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3 years ago

Does the distance between two charged objects affect the force of attraction?

Physics

1 answer:

miv72 [106K]3 years ago

5 0

Answer:

Both forces act along the line joining the objects like masses or charges. And both forces are inversely proportional to the square of the distance between the objects, this is known as the inverse-square law.

Explanation:

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A bicycle is traveling north at 5.0 m/s. The mass of the wheel, 2.0 kg, is uniformly distributed along the rim, which has a radi

andreev551 [17]

Answer:2

Explanation:

Given

Velocity of bicycle is 5 m/s towards north

radius of rim $r=20 cm$

mass of rim $m=2 kg$

Angular momentum $\vec{L}=I\cdot \vec{\omega }$

$I=mr^2=2\times 0.2^2=0.08 kg-m^2$

$\omega =\frac{v}{r}=\frac{5}{0.2}=25 rad/s$

$L=0.08\times 25=2kg-m^2/s$

direction of Angular momentum will be towards west

7 0

3 years ago

Two charges that are separated by one meter exert 1-n forces on each other. if the magnitude of each charge is doubled, the forc

Goshia [24]

The electrostatic force between the two charges is
$F=k_E \frac{q_1 q_2}{r^2}$
where q1 and q2 are the magnitudes of the two charges, and r the distance between them.

We can see from the formula that F is proportional to the product between the two charges:
$F \sim q_1 q_2$
so, if the magnitude of each charge is doubled, the new force will get a factor 4:
$F' \sim (2 q_1 )(2 q_2 )=4 q_1 q_2 =4 F$
So, the new force will be 4 times the original force:
$F' = 4 \cdot 1N= 4N$

5 0

3 years ago

A straightforward method of finding the density of an object is to measure its mass and then measure its volume by submerging it

Marina86 [1]

Answer:

Density of rock will be equal to $3.13g/cm^3$

Explanation:

It is given that mass of the rock $m=260gram$

Volume displaced bu rock $V=83cm^3$

We have to find the density of rock

Density is equal to ratio of mass and volume

Therefore density of rock $\rho =\frac{m}{V}$

$\rho =\frac{260}{83}=3.13g/cm^3$

So density of rock will be equal to $3.13g/cm^3$

3 0

3 years ago

A 100-g toy car moves along a curved frictionless track. At first, the car runs along a flat horizontal segment with an initial

Sliva [168]

Answer:

The final velocity of the car is 1.85 m/s

Explanation:

Hi there!

The initial kinetic energy of the toy car can be calculated as follows:

KE = 1/2 · m · v²

Where:

KE = kinetic energy.

m = mass.

v = velocity.

KE = 1/2 · 0.100 kg · (2.66 m/s)² = 0.354 J

The gain in altitude produces a gain in potential energy. This gain in potential energy is equal to the loss in kinetic energy. So let´s calculate the potential energy of the toy car after gaining an altitude of 0.186 m.

PE = m · g · h

Where:

PE = potential energy.

m = mass.

g = acceleration due to gravity.

h = height.

PE = 0.100 kg · 9.8 m/s² · 0.186 m = 0.182 J

The final kinetic energy will be: 0.354 J - 0.182 J = 0.172.

Using the equation of kinetic energy, we can obtain the velocity of the toy car after running up the slope:

KE = 1/2 · m · v²

0.172 J = 1/2 · 0.100 kg · v²

2 · 0.172 J / 0.100 kg = v²

v = 1.85 m/s

The final velocity of the car is 1.85 m/s

3 0

3 years ago

Consider two positively charged particles, one of charge q0 (particle 0) fixed at the origin, and another of charge q1 (particle

charle [14.2K]

Answer:

F12= (kq1q2/r12 squared)*e12

Explanation:

F12→ is the force q 1 makes on q 2 and e12 is the unit vector from q1 toward q2.

8 0

3 years ago