1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
alukav5142 [94]
3 years ago
9

A neutron consists of one "up" quark of charge +2e/3 and two "down" quarks each having charge -e/3. If we assume that the down q

uarks are 4.6 × 10-15 m apart inside the neutron, what is the magnitude of the electrostatic force between them?
Physics
1 answer:
Anon25 [30]3 years ago
8 0

Answer:

The magnitude of the electrostatic force is 120.85 N

Explanation:

We can use Coulomb's law to find the electrostatic force between the down quarks.

In scalar form, Coulomb's law states that for charges q_1 and q_2 separated by a distance d, the magnitude of the electrostatic force F between them is:

F = k \frac{|q_1q_2|}{d^2}

where k is Coulomb's constant.

Taking the values:

d = 4.6 \ 10^{-15} m

q_1 = q_2 = - \frac{e}{3} = - \frac{1.6 \ 10^{-19} \ C}{3}

and knowing the value of the Coulomb's constant:

k = 8.99 \ 10 ^{9} \frac{N m^2}{C^2}

Taking all this in consideration:

F = 8.99 \ 10 ^{9} \frac{N m^2}{C^2} \frac{ (- \frac{1.6 \ 10^{-19} \ C}{3} ) ^2}{(4.6 \ 10^{-15} m)^2}

F = 120.85  \ N

You might be interested in
Which statements accurately describe the Doppler effect?
chubhunter [2.5K]

The doppler effect is the increase or decrease in the frequency of sound, light, or other waves as the source and observer move toward or away from each other.

5 0
4 years ago
Read 2 more answers
Two cars are facing each other. Car A is at rest while car B is moving toward car A with a constant velocity of 20 m/s. When car
lapo4ka [179]

Answer:

Let's define t = 0s (the initial time) as the moment when Car A starts moving.

Let's find the movement equations of each car.

A:

We know that Car A accelerations with a constant acceleration of 5m/s^2

Then the acceleration equation is:

A_a(t)  = 5m/s^2

To get the velocity, we integrate over time:

V_a(t) = (5m/s^2)*t + V_0

Where V₀ is the initial velocity of Car A, we know that it starts at rest, so V₀ = 0m/s, the velocity equation is then:

V_a(t) = (5m/s^2)*t

To get the position equation we integrate again over time:

P_a(t) = 0.5*(5m/s^2)*t^2 + P_0

Where P₀ is the initial position of the Car A, we can define P₀ = 0m, then the position equation is:

P_a(t) = 0.5*(5m/s^2)*t^2

Now let's find the equations for car B.

We know that Car B does not accelerate, then it has a constant velocity given by:

V_b(t) =20m/s

To get the position equation, we can integrate:

P_b(t) = (20m/s)*t + P_0

This time P₀ is the initial position of Car B, we know that it starts 100m ahead from car A, then P₀ = 100m, the position equation is:

P_b(t) = (20m/s)*t + 100m

Now we can answer this:

1) The two cars will meet when their position equations are equal, so we must have:

P_a(t) = P_b(t)

We can solve this for t.

0.5*(5m/s^2)*t^2 = (20m/s)*t + 100m\\(2.5 m/s^2)*t^2 - (20m/s)*t - 100m = 0

This is a quadratic equation, the solutions are given by the Bhaskara's formula:

t = \frac{-(-20m/s) \pm \sqrt{(-20m/s)^2 - 4*(2.5m/s^2)*(-100m)}  }{2*2.5m/s^2} = \frac{20m/s \pm 37.42 m/s}{5m/s^2}

We only care for the positive solution, which is:

t = \frac{20m/s + 37.42 m/s}{5m/s^2} = 11.48 s

Car A reaches Car B after 11.48 seconds.

2) How far does car A travel before the two cars meet?

Here we only need to evaluate the position equation for Car A in t = 11.48s:

P_a(11.48s) = 0.5*(5m/s^2)*(11.48s)^2 = 329.48 m

3) What is the velocity of car B when the two cars meet?

Car B is not accelerating, so its velocity does not change, then the velocity of Car B when the two cars meet is 20m/s

4)  What is the velocity of car A when the two cars meet?

Here we need to evaluate the velocity equation for Car A at t = 11.48s

V_a(t) = (5m/s^2)*11.48s = 57.4 m/s

7 0
3 years ago
Which remains the same as the distance of an object from Earth changes?
Gemiola [76]

Answer:

MASS

Explanation:

weight has to do with the gravitational pull, as distance from earth decreases, so does the gravitational pull. meaning, the size of the force, the pull, and the weight would all change. mass stays the same (in this sense. if you gain weight on earth you will gain mass as well, but if you leave earth your weight will lessen and your mass will stay the same.)

hope this helped.

5 0
4 years ago
Read 2 more answers
A 1542 kg car has a speed of 13.4 m/s when
My name is Ann [436]

provide formulas to help you more

3 0
3 years ago
What do radio waves and microwaves have in common?
Rom4ik [11]

All the options are wrong is there another?

6 0
4 years ago
Read 2 more answers
Other questions:
  • What is the acceleration due to gravity at an altitude of 116?
    7·1 answer
  • How many electrons can the first shell hold
    14·1 answer
  • PLEASE HELP!!!!! Will give brainliest to best answer
    5·1 answer
  • What are the cells in a cell phone system
    8·1 answer
  • A woman makes 50% more than her husband. Together they make $2500 each month? How much does the woman earn each month?a) $950b)
    6·1 answer
  • Ex 10: My dog runs at 6 m/s for 18 meters. How long did she run for?
    14·1 answer
  • Hydroelectric power uses: wind water coal the sun
    13·1 answer
  • Complete the route of the blood from where it entered the heart and leaves the heart
    12·1 answer
  • A level of
    7·1 answer
  • The James Webb Space Telescope is positioned around 1.5 million kilometres from the Earth on the side facing away from the Sun.
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!