Samuel throws a ball into the air 20 meters. It reaches the top of its flight at 2

Two charges, each 9 µC, are on the x axis, one at the origin and the other at x = 8 m. Find the electric field on the x axis at

bearhunter [10]

a) Electric field at x = -2 m: 21,060 N/C to the left

b) Electric field at x = 2 m: 18,000 N/C to the right

c) Electric field at x = 6 m: 18,000 N/C to the left

d) Electric field at x = 10 m: 21,060 N/C to the right

e) Electric field is zero at x = 4 m

Explanation:

a)

The electric field produced by a single-point charge is given by

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

where:

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

q is the magnitude of the charge

r is the distance from the charge

Here we have two charges of

$q=9\mu C = 9\cdot 10^{-6} C$

each. Therefore, the net electric field at any point in the space will be given by the vector sum of the two electric fields. The two charges are both positive, so the electric field points outward of the charge.

We call the charge at x = 0 as $q_0$ , and the charge at x = 8 m as $q_8$ .

For a point located at x = -2 m, both the fields $E_0$ and $E_8$ produced by the two charges point to the left, so the net field is the sum of the two fields in the negative direction:

$E=-\frac{kq_0}{(0-x)^2}-\frac{kq_8}{(8-x)^2}=-kq(\frac{1}{(-2)^2}+\frac{1}{(8-(-2))^2})=-21060 N/C$

b)

In this case, we are analyzing a point located at

x = 2 m

The field produced by the charge at x = 0 here points to the right, while the field produced by the charge at x = 8 m here points to the left. Therefore, the net field is given by the difference between the two fields, so:

$E=\frac{kq_0}{(0-x)^2}-\frac{kq_8}{(8-x)^2}=kq(\frac{1}{(2)^2}-\frac{1}{(8-2)^2})=18000 N/C$

And since the sign is positive, the direction is to the right.

c)

In this case, we are considering a point located at

x = 6 m

The field produced by the charge at x = 0 here points to the right again, while the field produced by the charge at x = 8 m here points to the left. Therefore, the net field is given by the difference between the two fields, as before; so:

$E=\frac{kq_0}{(0-x)^2}-\frac{kq_8}{(8-x)^2}=kq(\frac{1}{(6)^2}-\frac{1}{(8-6)^2})=-18000 N/C$

And the negative sign indicates that the electric field in this case is towards the left.

d)

In this case, we are considering a point located at

x = 10 m

This point is located to the right of both charges: therefore, the field produced by the charge at x = 0 here points to the right, and the field produced by the charge at x = 8 m here points to the right as well. Therefore, the net field is given by the sum of the two fields:

$E=\frac{kq_0}{(0-x)^2}+\frac{kq_8}{(8-x)^2}=kq(\frac{1}{(10)^2}+\frac{1}{(8-(10))^2})=21060 N/C$

And the positive sign means the field is to the right.

e)

We want to find the point with coordinate x such that the electric field at that location is zero. This point must be in between x = 0 and x = 8, because that is the only region where the two fields have opposite directions. Therefore, te net field must be

$E=\frac{kq_0}{(0-x)^2}-\frac{kq_8}{(8-x)^2}=kq(\frac{1}{(-x)^2}-\frac{1}{(8-x)^2})=0$

This means that we have to solve the equation

$\frac{1}{x^2}-\frac{1}{(8-x)^2}=0$

Re-arranging it,

$\frac{1}{x^2}-\frac{1}{(8-x)^2}=0\\\frac{(8-x)^2-x^2}{x^2(8-x)^2}=0$

So

$(8-x)^2-x^2=0\\64+x^2-16x-x^2=0\\64-16x=0\\64=16x\\x=4 m$

So, the electric field is zero at x = 4 m, exactly halfway between the two charges (which is reasonable, because the two charges have same magnitude)

Learn more about electric fields:

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3 years ago

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A massless string connects a 1.00 kg mass to a 3.00 kg cart which is resting on a frictionless horizontal surface. The mass hang

Romashka-Z-Leto [24]

Both masses will have the same acceleration. The cart accelerates to the right with a magnitude of 4.9 m/ $s^{2}$ . The correct answer is 4.90 m/ $s^{2}$

Given that a massless string connects a 1.00 kg mass to a 3.00 kg cart which is resting on a frictionless horizontal surface.

Let M = 1kg and m = 3 kg

Since the horizontal surface is frictionless, the tension in the string will be the same. when the mass is hanged over a frictionless pulley, the tension will also be the same.

When the mass is released, the cart accelerates to the right can be calculated from Newton' second law of motion. That is,

M( g + a) = m(g - a)

1(9.8 + a) = 3( 9.8 - a)

9.8 +a = 29.4 - 3a

collect the like terms

4a = 19.6

a = 19.6/4

a = 4.9 m/ $s^{2}$

Therefore, the cart accelerates to the right with a magnitude of 4.9 m/ $s^{2}$ . The correct answer is 4.90 m/ $s^{2}$

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3 years ago

The term that describes the direction closest to the point of origin is:

andreyandreev [35.5K]

The term that describes the direction closest to the point of origin is Proximal. Dorsal is the directional term for the movement toward the back of the body. Cephalic is the term that describes the movement towards the top of the body. Ventral on the other hand describes the movement toward the front of the body.

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3 years ago

A car (mass 1200 kg, speed 100 km/h) and a truck (mass 2800 kg, speed 50 km/h) are moving in the same direction along a highway.

Sloan [31]

Answer:

Speed of the wreck after the collision is 65 km/h

Explanation:

When a car hits truck and sticks together, the collision would be totally inelastic. Since the both the vehicles locked together, they have the same final velocity.

Mass of car $m_{1}=1200 kg$

Mass of truck $m_{2}=2800 kg$

Initial speed of the car $u_{1}=100 km /h$

Initial speed of the truck $u_{2}=50 km /h$

The final velocity of the wreck will be

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}$

since final speed are same, $v_{1}=v_{2}=v$

$m_{1}u_{1}+m_{2}u_{2}=(m_{1}+m_{2})v$

$1200\times 100 +2800 \times 50 =(1200+2800)v\\v=\frac{260000}{4000} \\v=65 km/h$

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3 years ago

Explain how Newton’s second law of motion affects a game of Tug of War

notka56 [123]

Answer: Newton's second law of motion is F = ma or force is equal to the mass the reason this applies to Newton's third law is because When the game starts, both the sides are pulling the rope and neither side is moving. The force on the rope is the same on each side This is Newton's second law of motion which equates force as mass times acceleration.

Explanation:

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3 years ago

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