Ask question

Ask question

All categories

3 years ago

5

An electron and a positron are located 15 m away from each other and held fixed by some mechanism. The positron has the same mas

s and the same magnitude of charge as those of the electron, but its charge is positive. The electron and the positron are released at the same time by the mechanism. The electron and the positron begin to speed up towards each other. What velocities should they have when they are 2 m away from each other

1 answer:

ICE Princess25 [194]3 years ago

8 0

Answer:

Explanation:

Electrical potential energy will be converted into kinetic energy .

Electrical potential energy when distance was 15 m .

E₁ = 9 x 10⁹ x - q² /d where q is magnitude of charge on electron or positron

E₁ = 9 x 10⁹ x - ( 1.6 x 10⁻¹⁹ )² /15

= - 1.536 x 10⁻²⁹ J .

Electrical potential energy when distance was 2 m .

E₁ =9 x 10⁹ x - q² /d where q is magnitude of charge on electron or positron

E₁ = 9 x 10⁹ x - ( 1.6 x 10⁻¹⁹ )² /2

= -11.52 x 10⁻²⁹ J .

Decrease in energy = (11.52 - 1.536 ) x 10⁻²⁹

= 9.984 x 10⁻²⁹ J .

This energy will be converted into kinetic energy and they will be distributed equally in each .

Energy of each = 9.984 x 10⁻²⁹ /2

= 4.992 x 10⁻²⁹ J .

1/2 m v² = 4.992 x 10⁻²⁹ , m is mass of electron

.5 x 9.1 x 10⁻³¹ v² = 4.992 x 10⁻²⁹

v² = 109.71

v = 10.47 m/s .

You might be interested in

Humans protect themselves from natural disasters by all of the following except

jeyben [28]

I'm pretty sure it's sunscreen

4 0

3 years ago

How far does a runner run if he runs for 60 seconds at 5 m/s

mylen [45]

Answer: 20 miles

Explanation:

4 0

3 years ago

A hole is punched at A in a plastic sheet by applying a 660-N force P to end D of lever CD, which is rigidly attached to the sol

olasank [31]

Answer:

hello your question lacks some data and required diagram

G = 77 GPa, т all = 80 MPa

answer : required diameter = 252.65 * 10-^3 m

Explanation:

Given data :

force ( P ) = 660 -N force

displacement = 15 mm

G = 77 GPa

т all = 80 MPa

i) Determine the required diameter of shaft BC

considering the vertical displacement ( looking at handle DC from free body diagram )

D' = 0.3 sin∅ , where D = 0.015

hence ∅ = 2.8659°

calculate the torque acting at angle ∅ of CD on the shaft BC

Torque = 660 * 0.3 cos∅

= 660 * 0.3 * cos 2.8659 = 198 * -0.9622 = 190.5156 N

hello attached is the remaining part of the solution

4 0

3 years ago

According to Newton's first law of motion, what is required to make an object slow down? Enter your answer in the space provided

LenaWriter [7]

<u>Answer:</u>

According to newton's first law of motion, friction is required to make an object slow down.

<u>Explanation:</u>

According to the Newton's first law of motion, for an object to change its velocity (either a change in the magnitude or the direction), there must be a cause to it which is defined as a net external force.

For example, an object which is sliding across a table or floor slows down due to the net force of friction that is acting on that object.

5 0

3 years ago

A projectile is thrown with velocity v at an angle θ with horizontal. When the projectile is at a height equal to half of the ma

raketka [301]

Let, the maximum height covered by projectile be $\sf{H_m}$

$\purple{ \longrightarrow \bf{h_m = \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2g} }}$

Projectile is thrown with a velocity = v
Angle of projection = θ

Velocity of projectile at a height half of the maximum height covered be $\sf{v_0}$

$\qquad$ ______________________________

Then –

$\qquad$ $\pink{ \longrightarrow \bf{ \dfrac{h_m}{2} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }}$

$\qquad$ $\longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2g} \times \dfrac{1}{2} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }$

$\qquad$ $\longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{4g} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }$

$\qquad$ $\longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2} = {v_0}^{2} \: {sin}^{2} \theta }$

$\qquad$ $\longrightarrow \sf{ \dfrac{ {v}^{2} }{2} = {v_0}^{2} }$

$\qquad$ $\longrightarrow \bf{v_0 = \sqrt{ \dfrac{ {v}^{2} }{2} } = \dfrac{v}{ \sqrt{2} } }$

Now, the vertical component of velocity of projectile at the height half of $\sf{h_m}$ will be –

$\qquad$ $\longrightarrow \bf{v_{(y)}=v_0 \: sin \theta }$

$\qquad$ $\longrightarrow \bf{v_{(y)} = \dfrac{v}{ \sqrt{2} } \: sin \theta = \dfrac{v \: sin \: \theta}{ \sqrt{2} } }$

Therefore, the vertical component of velocity of projectile at this height will be–

☀️ $\qquad$ $\pink {\bf{ \dfrac{v \: sin \: \theta}{ \sqrt{2} }} }$

6 0

3 years ago

Read 2 more answers