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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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Read 2 more answers
You throw a 20-N rock vertically into the air from ground level. You observe that when it is a height 14.8m above the ground, it
VladimirAG [237]

Answer:

(A) The speed just as it left the ground is 30.25 m/s

(B) The maximum height of the rock is 46.69 m

Explanation:

Given;

weight of rock, w = mg  = 20 N

speed of the rock at 14.8 m, u = 25 m/s

(a) Apply work energy theorem to find its speed just as it left the ground

work = Δ kinetic energy

F x d = ¹/₂mv² - ¹/₂mu²

mg x d = ¹/₂m(v² - u²)

g x d = ¹/₂(v² - u²)

gd = ¹/₂(v² - u²)

2gd = v² - u²

v² = 2gd  + u²

v² = 2(9.8)(14.8) + (25)²

v² = 915.05

v = √915.05

v = 30.25 m/s

B) Use the work-energy theorem to find its maximum height

the initial velocity of the rock = 30.25 m/s

at maximum height, the final velocity = 0

- mg x H = ¹/₂mv² - ¹/₂mu²

- mg x H = ¹/₂m(0) - ¹/₂mu²

- mg x H = - ¹/₂mu²

2g x H = u²

H = u² / 2g

H = (30.25)² / 2(9.8)

H = 46.69 m

4 0
3 years ago
14.A 90 kg quarterback gets tackled by being hit by a 120 kg lineman backwards
quester [9]

The acceleration of the quarterback and the lineman is 5.55m/s² and 4.16m/s² respectively in the same direction.

As, we know, the 120 Kg lineman is moving with a force of 500N.

His net acceleration will be in the same direction as his motion.

It is already known that, If M is the mass of the body and a is the acceleration of the body, then the force F on the body can be calculated by using the formula,

F = Ma.

The weight of the quarterback is 90 Kg. He is being hit by a force of 500N.

So, the acceleration can be calculated using the formula,

500N = 90kg x a

a = 5.55 m/s².

Now, the weight if the lineman is 120kg, the force applied by him is 500N.

So, from the formula, his acceleration A will be,

500N = 120Kg x A

A = 4.16 m/s².

both of them will have acceleration in the same direction,

To know more about Force, visit,

brainly.com/question/25239010

#SPJ9

8 0
1 year ago
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