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Gre4nikov [31]
4 years ago
7

Two massive, positively charged particles are initially held a fixed distance apart. When they are moved farther apart, the magn

itude of their mutual gravitational force changes by a factor of n. Which of the following indicates the factor by which the magnitude of their mutual electrostatic force changes?
a. 1/n2
b. 1/n
c. n
d. n2
Physics
1 answer:
Aneli [31]4 years ago
6 0

c. n

The magnitude of the gravitational force is given by Newton's law of universal gravitation:

F_g=-G\frac{m_1m_2}{d^2}

Here G is the gravitational constant, m_1, m_2 are the masses of the particles and d is the distance between them.

The magnitude of the electrostatic force is given by Coulomb's law:

F_e=k\frac{q_1q_2}{d^2}

Here k is the Coulomb constant, q_1, q_2 are the chargues of the particles and d is the distance between them.

As can be seen, both forces are inversely proportional to the square of the distance. Thus, the factor is the same in both cases.

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sergey [27]

Answer:0.502kg

Explanation:

F4om the relation

Power x time = mass x latent heat of vapourization

P.t=ML

1260 * 15 *60 = M * 22.6 * 10^5

M= 1134000/(22.6 *10^5)

M=0.502kg=502g

3 0
3 years ago
A motorcycle is traveling along a highway at 26 m/s. How far does the motorcycle travel in 15 s?​
leva [86]
26 m/s for fifteen seconds. distance = rate times time, so distance = 26 m/s * 15 seconds. this gives you distance = 390 meters.
8 0
3 years ago
A thin ring of radius 73 cm carries a positive charge of 610 nC uniformly distributed over it. A point charge q is placed at the
kow [346]

Answer:

q = - 93.334 nC

Explanation:

GIVEN DATA:

Radius of ring  73 cm

charge on ring 610 nC

ELECTRIC FIELD p FROM CENTRE IS AT 70 CM

E  =  2000 N/C

Electric field due tor ring is guiven as

E = \frac{KQx}{[x^2+ R^2]^{3/2}}

E = \frac{9\time 10^9 \times 610\times 10^[-9} 0.70}{(0.70^2 + 0.73^2)^{3/2}}

E1 = 3714.672 N/C

electric field due to point charge q

E  =\frac[kq}{x^2}

E = \frac{9\times 10^9 \times q}{0.70^2}

E2 = 1.837\times 10^{10}\times q

now the eelctric charge at point P is

E = E1 + E22000 =  3714.672 + 1.837\times 10[10} \times q

solving for q

q = - 93.334 nC

7 0
4 years ago
A mechanic pushes a 2.30 ✕ 103-kg car from rest to a speed of v, doing 4,800 J of work in the process. During this time, the car
Illusion [34]

The horizontal force applied is  160 N while the velocity is  2.03 m/s.

<h3>What is the speed of the car?</h3>

The work done by the car is obtained as the product of the force and the distance;

W = F x

F = ?

x = 30.0 m

W = 4,800 J

F = 4,800 J/30.0 m

F = 160 N

But F = ma

a = F/m

a = 160 N/2.30 ✕ 10^3-kg

a= 0.069 m/s

Now;

v^2 = u^2 + 2as

u = 0/ms because the car started from rest

v = √2as

v = √2 * 0.069 * 30

v = 2.03 m/s

Learn more about force and work:brainly.com/question/758238

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5 0
2 years ago
Which of the following objects is accelerating?
jenyasd209 [6]

Answer:

A

Explanation:

Constant speed (without change in direction) is not accelerating. If you are slowing down, speeding up, or changing direction, you are accelerating

4 0
3 years ago
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