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3 years ago

Across what potential difference does an electron have to be accelerated in order to reach the speed v=3e7 m/s?

1 answer:

4 0

When an electron is accelerated through potential difference then the speed that it attain will be explained by energy conservation

here by energy conservation we can say that

change in kinetic energy of electron = electrostatic potential energy gained through given potential difference

kinetic energy is given as

$KE = \frac{1}{2}mv^2$

electrostatic potential energy is given as

$PE = qV$

now by energy conservation

$qV = \frac{1}{2}mv^2$

given that for electron

$m = 9.1 * 10^{-31} kg$

$v = 3 * 10^7 m/s$

$q = 1.6 * 10^{-19} C$

now by plug in values

$1.6 * 10^{-19} * V = \frac{1}{2}*9.1* 10^{-31} *(3*10^7)^2$

$V = \frac{4.095 * 10^{-16}}{1.6 * 10^{-19}}$

$V = 2559.4 Volts$

So here it is accelerated through potential difference of 2559.4 Volts

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If two objects are moved closer together so that the distance between them is one-third the original distance, what is the gravi

r-ruslan [8.4K]

Answer:
nine times the original gravitational force
Explanation:
The rule of the gravitational force between two bodies is shown in the attached images.
The parameters in the rule are:
Fg : the gravitational force between the two bodies
G : universal gravitational constant
m1 and m2 : the masses of the two bodies
r : the distance between the two bodies

From the given rule, we can notice that:
The force of attraction between the two bodies is inversely proportional to the square of the distance between them.
This means that:
As the distance decreases to 1/3 its original value, the gravitational force would increase by a factor of 9.

Hope this helps :)

3 0

3 years ago

Read 2 more answers

A student on a tower 49 m height drops a stone. One second later he throws a second stone after the first. They both hit the gro

ohaa [14]

By applying the second equation of motion, the speed at which he threw the second stone is equal to 12.10 m/s.

<h3>How to determine the speed?</h3>

First of all, we would calculate the time taken by the first stone to reach a height of 49 meters by applying the second equation of motion as follows:

S = ut + ½gt²

49 = 0(t) + ½ × 9.8 × t²

49 = 4.9t²

t² = 49/4.9

t = √10

t = 3.16 seconds.

Now, we can determine the speed at which he threw the second stone:

<u>Note:</u> Time = 3.16 - 1 = 2.16 seconds.

S = ut + ½gt²

49 = u(2.16) + ½ × 9.8 × 2.16²

49 = 2.16u + 22.86

2.16u = 49 - 22.86

u = 26.14/2.16

u = 12.10 m/s.

Read more on initial speed here: brainly.com/question/19365526

#SPJ1

6 0

2 years ago

If C moves to position 5 (60 meters), what is its average velocity during these 5 seconds?

Hoochie [10]

Velocity:

60 meters/ 5 seconds= 12 meters/second

Velocity=12meters/second

(60 meters divided by 5 seconds equals 12 meters per second)

Hope this helps!

3 0

3 years ago

Read 2 more answers

A proton is traveling horizontally to the right at 1.8 × 106 m/s. (a) Find the magnitude and direction of the weakest electric f

Mars2501 [29]

Answer:

528398.4375 N/C opposite to the direction of the proton

$3.56\times 10^{-8}\ s$

288.24609375 N/C in the same direction of the motion of the electron

Explanation:

t = Time taken

u = Initial velocity = 0

v = Final velocity = $1.8\times 10^{6}\ m/s$

s = Displacement = 3.2 cm

a = Acceleration

Mass of electron = $9.11\times 10^{-31}\ kg$

Mass of electron = $1.67\times 10^{-27}\ kg$

q = Charge of particle = $1.6\times 10^{-19}\ C$

$v^2-u^2=2as\\\Rightarrow a=\dfrac{v^2-u^2}{2s}\\\Rightarrow a=\dfrac{0^2-(1.8\times 10^6)^2}{2\times 0.032}\\\Rightarrow a=-5.0625\times 10^{13}\ m/s^2$

Electric field is given by

$E=\dfrac{ma}{q}\\\Rightarrow E=\dfrac{1.67\times 10^{-27}\times -5.0625\times 10^{13}}{1.6\times 10^{-19}}\\\Rightarrow E=−528398.4375\ N/C$

The electric field is 528398.4375 N/C opposite to the direction of the proton

$v=u+at\\\Rightarrow t=\dfrac{v-u}{a}\\\Rightarrow t=\dfrac{0-1.8\times 10^6}{-5.0625\times 10^{13}}\\\Rightarrow t=3.56\times 10^{-8}\ s$

The time taken is $3.56\times 10^{-8}\ s$

$E=\dfrac{ma}{q}\\\Rightarrow E=\dfrac{9.11\times 10^{-31}\times -5.0625\times 10^{13}}{-1.6\times 10^{-19}}\\\Rightarrow E=288.24609375\ N/C$

The electric field is 288.24609375 N/C in the same direction of the motion of the electron

7 0

4 years ago

An electromagnetic am band radio wave could have a wavelength of?

MaRussiya [10]

The frequency of choice-3 is

(3 x 10⁸ m/s) / (500 m) = 600,000 Hz.

That number is the 600 near the bottom of the AM radio dial.

6 0

3 years ago