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Bad White [126]
3 years ago
13

A particle with charge 3.01 µC on the negative x axis and a second particle with charge 6.02 µC on the positive x axis are each

a distance 0.0429 m from the origin. Where should a third particle with charge 9.03 µC be placed so that the magnitude of the electric field at the origin is zero?
Physics
1 answer:
ra1l [238]3 years ago
3 0

Answer:

The third particle should be at 0.0743 m from the origin on the negative x-axis.

Explanation:

Let's assume that the third charge is on the negative x-axis. So we have:

E_{1}+E_{3}-E_{2}=0

We know that the electric field is:

E=k\frac{q}{r^{2}}

Where:

  • k is the Coulomb constant
  • q is the charge
  • r is the distance from the charge to the point

So, we have:

k\frac{q_{1}}{r_{1}^{2}}+k\frac{q_{3}}{r_{3}^{2}}-k\frac{q_{2}}{r_{2}^{2}}=0

Let's solve it for r(3).

\frac{3.01}{0.0429^{2}}+\frac{9.03}{r_{3}^{2}}-\frac{6.02}{0.0429^{2}}=0

r_{3}=0.0743\:  

Therefore, the third particle should be at 0.0743 m from the origin on the negative x-axis.

I hope it helps you!

 

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A 12,000kg. railroad car is traveling at +2m/s when it
ivann1987 [24]

<u>Answer:</u>

The final velocity of the two  railroad cars is 1.09 m/s

<u>Explanation:</u>

Since we are given that the two cars lock together it shows that the collision is inelastic in nature. The final velocity due to inelastic collision is given by  

\mathrm{V}=\frac{V 1 M 1+V 2 M 2}{M 1+M 2}

where

V= Final velocity

M1= mass of the first object in kgs = 12000

M2= mas of the second object in kgs = 10000

V1= initial velocity of the first object in m/s = 2m/s

V2= initial velocity of the second object in m/s = 0 (given at rest)

Substituting the given values in the formula we get

V = 2×12000 + 0x100012000 + 10000= 2400022000= 1.09 m/s  

\mathrm{V}=\frac{2 \times 1200+0 \times 1000}{12000+10000}=\frac{24000}{22000}=1.09 \mathrm{m} / \mathrm{s}

Which is the final velocity of the two  railroad cars

8 0
3 years ago
What does each letter stand for in the formula p=v x i?
Nataly_w [17]
I only know P and V and P is pressure and V is volume 
6 0
3 years ago
A ball is thrown horizontally from the top of a 55 m building and lands 150 m from the base of the building. Ignore air resistan
PtichkaEL [24]

Answer:

a) t =3.349 s

b) V_x,i = 44.8 m/s

c) V_y,f = 32.85 m/s

d)  V = 55.55 m/s

Explanation:

Given:

- Total throw in x direction x(f) = 150 m

- Total distance traveled down y(f) = 55 m

Find:

a) How long is the rock in the air in seconds.  

b) What must have been the initial horizontal component of the velocity, in meters per second?

c) What is the vertical component of the velocity just before the rock hits the ground, in meters per second?

d) What is the magnitude of the velocity of the rock just before it hits the ground, in meters per second?

Solution:

- Use the second equation of motion in y direction:

                                 y(f) = y(0) + V_y,i*t + 0.5*g*t^2

- V_y,i = 0 (horizontal throw)

                                 55 = 0 + 0 + 0.5*(9.81)*t^2

                                 t = sqrt ( 55 * 2 / 9.81 )

                                 t =3.349 s

- Use the second equation of motion in x direction:

                                 x(f) = x(0) + V_x,i*t

                                 150 = 0 + V_x,i*3.349

                                  V_x,i = 150 / 3.349 = 44.8 m/s

- Use the first equation of motion in y direction:

                                 V_y,f = V_y,i + g*t

                                 V_y,f = 0 + 9.81*3.349

                                 V_y,f = 32.85 m/s

- The magnitude of velocity of ball when it hits the ground is:

                                 V^2 = V_y,f^2 + V_x,i^2

                                 V = sqrt (32.85^2 + 44.8^2)

                                 V = 55.55 m/s

5 0
3 years ago
Of all the planets in our solar system, Jupiter has the greatest gravitational strength. If a 1.5 kg pair of running shoes would
Andre45 [30]

Answer:

gₓ = 23.1 m/s²

Explanation:

The weight of an object is on the surface of earth is given by the following formula:

W = mg

where,

W = Weight of the object on surface of earth

m = mass of object

g = acceleration due to gravity on the surface of earth = strength of gravity on the surface of earth

Similarly, the weight of the object on Jupiter will be given as:

W_{x} = mg_{x}

where,

Wₓ = Weight of the object on surface of Jupiter = 34.665 N

m = mass of object = 1.5 kg

gₓ = acceleration due to gravity on the surface of Jupiter = strength of gravity on the surface of Jupiter = ?

Therefore,

34.65 N = (1.5 kg)g_{x}

g_{x} = \frac{34.65 N}{1.5 kg}

<u>gₓ = 23.1 m/s²</u>

7 0
3 years ago
An 4-kg ball experiences a force and accelerates at a rate of 1.5 m/s.
ANTONII [103]

Answer:

6.0 N

Explanation:

The strength of a force is expressed as the magnitude of the force in Newton.

The formula to apply here is :

Force= mass * acceleration

F=ma

Mass, m = 4 kg

Acceleration = 1.5 m/s²

Force= 4 *1.5 = 6.0 N

4 0
3 years ago
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