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Volgvan
3 years ago
13

A 1500-W heater is connected to a 120-V line. How much heat energy does it produce in 2.0 hours? Hint: how are power and energy

related?
Physics
1 answer:
ELEN [110]3 years ago
3 0
Power is the rate of work done or produced by a system. It will have units of Watts which is equivalent to Joules per second. To determine the work or energy produced by the system, we multiply power with time. We do as follows:

Energy = Power x time
Energy = 1500 J/s (2.0 hr) (3600 s / 1 hr) 
Energy = 10800000 J
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Kanna Kamui

Explanation:

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Which of the following is occurring while a satellite is in orbit around Earth? O It is continuously pulling away from Earth It
hoa [83]

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Explanation:

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3 years ago
A) A real object 4.0 cm high stands 30.0 cm in front of a converging lens of focal length 23 cm. Find the image distance, the im
vfiekz [6]

Explanation:

Given that,

Height of object = 4.0 cm

Distance of the object u= -30.0 cm

Focal length = 23 cm

We need to calculate the image distance

Using lens's formula

\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}

Where, u = object distance

v = image distance

f = focal length

Put the value into the formula

\dfrac{1}{v}-\dfrac{1}{-30}=\dfrac{1}{23}

\dfrac{1}{v}=\dfrac{1}{23}-\dfrac{1}{30}

\dfrac{1}{v}=\dfrac{7}{690}

v=98.57\ cm

We need to calculate the height of the image

Using formula of height

\dfrac{h'}{h}=\dfrac{-v}{u}

Where, h' = height of image

h = height of object

\dfrac{h'}{4}=\dfrac{-98.57}{-30}

h'=\dfrac{98.57\times4}{30}

h'=13.14\ cm

The image is real and inverted.

(b). Now, object distance u = 13.0 cm

We need to calculate the image distance

Using lens's formula

\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}

Put the value into the formula

\dfrac{1}{v}=\dfrac{1}{23}-\dfrac{1}{13.0}

\dfrac{1}{v}=-\dfrac{10}{299}

v=-29.9\ cm

We need to calculate the height of the image

\dfrac{h'}{4}=\dfrac{-(-29.9)}{-13.0}

h=-\dfrac{29.9\times4}{13.0}

h=-9.2\ cm

The image is virtual and erect.

Hence, This is required solution.

6 0
3 years ago
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gtnhenbr [62]

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