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QveST [7]
3 years ago
14

A lightbulb has a power of 100 W and is used for 4 hours. A microwave has a power of 1200 W and is used for 5 minutes. How much

energy is used by the lightbulb? How much energy is used by the microwave? The lightbulb has an efficiency of 1.8%. How much heat energy does the lightbulb produce in 4 hours?
Physics
1 answer:
ladessa [460]3 years ago
4 0

Answer:

a. E=1440KJ

b. E=360KJ

c. E=1.8 J

Explanation:

I have the power (Watts) is expressed as Energy (Joules) / Time (seconds), also I have to1hour * \frac{60min}{1hour}*\frac{60s}{1min}\\1h=3600s

so, for the lightbulb

100W= \frac{Energy}{3600s*4}\\Energy=100W*14400s\\ E=1440000J =1440KJ

Analogously, for the microwave

1200W= \frac{Energy}{60s*5}\\Energy=1200W*300s\\ E=360000J =360KJ

Now, I have to express the efficiency as Heat energy / power * 100 so1.8= \frac{Heat Energy}{100W}*100\\Heat Energy=1.8J

Done

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32.46m/s

Explanation:

Hello,

To solve this exercise we must be clear that the ball moves with constant acceleration with the value of gravity = 9.81m / S ^ 2

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When performing a mathematical demonstration, it is found that the equations that define this movement are the follow

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solving for Vf

Vf=\sqrt{Vo^2+2gX}\\Vf=\sqrt{7.3^2+2(9.81)(51)}=32.46m/s

the speed with the ball hits the ground is 32.46m/s

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A bat can detect small objects, such as an insect, whose size is approximately equal to one wavelength of the sound the bat make
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6.136 mm

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smallest insect bat can hear  will be equal to the wavelength  of the sound the bat make.

\lambda = \dfrac{v}{f}

\lambda = \dfrac{343}{5.59\times 10^4}

\lambda = 6.136\times 10^{-3}}

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The relative density of the second liquid is 7.

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\rho_{2}=7\times \rho _{1}

Since the first liquid is water thus \rho _{1}=1gm/cm^3

Thus the relative density of the second liquid is 7.

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