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Pie
3 years ago
11

Kate is working on a project in her tech education class. She plans to assemble a fan motor. Which form of energy does the motor

convert most of its electric energy into?
A.light energy
B. motion energy
C. sound energy
D. thermal energy
Physics
1 answer:
grin007 [14]3 years ago
4 0

The job that the fan is designed and built to do is to convert the electrical energy it uses into the kinetic (motion) energy of moving air.

I can't really guarantee that it accomplishes that with MOST of the electrical energy it uses, because I don't know how efficient your fan is. For example, if it's a really old fan, and one blade has the end broken off, and a lot of dust and mosquitoes have gotten into the motor, and it shakes and vibrates and makes a lot of noise when it's running, then it's converting a lot of the electrical energy into thermal energy (it gets hot when it runs) and some into sound energy too.

If you can live without the word "most" in the question, then we can assume that the fan is well designed and running like a top, and the answer is definitely choice-B .

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The average mass of a car in the US is 1.440 x 10^6 g. Express this mass in kg.
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Answer:

Average mass of acar in the US (in kg) = 1440 kg

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Mass in kg:

\rm 1 \: g =  {10}^{ - 3}  \: kg \\  \\  \rm 1.440 \times 10^6 \ g = 1.440 \times 10^6  \times  {10}^{ - 3} \ kg \\  \\  \rm = 1.440 \times 10^{6 - 3} \ kg \\  \\  \rm = 1.440 \times 10^3 \ kg \\  \\ \rm = 1.440 \times 1000 \ kg \\  \\ \rm = 1440 \ kg

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2 years ago
Lasers are classified according to the eye-damage danger they pose. Class 2 lasers, including many laser pointers, produce visib
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Answer:

<em>a) 318.2 W/m^2</em>

<em>b) 2.5 x 10^-4 J</em>

<em>c) 1.55 x 10^-8 v/m</em>

<em></em>

Explanation:

Power of laser P = 1 mW = 1 x 10^-3 W

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If beam diameter = 2 mm = 2 x 10^-3 m

then

cross-sectional area of beam A = \pi d^{2} /4 = (3.142 x (2*10^{-3} )^{2})/4

A = 3.142 x 10^-6 m^2

a) Intensity I = P/A

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I = ( 1 x 10^-3)/(3.142 x 10^-6) = <em>318.2 W/m^2</em>

<em></em>

b) Total energy delivered E = Pt

where P is the power of the beam

t is the exposure time

E = 1 x 10^-3 x 250 x 10^-3 = <em>2.5 x 10^-4 J</em>

<em></em>

c) The peak electric field is given as

E = \sqrt{2I/ce_{0} }

where I is the intensity of the beam

E is the electric field

c is the speed of light = 3 x 10^8 m/s

e_{0} = 8.85 x 10^9 m kg s^-2 A^-2

E = \sqrt{2*318.2/3*10^8*8.85*10^9}  = <em>1.55 x 10^-8 v/m</em>

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Answer:

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4 0
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