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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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Answer:

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Explanation:

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7 0
3 years ago
A rifle bullet with mass 8.00 g and initial horizontal velocity 280 m/s strikes and embeds itself in a block with mass 0.992 kg
anyanavicka [17]

Answer:

0.4113772 s

Explanation:

Given the following :

Mass of bullet (m1) = 8g = 0.008kg

Initial horizontal Velocity (u1) = 280m/s

Mass of block (m2) = 0.992kg

Maxumum distance (x) = 15cm = 0.15m

Recall;

Period (T) = 2π√(m/k)

According to the law of conservation of momentum : (inelastic Collison)

m1 * u1 = (m1 + m2) * v

Where v is the final Velocity of the colliding bodies

0.008 * 280 = (0.008 + 0.992) * v

2.24 = 1 * v

v = 2.24m/s

K. E = P. E

K. E = 0.5mv^2

P.E = 0.5kx^2

0.5(0.992 + 0.008)*2.24^2 = 0.5*k*(0.15)^2

0.5*1*5.0176 = 0.5*k*0.0225

2.5088 = 0.01125k

k = 2.5088 / 0.01125

k = 223.00444 N/m

Therefore,

Period (T) = 2π√(m/k)

T = 2π√(0.992+0.008) / 233.0444

T = 2π√0.0042910

T = 2π * 0.0655059

T = 0.4113772 s

6 0
3 years ago
When there is a change of state, such as a solid to liquid or liquid to gas, heat energy can be added without a temperature chan
Romashka-Z-Leto [24]
I’m pretty sure the answer is C. Any change of state or movement requires energy
3 0
3 years ago
How high can a roller coaster be? A roller coaster car is to roll down a frictionless ramp into a loop of a radius r. If the rid
gizmo_the_mogwai [7]

Answer:

h=4r

Explanation:

To solve the problem it is necessary to apply the energy conservation equations for the roller coaster.

The energy conservation equations warn that:

\Delta KE = \Delta PE

Where,

\Delta KE = \frac{1}{2} mv^2 \rightarrow Kinetic Energy

\Delta PE = mgh \rightarrow Potential Energy

Equating,

\frac{1}{2}mv^2 = mgh

Re-arrange for V,

V^2 = 2gh

For balance of forces, according to the announcement, those who are on a roller coaster can withstand up to a maximum of 9g.

Therefore, considering the centripede speed and the speed of the fall, we obtain that,

F_w+F_a = F_t

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The centripetal acceleration is given by the equation

a = \frac{V^2}{r}

Where

V = Tangencial velocity

r = Radius

Then replacing in the equation of Force,

mg + m\frac{V^2}{r} = 9mg

mg + m \frac{(2gh)}{r} = 9mg

1+\frac{2h}{r} = 9

h= \frac{8r}{2}

h= 4r

Therefore the maximum height of the incline if the cars starts from the rest is 4 times the raidus of the inclination

5 0
3 years ago
When you push a box with 20N of force what force does the box apply back on you ?
Elden [556K]
Newton’s 3rd Law: for every action there is an equal but opposite reaction.
Assuming that the box is motionless, we can say the force is -20N
3 0
3 years ago
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