Diffusion\ rate\ of unknown\ gas }{CO_{2}\ diffusion\ rate} = \frac{\sqrt{CO_{2\ molar\ mass}} }{\sqrt{Unknown\ gas\ molercular\ mass } }

where,

Diffusion rate of unknown gas = 155 mL/s

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Now placing these values to the above formula

\frac{155mL/s}{102mL/s} = \frac{\sqrt{44 g mol^{-1}} }{\sqrt{M_{unknown}} } \\\\ 1.519 = \frac{\sqrt{44 g mol^{-1}} }{\sqrt{M_{unknown}} } \\\\ {\sqrt{M_{unknown}} } = \frac{\sqrt{44 g mol^{-1}}}{1.519} \\\\ {\sqrt{M_{unknown}} } = \frac{44 g mol^{-1}}{(1.519)^{2}}

After solving this, the molecular mass of the unknown gas is

= 19.07 g mol^-1

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