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3 years ago

[Please help :^)]

Physics

2 answers:

Contact [7]3 years ago

5 0

Sorry wish I could help

Serhud [2]3 years ago

3 0

Answer:

the potentail of kinetic and potential energy

Explanation:

first explain the concept of kinetic energy (what it is and what its used for) and give examples (cars, a basketball thrown across a hall, and airplane), and do the same with potential energy (the energy an object stores, example: a streched rubber band)

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You are at a furniture store and notice that a Grandfather clock has its time regulated by a physical pendulum that consists of

artcher [175]

Answer:

The distance is 1.026 m.

Explanation:

mass of rod, M = 1.23 kg

Length, L = 1.25 m

mass, m = 10 kg

Time period, T = 2 s

Let the distance is d.

The formula of the time period is given by

$T = 2\pi\sqrt\frac{\frac{1}{3}ML^2+md^2}{(M +m)g}\\\\2\times 2 = 4\pi^2\times \frac{\frac{1}{3}\times1.23\times1.25\times 1.25+ 10d^2}{(1.23 + 10)\times9.8}\\\\11.16 = 0.64 + 10d^2\\\\d= 1.026 m$

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3 years ago

Ten seconds after an electric fan is turned on, the fan rotates at 300 rev/min. its average angular acceleration is

Tcecarenko [31]

Check the attached file for the answer.

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4 years ago

I need to know which ones are physical or chemical changes.

andrezito [222]

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I believe this is right but double check to make sure :)

7 0

3 years ago

What is an electrical charge ?

aliya0001 [1]

An Electric charge is the physical property of matter that causes it to experience a force when placed in an electromagnetic field.

I hope this helped you with your problem.

8 0

4 years ago

The amount of power required to move an object can be increased without changing the

VLD [36.1K]

Answer:

Yes it is possible to increase the power with out changing the amount of work.

Explanation:

The power is defined by the amount of power divided by the time. This time is the one needed to do the work. We can understand this issue by analyzing an example with numeric values.

Work = 500 [J]

Time = 5 [s]

Power will be:

$Power=\frac{500}{5} \\Power=100 []watt]\\$

Now if we change the time to 2 seconds:

$Power = 500 [J]/2[s]\\Power = 250 [watt]\\$

As we can see, the power was increased without the need to change the work.

3 0

4 years ago