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2 years ago

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If you were to separate all of the electrons and protons in 1.00 g (0.001 kg) of matter, you’d have about 96,000 C of positive c

harge and the same amount of negative charge. If you placed these charges 8.00 m apart, how strong would the attractive forces between them be?

1 answer:

natima [27]2 years ago

3 0

Answer:

The attractive force between them is $1.296 \times 10^{18}$ N

Explanation:

Given:

Charge $q = 96000$ C

Distance between two charges $r = 8$ m

According to the coulomb's law,

$F = \frac{kq^{2} }{r^{2} }$

Where $k = 9 \times 10^{9}$ = force constant.

$F = \frac{9 \times 10^{9} \times (96000)^{2} }{8^{2} }$

$F = 1.296 \times 10^{18}$ N

Therefore, the attractive force between them is $1.296 \times 10^{18}$ N

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At 213.1 K a substance has a vapor pressure of 45.77 mmHg. At 243.7 K it has a vapor pressure of 193.1 mm Hg. Calculate its heat

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Answer:

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Explanation:

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2 years ago

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1 year ago

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2 years ago

The wheels of the locomotive push back on the tracks with a constant net force of 7.50 × 105 N, so the tracks push forward on th

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Answer:

The freight train would take 542.265 second to increase the speed of the train from rest to 80.0 kilometers per hour.

Explanation:

Statement is incomplete. Complete description is presented below:

<em>A freight train has a mass of </em> $1.83\times 10^{7}\,kg$ <em>. The wheels of the locomotive push back on the tracks with a constant net force of </em> $7.50\times 10^{5}\,N$ <em>, so the tracks push forward on the locomotive with a force of the same magnitude. Ignore aerodynamics and friction on the other wheels of the train. How long, in seconds, would it take to increase the speed of the train from rest to 80.0 kilometers per hour?</em>

If locomotive have a constant net force ( $F$ ), measured in newtons, then acceleration ( $a$ ), measured in meters per square second, must be constant and can be found by the following expression:

$a = \frac{F}{m}$ (1)

Where $m$ is the mass of the freight train, measured in kilograms.

If we know that $F = 7.50\times 10^{5}\,N$ and $m = 1.83\times 10^{7}\,kg$ , then the acceleration experimented by the train is:

$a = \frac{7.50\times 10^{5}\,N}{1.83\times 10^{7}\,kg}$

$a = 4.098\times 10^{-2}\,\frac{m}{s^{2}}$

Now, the time taken to accelerate the freight train from rest ( $t$ ), measured in seconds, is determined by the following formula:

$t = \frac{v-v_{o}}{a}$ (2)

Where:

$v$ - Final speed of the train, measured in meters per second.

$v_{o}$ - Initial speed of the train, measured in meters per second.

If we know that $a = 4.098\times 10^{-2}\,\frac{m}{s^{2}}$ , $v_{o} = 0\,\frac{m}{s}$ and $v = 22.222\,\frac{m}{s}$ , the time taken by the freight train is:

$t = \frac{22.222\,\frac{m}{s}-0\,\frac{m}{s} }{4.098\times 10^{-2}\,\frac{m}{s^{2}} }$

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The freight train would take 542.265 second to increase the speed of the train from rest to 80.0 kilometers per hour.

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1 year ago