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3 years ago

Two protons are 0.00500 m apart. Find the force between them.

Physics

1 answer:

Lesechka [4]3 years ago

5 0

Answer:

Force, $F=9.21\times 10^{-24}\ N$

Explanation:

Charge on proton is $+1.6\times 10^{-19}\ C$ .

The electrical force between charges is given by the product of charges and divided by square of distance between them. So, electric force is :

$F=\dfrac{kq^2}{r^2}\\\\F=\dfrac{9\times 10^9\times (1.6\times 10^{-19})^2}{(0.00500 )^2}\\\\F=9.21\times 10^{-24}\ N$

So, force between protons is $9.21\times 10^{-24}\ N$ .

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Where do you think the government should put the greatest support: solar energy, wind energy, clean coal, oil exploration, gas e

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The government should put it support in a combination of sources, as no source in the present can fully provide all energy requirements.

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3 years ago

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Which of the following BEST summarizes the relationship between groups and culture and critical thinking?

slamgirl [31]

Answer:

Groups and culture helps in influencing our values,ethics and beliefs. This influence should always be questioned through the process of thinking critically.

This best summarizes the relationship between groups and culture and critical thinking.

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2 years ago

You yell into a canyon. You hear the echo in 3 seconds. How long did the sound of your voice travel before bouncing off a cliff?

Lilit [14]

Answer:

The sound travelled 516 meters before bouncing off a cliff.

Explanation:

The sound is an example of mechanical wave, which means that it needs a medium to propagate itself at constant speed. The time needed to hear the echo is equal to twice the height of the canyon divided by the velocity of sound. In addition, the speed of sound through the air at a temperature of 20 ºC is approximately 344 meters per second. Then, the height of the canyon can be derived from the following kinematic formula:

$2\cdot h = v\cdot t$ (1)

Where:

$h$ - Height, measured in meters.

$v$ - Velocity of sound, measured in meters per second.

$t$ - Time, measured in seconds.

If we know that $v = 344\,\frac{m}{s}$ and $t = 3\,s$ , then the height of the canyon is:

$h = \frac{v\cdot t}{2}$

$h = \frac{\left(344\,\frac{m}{s} \right)\cdot (3\,s)}{2}$

$h = 516\,m$

The sound travelled 516 meters before bouncing off a cliff.

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2 years ago

Cloudy nights are usually warm because

almond37 [142]

Answer:

When you have a cloudy night, the clouds prevent heat from escaping through the atmosphere and into space.

The clouds act like a blanket and trap the heat it, and that is why every time you have a cloudy night, it is always warmer than a clear night.

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2 years ago

magine an astronaut on an extrasolar planet, standing on a sheer cliff 50.0 m high. She is so happy to be on a different planet,

Mama L [17]

Answer:

$\Delta t=(\frac{20}{g'}+\sqrt{\frac{400}{g'^2}+\frac{100}{g'} } )-(\frac{20}{g}+\sqrt{\frac{400}{g^2}+\frac{100}{g} } )$

Explanation:

Given:

height above which the rock is thrown up, $\Delta h=50\ m$

initial velocity of projection, $u=20\ m.s^{-1}$

let the gravity on the other planet be g'

The time taken by the rock to reach the top height on the exoplanet:

$v=u+g'.t'$

where:

$v=$ final velocity at the top height = 0 $m.s^{-1}$

$0=20-g'.t'$ (-ve sign to indicate that acceleration acts opposite to the velocity)

$t'=\frac{20}{g'}\ s$

The time taken by the rock to reach the top height on the earth:

$v=u+g.t$

$0=20-g.t$

$t=\frac{20}{g} \ s$

Height reached by the rock above the point of throwing on the exoplanet:

$v^2=u^2+2g'.h'$

where:

$v=$ final velocity at the top height = 0 $m.s^{-1}$

$0^2=20^2-2\times g'.h'$

$h'=\frac{200}{g'}\ m$

Height reached by the rock above the point of throwing on the earth:

$v^2=u^2+2g.h$

$0^2=20^2-2g.h$

$h=\frac{200}{g}\ m$

The time taken by the rock to fall from the highest point to the ground on the exoplanet:

$(50+h')=u.t_f'+\frac{1}{2} g'.t_f'^2$ (during falling it falls below the cliff)

here:

$u=$ initial velocity= 0 $m.s^{-1}$

$\frac{200}{g'}+50 =0+\frac{1}{2} g'.t_f'^2$

$t_f'^2=\frac{400}{g'^2}+\frac{100}{g'}$

$t_f'=\sqrt{\frac{400}{g'^2}+\frac{100}{g'} }$

Similarly on earth:

$t_f=\sqrt{\frac{400}{g^2}+\frac{100}{g} }$

Now the required time difference:

$\Delta t=(t'+t_f')-(t+t_f)$

$\Delta t=(\frac{20}{g'}+\sqrt{\frac{400}{g'^2}+\frac{100}{g'} } )-(\frac{20}{g}+\sqrt{\frac{400}{g^2}+\frac{100}{g} } )$

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3 years ago