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Anestetic [448]
3 years ago
8

Who performed the coin and feather experiment?​

Physics
2 answers:
Pavlova-9 [17]3 years ago
8 0

Answer:

Robert Boyle

Explanation:

Robert Boyle performed the experiment of Galileo

Please give me brainliest please

Leokris [45]3 years ago
4 0

Answer: Robert Boyle

Explanation:

Galileo's experiment and coin and feather experiment performed by Robert Boyle.

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Drag the tiles to the boxes to form correct pairs. Match each hypothesis for how the Moon formed with the statement that best de
Roman55 [17]

Answer:

giant impact theory

Explanation:

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3 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
What will happen to a negatively charged particle placed at rest between points that have electric potentials 0V and 9V? Explain
finlep [7]

Hey there!

a) The electric field is in direction of decreasing value of potential so the electric field here will points towards 0v from 9 v. But the charge on particle is negative so the force will be towards the 9v point. and so the charge will move towards the 9 V point.

b) Electric potential is a location-dependent quantity that expresses the amount of potential energy per unit of charge at a specified location while.the electric potential difference is the difference in electric potential (V) between the final and the initial location when work is done upon a charge to change its potential energy. Electric potential is always a relative quantity because no one can find out absolute potential at a point, so in reality only potential difference exists. We can assume potential at certain point (say at infinity it is zero) only then we are able to define potential at every point.

c) The knowlendge of electric potential helps us in finding the work done on the particle which is used in work energy theorem to find out the chanrge in kinetic energy and other stuff.

Hope this helps!

8 0
3 years ago
I have a problem in the questions
Taya2010 [7]
WHTATATATTSSHSJSD ....... 22
5 0
3 years ago
The series circuit shown below, what would happen if one of the light bulbs and its attached wire segment were removed?
rewona [7]

Answer:

The remaining lights would shine with the same brightness.

Explanation:

6 0
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