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3 years ago

When the distance between two charges is halved, the electrical force between them?

1 answer:

3 0

If the distance between two charges is halved, the electrical force between them increases by a factor 4.

In fact, the magnitude of the electric force between two charges is given by:
$F= k \frac{q_1 q_2}{r^2}$
where
k is the Coulomb's constant
q1 and q2 are the two charges
r is the separation between the two charges

We see that the magnitude of the force F is inversely proportional to the square of the distance r. Therefore, if the radius is halved:
$r'= \frac{r}{2}$
the magnitude of the force changes as follows:
$F'=k \frac{q_1 q_2}{r'^2}=k \frac{q_1 q_2}{( \frac{r}{2})^2 }=k \frac{q_1 q_2}{ \frac{r^2}{4} } =4k \frac{q_1 q_2}{r^2}=4 F$
so, the force increases by a factor 4.

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According to Newton’s law of universal gravitation, which statements are true?

andreyandreev [35.5K]

Before we solve this, we should know this fact:

According to Newton's Law of Gravitation, the force between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. The force acts along the line joining the centres of the two objects. It can be shown by this:

$F ∝ \frac{Mm}{ {d}^{2} }$

Now, let us check all the options.

A. As we move to higher altitudes, the force of gravity on us decreases.

This statement is true.

The force of gravity is inversely proportional to the square of distance from the centre of the earth. If, we go up the surface of the earth, the distance from the centre of the earth increases and hence the value of force of gravity decrease. So, force of gravity decreases with altitude.

B. As we move to higher altitudes, the force of gravity on us increases.

This statement is false.

We have already got the result in option A. that the force of gravity decreases with altitude. It never increases with altitude.

C. As we gain mass, the force of gravity on us decreases.

This statement is false.

The force of gravity is directly proportional to the product of the masses. So, if increase our mass, then the force of gravity will also increase and if we decrease our mass, then the force of gravity decreases.

D. As we gain mass, the force of gravity on us increases.

This statement is true.

As mentioned earlier in option C., the force of gravity is directly proportional to the product of the masses of the earth and another object. So, as we gain mass, the force of gravity on us increases.

E. As we move faster, the force of gravity on us increases.

This statement is true.

Here, we have to consider a different formula. According to Newton's Second Law,

F = ma, where F is the force, m is the mass and a is the acceleration.

In other words,

F ∝ a, i.e., force is directly proportional to acceleration.

We know, acceleration is the rate of change of velocity of an body within a time period.

So, if speed is increased, then acceleration will also be greater, which results in the increase of force. So, as we move faster, the force of gravity on us increases.

Answers:

A: As we move to higher altitudes, the force of gravity on us decreases.

D: As we gain mass, the force of gravity on us increases.

E: As we move faster, the force of gravity on us increases.

Hope you could understand.

If you have any query, feel free to ask.

7 0

2 years ago

A hiker walks 20.51 m at 33.16 degrees. What is the Y component of his displacement?

Serhud [2]

Answer:

The y component of his displacement is 11.22 meters

Explanation:

Components of the displacement

The displacement is a vector because it has a magnitude and a direction. Let's suppose a displacement has a magnitude r and a direction θ, measured with respect to the positive x-direction. The horizontal component of the displacement is calculated by:

$x=r\cos\theta$

The vertical component is calculated by:

$y=r\sin\theta$

The hiker has a displacement with magnitude r = 20.51 m at an angle of 33.16 degrees. Substituting in the above equation:

$y=20.51\sin(33.16^\circ)$

$y=11.22\ m$

The y component of his displacement is 11.22 meters

7 0

3 years ago

A classroom has 24 fluorescent bulbs, each of which is 32 W. how much energy does it take to light the room for a minute?(unit=J

cestrela7 [59]

Answer:

Energy= 46.08KJ

Explanation:

Given that the power needed to light each bulb is 32W

We know that Power = $\frac{energy}{time}$

The energy needed to light one bulb= $power*time$

Given time = 1minute = 60 seconds

Energy = $32W*60sec$ =1920J

Therefore energy needed to light one bulb is 1920J

The energy needed to light 24 bulbs = $1920*24$ =46080J=46.08KJ

3 0

3 years ago

Read 2 more answers

A piston of volume 0.1 m3 contains two moles of a monatomic ideal gas at 300K. If it undergoes an isothermal process and expands

seropon [69]

Answer:

the work is done by the gas on the environment -is W= - 3534.94 J (since the initial pressure is lower than the atmospheric pressure , it needs external work to expand)

Explanation:

assuming ideal gas behaviour of the gas , the equation for ideal gas is

P*V=n*R*T

where

P = absolute pressure

V= volume

T= absolute temperature

n= number of moles of gas

R= ideal gas constant = 8.314 J/mol K

P=n*R*T/V

the work that is done by the gas is calculated through

W=∫pdV= ∫ (n*R*T/V) dV

for an isothermal process T=constant and since the piston is closed vessel also n=constant during the process then denoting 1 and 2 for initial and final state respectively:

W=∫pdV= ∫ (n*R*T/V) dV = n*R*T ∫(1/V) dV = n*R*T * ln (V₂/V₁)

since

P₁=n*R*T/V₁

P₂=n*R*T/V₂

dividing both equations

V₂/V₁ = P₁/P₂

W= n*R*T * ln (V₂/V₁) = n*R*T * ln (P₁/P₂ )

replacing values

P₁=n*R*T/V₁ = 2 moles* 8.314 J/mol K* 300K / 0.1 m3= 49884 Pa

since P₂ = 1 atm = 101325 Pa

W= n*R*T * ln (P₁/P₂ ) = 2 mol * 8.314 J/mol K * 300K * (49884 Pa/101325 Pa) = -3534.94 J

5 0

3 years ago

What is the period of a wave if 3 waves passes every 6 seconds

Naya [18.7K]

Time period = time/no. of waves = 6/3 = 2s

4 0

3 years ago