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4 years ago

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6) Two small balls, A and B, attract each other gravitationally with a force of magnitude F. If we now double both masses and th

e separation of the balls, what will now be the magnitude of the attractive force on each one?

1 answer:

Butoxors [25]4 years ago

4 0

Answer:

The magnitude of the attractive force is unchanged

Explanation:

Newton's Law of Universal Gravitation: The law states that the force of attraction between two mass m₁ and m₂ is directly proportional to the product of the masses and inversely proportional to the square of the distance (d) between them.

Mathematically it is represented as

F = Gm₁m₂/d₂ ............................. Equation 1

Where G = universal Constant, F = Force of attraction or repulsion, m₁ = mass of the first body, m₂ = mass of the second body, d = distance between the masses.

When both masses and the separation of the ball are doubled,

I.e

Fₙ = G(2m₁)(2m₂)/(2d)²

Where Fₙ = the new gravitational force, when both masses and distance are separation are doubled.

Fₙ = 4Gm₁m₂/4d²

Fₙ = Gm₁m₂/d²........................... Equation 2

Comparing Equation 1 and equation 2,

Fₙ = F

Therefore the magnitude of the attractive force between the balls when their masses and separation are doubled remains the same.

I.e The magnitude of the attractive force is unchanged,

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Assume the average value of the vertical component of Earth's magnetic field is 42 μT (downward) in some region that has an area

Oliga [24]

Answer:

The net magnetic flux through the rest of Earth's surface is $5.782\times10^{-2}\ T$

Explanation:

Given that,

Magnetic field = 42 μT

Area $A=3.71\times10^{5}\ km^2$

$A=3.71\times10^{11}\ m^2$

We need to calculate the flux per unit area

$flux\ per\ unit\ area=\dfrac{42\times10^{-6}}{3.71\times10^{11}}$

$flux\ per\ unit\ area=1.132\times10^{-16}\ T/m^2$

We need to calculate the total earth's surface area

$A'=4\pi r^2$

$A'=4\times\pi\times(6.3781\times10^{6})^2$

$A'=5.1120\times10^{14}\ m^2$

We need to calculate the rest of earth's area

$A''=A-A'$

Put the value into the formula

$A''=5.1120\times10^{14}-3.71\times10^{11}$

$A''=5.10829\times10^{14}\ m^2$

We need to calculate the net magnetic flux through the rest of Earth's surface

$B'=5.10829\times10^{14}\times1.132\times10^{-16}$

$B'=5.782\times10^{-2}\ T$

Hence, The net magnetic flux through the rest of Earth's surface is $5.782\times10^{-2}\ T$

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4 years ago

- How long does it take a packet of length 1,000 bytes to propagate over a link of distance 2,500 km, propagation speed 2.5 *10^

Sunny_sXe [5.5K]

Answer:

It will take 0.01 s or 10 ms

Solution:

As per the question:

Length of the packet, L = 1,000 bytes = $1000\times 8 = 8000 bits$

Distance, d = 2500 km = $2.5\times 10^{6}\ m$

Speed of propagation, s = $2.5\times 10^{8}\ m/s$

Transmission rate, R = 2 Mbps

Now,

Propagation time, t can be calculated as:

$t = \frac{d}{s} = \frac{2.5\times 10^{6}}{2.5\times 10^{8}} = 0.01\ s$

t = 10 ms

In general, propagation time, t is given by:

$t = \frac{link\ distance}{Propagation\ speed}$

No, this delay is independent of the length of the packet.
No, this delay is independent of the rate of transmission.

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4 years ago

How can physicists use Newton’s laws of motion and universal gravitation to predict objects’ motion?

koban [17]

Answer:

Newtons law of universal gravitation is the phenomena in which Newton said that every particle in environment will attract other particles in the space.

Explanation:

Newton law of motion and gravitation attract forcefully which is direct proportional to the masses of the particles and inversely proportional to the square of the distances between the centers. Other physicist used Newtons law in research related to motion or to find out the distance between earth and sun. Because this law is about the mass of the objects. When the mas of an object is doubled, the force between the objects get too doubled. When the mass of the two objects is doubled the gravitational pull between these two objects gets doubled. Sir Isaac Newton first gave this law when see fall down an apple from tree.

Formula is : F = G Mm/r2 Here G is gravitational pull that is constant.

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4 years ago

A 31.0 kW electric motor drives a vehicle at an average velocity of 9.20 m/s. Assuming a constant force is applied in the direct

babunello [35]

The magnitude of the applied force is 3369.57 N.

<h3>What is Force?</h3>

Force can be defined as the product of mass and acceleration.

To calculate the magnitude of the applied force, we use the formula below.

Formula:

P = Fv................ Equation 1

Where:

P = Power of the electric motor
F = Applied force
v = Velocity

Make F the subject of the equation

F = P/v................. Equation 2

From the question.

Given:

P = 31 kW = 31000 W
v = 9.2 m/s

Substitute these values into equation 2

F = 31000/9.2
F = 3369.57 N

Hence, the magnitude of the applied force is 3369.57 N.

Learn more about force here: brainly.com/question/12970081

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3 years ago

4. A car with a mass of 2,200 kg can accelerate from 0 to 30 m/s in 6 seconds.

wlad13 [49]

Answer:

The engine generates a force of 11000 Newtons.

Explanation:

To determine the force, first you will need to calculate the acceleration. Acceleration is change in velocity over change in time, so in this case:

$a = \frac{\Delta v}{\Delta t}=\frac{(30-0)\frac{m}{s}}{(6 - 0 )s}=5 \frac{m}{s^2}$

Force is then determined as

$F=ma = 2200kg\cdot 5\frac{m}{s^2}=11000N$

This is the forward force, i.a. one acting in the direction of the car's acceleration. we are told that there is no inefficiency in the engine and transmission, therefore this same force is generated by the engine.

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3 years ago