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lys-0071 [83]
3 years ago
15

Which force is responsible for allowing carbon-14 to become stable? *

Physics
1 answer:
Bas_tet [7]3 years ago
3 0

Answer:weak nuclear

Explanation:

There weak nuclear force is responsible for allowing carbon-14 to become stable

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Come se in so mi tiempo si
6 0
3 years ago
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:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

6 0
3 years ago
Read 2 more answers
Can someone help me fill in the two boxes.
puteri [66]

Answer:

Explanation:

I believe what you're looking for is :

Opposites

Density

Could be wrong but I think so.

7 0
3 years ago
The strength of an electrical field decreases as you move farther from the source.<br> True or False
slamgirl [31]
It is True it decreases



4 0
3 years ago
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The speedometer of your automobile shows that you are proceedig at a steady speed of 80 although your automobile is in accelerat
nika2105 [10]

Answer:

<em>The engine must provide power to compensate friction</em>

Explanation:

<u>Accelerated Motion </u>

Newton's first law states that a body will keep its speed or state of rest while no external net force is applied. Our automobile is running through a road which surface exerts a frictional force on the wheels. If we didn't use fuel, the automobile will eventually stop, because the unbalanced friction force causes deceleration. To compensate that force, we must provide power to the engine. When our speedometer shows a constant speed, the net force is zero, but the car needs to accelerate some, just to compensate the friction force.

The total acceleration keeps being zero, but the engine is actually doing work against friction.

7 0
3 years ago
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