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3 years ago

What is the magnitude of the electrostatic force

Physics

2 answers:

Tresset [83]3 years ago

7 0

Answer:

2.3 x 10^-12

Explanation:

Plug into fe=kq1q2/r^2

ozzi3 years ago

5 0

The electron charge is called the Elementary Charge, given by:

$e=1.6*10^{-19}$

The Eletrostatic Force bewteen two changes is giver by Coulomb Law:

$F= \frac{Qqk}{d^2}$

Substituing the unknowns, we have:

$F= \frac{(1.6*10^{-19})^2}{(10^{-8})^2} \\ F=(1.6*10^{-11})^2 \\ \boxed {F=2.56*10^{-22}N}$

If you notice any mistake in my english, please let me know, because i am not native.

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Two satellites are in circular orbits around Earth. Satellite A has speed vA. Satellite B has an orbital radius nine times that

Pani-rosa [81]

Answer:

option B

Explanation:

given,

Satellite B has an orbital radius nine times that of satellite A.

R' = 9 R

now, orbital velocity of the satellite A

........(1)

now, orbital velocity of satellite B

from equation 1

hence, the correct answer is option B

8 0

2 years ago

If a ball is thrown into the air with a velocity of 36 ft/s, its height (in feet) after t seconds is given by y = 36t − 16t2. fi

Juliette [100K]

Height (y) = 36t - 16t^2, where t = time in seconds (s).
Our height (y) after 1s = 36(1) - 16(1)^2
y = 36 - 16 = 20 ft
So it reached a height of 20 ft during that 1 second, which means that at that 1 second it had a velocity of 20ft/s, since v = d(distance)/t = 20ft/1s

6 0

3 years ago

A transformer is to be used to provide power for a computer disk drive that needs 6.0 V (rms) instead of the 120 V (rms) from th

Ipatiy [6.2K]

Answer:

$N_{2}$ =20 turns

Explanation:

The given case is a step down transformer as we need to reduce 120 V to 6 V.

number of turns on primary coil N_{P}= 400

current delivered by secondary coil I_{S}= 500 mA

output voltage = 6 V (rms)

we know that

$I_{p}=\frac{V_{out}}{V_{in}\times I_{s}}$

putting values we get

$I_{p}=\frac{6}{120\times 0.5}$

$I_{p}= 0.1 A$

to calculate number of turns in secondary

$\frac{N_{2}}{400} =\frac{6}{120}$

therefore, $N_{2}$ =20 turns

5 0

3 years ago

You may have noticed runaway truck lanes while driving in the mountains. These gravel-filled lanes are designed to stop trucks t

Sladkaya [172]

Answer:

0.767

Explanation:

The work done on the truck by the frictional drag force is given by

$W=-Fd$

where

F is the magnitude of the frictional force

d = 38.0 m is the maximum displacement allowed for the truck

The negative sign is due to the fact that the force of friction is opposite to the motion of the truck

The force of friction can also be written as:

$F=\mu mg$

where

$\mu$ is the coefficient of kinetic friction between the truck and the lane

m is the mass of the truck

g is the acceleration of gravity

So we can rewrite the work done as

$W=-\mu mg d$ (1)

According to the work-energy theorem, the work done by friction is equal to the change in kinetic energy of the truck:

$W=K_f - K_i = \frac{1}{2}mv^2-\frac{1}{2}mu^2$ (2)

where

v = 0 is the final velocity of the truck

u = 23.9 m/s is the initial velocity of the truck

By combining (1) and (2) we get

$-\frac{1}{2}mu^2 = -\mu mg d$

And solving for $\mu$ , we find the minimum coefficient of kinetic friction able to stop the truck in a distance d:

$\mu = \frac{u^2}{2gd}=\frac{23.9^2}{2(9.8)(38.0)}=0.767$

7 0

3 years ago

A planet exerts a gravitational force of magnitude 9e22 N on a star. If the planet were 2 times closer to the star (that is, if

Dmitrij [34]

To solve this problem we will use the related concepts in Newtonian laws that describe the force of gravitational attraction. We will use the given value and then we will obtain the proportion of the new force depending on the Radius. From there we will observe how much the force of attraction increases in the new distance.

Planet gravitational force

$F_p = 6*10^{22}N$