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3 years ago

Two resistors, R1 = 25.0 ohms and

Physics

1 answer:

Leni [432]3 years ago

3 0

Answer: 0.078

Explanation:

Given variables:

R1 = 25.0 , R2 = 64.0 , I1 = 0.200

Step 1: find V

0.200 = v/25

0.200 * 25 = v

5 = v

Step 2: Solve for I2

I = V/R

I = 5/64

I = 0.078125

Hope this helps!

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How much electrical energy is used by a 400 W toaster that is operating for 5 minutes?

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The electrical energy consumed by a toaster is 0.033 Kwh.

<u>Explanation</u>:

The power utilized by the toaster is 400 W.

$\text { The power utilized by the toaster is } 400 \mathrm{W} \text { in kilo-watts is } \frac{400}{1000}=0.4 \mathrm{Kw}$

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$\text { The toaster is operated for } 5 \text { Minutes in hours is } \frac{5}{60}=0.083 \text { hours. (One minute is } 60 \text { seconds) }$

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3 years ago

What does an atom look like under an electron microscope?

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If you look at a atom under a electron microscope you would use it

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3 years ago

During a solar eclipse, the Moon, Earth, and Sun all lie on the same line, with the Moon between the Earth and the Sun.

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Answer:

(a) $F_{sm} = 4.327\times 10^{20}\ N$

(b) $F_{em} = 1.983\times 10^{20}\ N$

Solution:

As per the question:

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Mass of Moon, $M_{m} = 7.34\times 10^{22}\ kg$

Mass of Sun, $M_{s} = 1.989\times 10^{30}\ kg$

Distance between the earth and the moon, $R_{em} = 3.84\times 10^{8}\ m$

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Now,

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$F_{G} = \frac{Gmm'_{2}}{r^{2}}$ (1)

Now,

(a) The force exerted by the Sun on the Moon is given by eqn (1):

$F_{sm} = \frac{GM_{s}M_{m}}{R_{sm}^{2}}$

$F_{sm} = \frac{6.67\times 10^{- 11}\times 1.989\times 10^{30}\times 7.34\times 10^{22}}{(1.5\times 10^{11})^{2}}$

$F_{sm} = 4.327\times 10^{20}\ N$

(b) The force exerted by the Earth on the Moon is given by eqn (1):

$F_{em} = \frac{GM_{s}M_{m}}{R_{em}^{2}}$

$F_{em} = \frac{6.67\times 10^{- 11}\times 5.972\times 10^{24}\times 7.34\times 10^{22}}{(3.84\times 10^{8})^{2}}$

$F_{em} = 1.983\times 10^{20}\ N$

$F_{se} = \frac{GM_{s}M_{m}}{R_{es}^{2}}$

$F_{se} = \frac{6.67\times 10^{- 11}\times 1.989\times 10^{30}\times 5.972\times 10^{24}}{((1.5\times 10^{11}))^{2}}$

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