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3 years ago

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When two point charges are a distance d apart, the magnitude of the electrostatic force between them is F. If the distance betwe

en the point charges is increased to 3d, the magnitude of the electrostatic force between the two charges will be

2 answers:

dedylja [7]3 years ago

8 0

Decreased by a factor of 9

EastWind [94]3 years ago

3 0

Answer:

The magnitude of the electrostatic force decreases by a factor 9

Explanation:

The electrostatic force between two charges is given by:

$F=k\frac{q_1 q_2}{d^2}$

where

k is the Coulomb's constant

q1 and q2 are the two charges

d is the distance between the two charges

We see that the magnitude of the force is inversely proportional to the square of the distance. If the distance is increased to 3d: d' = 3d, the new electrostatic force would be:

$F'=k\frac{q_1 q_2}{(d')^2}=k\frac{q_1 q_2}{(3d)^2}=\frac{1}{9} k\frac{q_1 q_2}{d^2}=\frac{F}{9}$

So, the electrostatic force decreases by a factor 9.

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A beam of electrons moving in the x-direction enters a region where a uniform 208-G magnetic field points in the y-direction. Th

GREYUIT [131]

Answer:

$1.26\cdot 10^7 m/s$

Explanation:

When a charged particle moves perpendicularly to a magnetic field, the force it experiences is:

$F=qvB$

where

q is the charge

v is its velocity

B is the strength of the magnetic field

Moreover, the force acts in a direction perpendicular to the motion of the charge, so it acts as a centripetal force; therefore we can write:

$qvB=m\frac{v^2}{r}$

where

m is the mass of the particle

r is the radius of the orbit of the particle

The equation can be re-arranges as

$v=\frac{qBr}{m}$

where in this problem we have:

$q=1.6\cdot 10^{-19}C$ is the magnitude of the charge of the electron

$B=208 G=208\cdot 10^{-4}T$ is the strength of the magnetic field

The beam penetrates 3.45 mm into the field region: therefore, this is the radius of the orbit,

$r=3.45 mm = 3.45\cdot 10^{-3} m$

$m=9.11\cdot 10^{-31} kg$ is the mass of the electron

So, the electron's speed is

$v=\frac{(1.6\cdot 10^{-19})(208\cdot 10^{-4})(3.45\cdot 10^{-3})}{9.11\cdot 10^{-31}}=1.26\cdot 10^7 m/s$

6 0

3 years ago

A car interior is heated to 48 C by the sun What type of energy transfer?

Aneli [31]

Answer:

Radiation heat energy transfer

Explanation:

The type of heat transfer from the Sun is radiation heat transfer, which is the transfer of heat through electromagnetic radiation

The distance of the Sun to the Earth is several million kilometers away, with the space between being composes of vacuum and the nuclear reaction in the Sun's core generates vast amount of electromagnetic radiation that is transferred all across the universe and reaches the Earth as visible light and radiant energy at the speed of light

The radiant energy transferred from the Sun heats up the Earth, including the car's interior.

7 0

3 years ago

12. A frain moves from rest to a speed of 25 m/s in 30.0 seconds. What is its acceleration?

ziro4ka [17]

initial velocity=u=0m/s
Final velocity=v=25m/s
Time=t=30s

$\\ \tt\longmapsto Acceleration=\dfrac{v-u}{t}$

$\\ \tt\longmapsto Acceleration=\dfrac{25-0}{30}$

$\\ \tt\longmapsto Acceleration=\dfrac{25}{30}$

$\\ \tt\longmapsto Acceleration=0.8m/s^2$

6 0

3 years ago

While a block slides forward 1.35 m, a force pulls back at a 135 direction, doing -17.8 J of work. what is the magnitude of the

Verdich [7]

The magnitude of the force is 18.6 N

Explanation:

The work done by a force on an object is given by

$W=Fd cos \theta$

where

F is the magnitude of the force

d is the displacement of the object

$\theta$ is the angle between the direction of the force and of the displacement

For the block in this problem, we have

$W=-17.8 J$ is the work done

d = 1.35 m is the displacement of the block

$\theta=135^{\circ}$ is the angle between the force and the displacement

Solving for F, we find the magnitude of the force:

$F=\frac{W}{d cos \theta}=\frac{-17.8}{(1.35)(cos 135)}=18.6 N$

Learn more about work here:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

8 0

4 years ago

Write equations for both the electric and magnetic fields for an electromagnetic wave in the red part of the visible spectrum th

NeTakaya

Answer:

Explanation:

General equation of the electromagnetic wave:

$E(x, t)= E_0sin[\frac{2\pi}{\lambda}(x-ct)+\phi ]$

where

$\phi =$ Phase angle, 0

c = speed of the electromagnetic wave, 3 × 10⁸

$\lambda =$ wavelength of electromagnetic wave, 698 × 10⁻⁹m

E₀ = 3.5V/m

Electric field equation

$E(x, t)= 3.5sin[\frac{2\pi}{6.98\times10^{-7}}(x-3\times 10^8t)]\\\\E(x, t)= 3.5sin[{9 \times 10^6}(x-3\times 10^8t)]\\\\E(x, t)= 3.5sin[{9 \times 10^6x-2.7\times 10^{15}t)]$

Magnetic field Equation

$B(x, t)= B_0sin[\frac{2\pi}{\lambda}(x-ct)+\phi ]$

Where B₀= E₀/c

$B_0 = \frac{E_0}{c} = \frac{3.5}{3\times10^8}=1.2 \times 10^{-8}T$

$B(x, t)= 1.2\times10^{-8}sin[\frac{2\pi}{6.98\times10^{-7}}(x-3\times 10^8t)]\\\\B(x, t)= 1.2\times10^{-8}sin[{9 \times 10^6}(x-3\times 10^8t)]\\\\B(x, t)= 1.2\times10^{-8}sin[{9 \times 10^6x-2.7\times 10^{15}t)]$

6 0

3 years ago