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Blizzard [7]
3 years ago
12

When you are running around a track, what kind of energy are you using?

Physics
1 answer:
Sergeeva-Olga [200]3 years ago
7 0

Answer:

kinetic energy

Explanation:

we are using chemical energy in our bodies to produce movement, whitch in turn convents to warmth.

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How much energy is used for boiling water in a 4 KW kettle for 20 minutes.​
Marta_Voda [28]

Answer:About 860,000 results (1.08 seconds)

kettle-power

Jun 18, 2014 — It's interesting to see how much energy heating and cooling systems use ... the kettle uses goes into the water it costs more to use electricity to boil ... 3 minutes is 1/20 of an hour, so 2.4kW for 3 minutes is equivalent to 2.4 / 20 = 0.12kWh. ... kettle should NOT clock over 1kw/h in the 4 minutes it takes to boil

Explanation:

5 0
3 years ago
Which of these is an example of electromagnetic induction? A current passing through a resistor uses power IR2. A current throug
Novosadov [1.4K]
Well as a pretty direct hint, recall that the word "induction" comes from the fact that electric fields are "induced" by (changing) magnetic fields
6 0
3 years ago
A potter’s wheel moves from rest to an angular speed of 0.40 rev/s in 37.5 s.
svlad2 [7]

The angular acceleration of the potter's wheel is 0.067 rad/s².

The given parameters:

  • Final angular speed, ωf = 0.4 rev/s
  • Time of motion, t = 37.5 s

<h3>What is angular acceleration?</h3>
  • Angular acceleration of an object is the rate of change of angular speed of the object.

The angular acceleration of the potter's wheel is calculated as follows;

\alpha = \frac{\Delta \omega }{t} \\\\\alpha = (0.4\ rev/s \times \frac{2 \pi \ rad}{1 \ rev} ) \times \frac{1}{37.5 \ s} \\\\\alpha = 0.067 \ rad/s^2

Thus, the angular acceleration of the potter's wheel is 0.067 rad/s².

Learn more about angular acceleration here: brainly.com/question/25129606

5 0
3 years ago
You do a certain amount of work on an object initially at rest, and all the work goes into increasing the object’s speed. If you
EleoNora [17]

Answer:

\sqrt{2}v

Explanation:

The work done on the object at rest is all converted into kinetic energy, so we can write

W=\frac{1}{2}mv^2

Or, re-arranging for v,

v=\sqrt{\frac{2W}{m}}

where

v is the final speed of the object

W is the work done

m is the object's mass

If the work done on the object is doubled, we have W' = 2W. Substituting into the previous formula, we can find the new final speed of the object:

v'=\sqrt{\frac{2W'}{m}}=\sqrt{\frac{2(2W)}{m}}=\sqrt{2}\sqrt{\frac{2W}{m}}=\sqrt{2}v

So, the new speed of the object is \sqrt{2}v.

3 0
4 years ago
<img src="https://tex.z-dn.net/?f=%20%5Cunderline%7B%20%5Ctext%7BQUESTIONS%7D%7D%20%3A%20" id="TexFormula1" title=" \underline{
Musya8 [376]

Answer:

Solution given:

<u>any three applications of electromagnetic </u><u>waves</u><u> </u><u>are</u><u>:</u>

  1. <u>Used</u><u> </u><u>for</u><u> </u><u>cooking</u><u> </u><u>purpose</u><u> </u><u>in</u><u> </u><u>microwave</u><u>.</u>
  2. <u>Used</u><u> </u><u>in</u><u> </u><u>RADAR</u><u> </u><u>system</u><u> </u><u>for</u><u> </u><u>aircraft</u><u> </u><u>navigation</u><u>.</u>
  3. <u>Used</u><u> </u><u>in</u><u> </u><u>communication</u><u>.</u>

2.

Solution given;

power [P]=100W

Current [I]=5A.

Voltage [V]=?

we have

P=IV

100=5*V

V=\frac{100}{5}=<u>2</u><u>0</u><u>V</u>

<u>the voltage of the battery used in the </u><u>car</u><u> </u><u>is</u><u> </u><u>2</u><u>0</u><u>V</u><u>.</u>

8 0
3 years ago
Read 2 more answers
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