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Assoli18 [71]
3 years ago
11

Imagine an isolated positive point charge Q (many times larger than the charge on a single proton). There is a charged particle

A (whose charge is much smaller than charge Q) at a distance from the point charge Q. On which of the following quantities does the magnitude of the electric field created by charge Q at particle A's position depend?a) the amount of the charge on the point charge Qb) the specific location of the charged particle A (while the distance between Q and A is fixed)c) the specific location of the point charge Q (while the distance between Q and A is fixed)d) the type of the charge on the charged particle Ae) the relative orientation between Q and A (while the distance between Q and A is fixed)f) the distance between the point charge Q and the charged particle Ag) the amount of the charge on the charged particle A
Physics
1 answer:
Delvig [45]3 years ago
4 0

Answer:

b) the specific location of the charged particle A

f) the distance between the point charge Q and the charged particle A

Explanation:

For this exercise we write the electric field

      E = k Q / r²

where r is the distance between charge Q and test charge A

Let's examine the different claims

a) do not depend. The charge of the individual is independent of the elective field

b) depends. The position of the particle is different if it has the same distance

c) do not depend. The particle Q must be considered the origin of the coordinate system

d) does not depend. The charge of a particle is independent of the field where it is

e) does not depend. The position of the particle is fixed

f) depends. The field is different for each distance

g) does not depend. The amount of electric charge is independent of the field

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To solve this problem we will apply the laws of Mersenne. Mersenne's laws are laws describing the frequency of oscillation of a stretched string or monochord, useful in musical tuning and musical instrument construction. This law tells us that the velocity in a string is directly proportional to the root of the applied tension, and inversely proportional to the root of the linear density, that is,

v = \sqrt{\frac{T}{\mu}}

Here,

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A rock falls from rest off a cliff and hits the ground in 2 seconds. If there is no air resistance, determine the rock's velocit
mihalych1998 [28]

Answer:

19.6m/s

Explanation:

A Rock falling off a cliff can be modeled as an object starting with  zero velocity moves with constant acceleration for certain period of time, for such motion following equation of motion can be used.

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here in our case v_{0} =0 because object starts off from rest and a = g =9.8m/s^2 is acceleration because of gravity ( Motion under gravity).

and of course t = 2 second.

Now by substituting all this information in equation of motion we get.

v(2s) = 0+9.8m/s^2 *2s = 19.6m/s

that would be the velocity of rock as it would hit the ground.

Note! We have assumed that there is no air resistance.

A rock falling off a cliff can be modeled as an object starting with zero velocity moves with constant acceleration for a certain period of time, for such motion following equation of motion can be used.

here in our case  because object starts off from rest and  is acceleration because of gravity ( Motion under gravity).

and of course t = 2 seconds.

Now by substituting all this information in equation of motion we get.

V = 19.6m/s

that would be the velocity of rock as it would hit the ground.

Note! We have assumed that there is no air resistance.

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