All categories

3 years ago

Calculate the force between charges of 55 × 108 C and 1 × 107C if they are 5 cm apart.?

Physics

1 answer:

lions [1.4K]3 years ago

6 0

First:

5 cm = 0,05 m

Now, the formula:

F = k * (qq') / r²

Replacing:

F = 9x10⁹ * (55x10⁸ * 1x10⁷) / 0,05²

F = 9x10⁹ * (5,5x10¹⁶ / 2,5x10⁻³)

F = 9x10⁹ * 2,2x10¹⁹

F = 1,98x10²⁹

The force between the charges is 1,98x10²⁹ Newtons.

You might be interested in

Jake has noticed his best friend Alison cheating at chess several times this week. If she gets caught, she will be kicked off th

tensa zangetsu [6.8K]

He should confront her about it and if after that point she continues report it to the chess team

8 0

3 years ago

Read 2 more answers

Unless indicated otherwise, assume the speed of sound in air to be v = 344 m/s. You have a stopped pipe of adjustable length clo

faltersainse [42]

Answer:

Length of pipe $= 0.057$ meter

Explanation:

Speed of a transverse wave on a string

$v = \sqrt{\frac{F}{\mu} }$

where F is the tension in string and $\mu$ is the mass per unit length

Thus,

$\mu = \frac{m}{L}$

Substituting the given values we get -

$\mu = \frac{7.25 * 10^{-3}}{0.62}\\mu = 0.0117 \frac{Kg}{m}$

Speed of a transverse wave on a string

$v = \sqrt{\frac{4510}{0.0117} } \\v = 620.86 \frac{m}{s}$

For third harmonic wave , frequency is equal to

$f = \frac{nv}{2L}$

Substituting the given values, we get -

$f = \frac{3 * 620.86}{2 * 0.62} \\f = 1502.08$

Length of pipe

$L = \frac{nv}{4 f}$

Substituting the given values we get

$n = 1$ for first harmonic wave

$L = \frac{344* 1}{4*1502.08} \\L = 0.057$

Length of pipe $= 0.057$ meter

5 0

3 years ago

What is kinetic energy? Question 1 options: energy that is due to motion energy that is stored the mass of an object the rate at

mario62 [17]

<span>Kinetic Energy means energy that a body possesses by virtue of being in motion. I hope this helps!!!</span>

8 0

3 years ago

If the action force is the boy pushing on the ball, what is the reaction force?

Brums [2.3K]

If the boy outputs enough weight upon the ball in response the ball will move.

Please vote my answer brainliest! Thanks.

7 0

3 years ago

A 21500 kg jet airplane is flying through some high winds. At some point in time, the airplane is pointing due north, while the

GalinKa [24]

Answer: $\theta =79.47 North\ of\ west$

Explanation:

Given

Mass of airplane =21500 kg

Force due to jet engines=35100 N

Force from wind is 14900 N at angle of 75 south of west

Resolving forces

$F_x=14900cos75=-3856.403 N$

$F_y=35100-14900sin75$

$F_y=35100-14392.29=20707.70 N$

Therefore acceleration in x direction

$a_x=\frac{F_x}{m}=\frac{-3856.403}{21500}=-0.179 m/s^2$

$a_y=\frac{F_y}{m}=\frac{20707.70}{21500}=0.963 m/s^2$

Direction of acceleration

$tan\theta =\frac{a_y}{a_x}=\frac{0.963}{0.179}=-5.38$

$\theta =79.47\ North\ of\ west$

5 0

3 years ago

Calculate the force between charges of 55 × 108 C and 1 × 107C if they are 5 cm apart.?​

Calculate the force between charges of 55 × 108 C and 1 × 107C if they are 5 cm apart.?