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3 years ago

A ball of mass 12 kg is released from a height of 2.5 m

Physics

1 answer:

VMariaS [17]3 years ago

4 0

Answer:

300 J

Explanation:

From conservation of energy;

Gravitational potential energy at top = Kinetic Energy at bottom

mgh = ½mv²

We are given;

Mass; m = 12 kg.

Height; h = 2.5 m

g = 10 m/s²

Thus;

mgh = 12 × 2.5 × 10 = 300 J

Since mgh = Kinetic Energy at bottom.

Thus; kinetic energy of the ball once it reach the ground = 300 J

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3 years ago

Which of the following is most likely to be an observation made by a physiologist

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3 years ago

A year 11 pupil with a mass of 55kg swinging back on their chair and falling off it at a speed of 0.6m/s. What is his kinetic en

posledela

Answer:

Uk = 9.9 J

Explanation:

To calculate the kinetic energie (Uk), you can make use of this formula:

Uk = 0.5 * m * v²

given m = 55 kg and v = 0.6 m/s

Substituting in the formula gives:

Uk = 0.5 * 55 * (0.6)²

Uk = 0.5 * 55 * 0.36

Uk = 9.9 J

Extra:

Now let's examine the formula in relation to the SI units. If you understand the following, it will give you great insight in how smart Phisics is inter twained by looking at formulas and their standard units. It will save you time in future to convert formulas, if you use the right standard units.

The formula for kinetic energie is:

Uk = 0.5 * m * v²

Standard SI unit for mass m is kg.

Standard SI unit for speed v is m/s.

So v * v = v² and therefore v² must have the standard SI unit of m²/s².

From the formula, you see that the unit of Uk must be kg*m²/s² and since Uk is normally given in J, these both forms must be the same !

The main unit for Uk is the Joule. Now please see the picture, which shows the relation between the J and other SI units. Please understand that you can construct your 'own' formulas based these units. Now here is the time saver:

Because almost always the right units are given in a question, or because sometimes you can look up a constant in a table with an exotic and seemingly complicated unit, but that says a lot about the formula which must have been some how involved!

By this, I hope you now understand the implication of using the right standard SI units and how that can help you figure out what formula is needed.

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3 years ago

What is the best use of an atomic model to explain the charge of the particles in thomson's beams

hoa [83]

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3 years ago