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3 years ago

A positively charged glass rod is bought close to a suspended metal needle. What

Physics

2 answers:

Darina [25.2K]3 years ago

7 0

Answer:

attracted

Explanation:

opposite charges attract each other when the rub against each other

EastWind [94]3 years ago

6 0

Answer:

This depends because the electrostatic force obeys the principle that states that force between both of the particles does not get affected by the charges of the other thus if the needle is getting attracted it possess negative charges( the opposite charge) .And if they repel it means they have the same charges ( positive charges).

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A 10.0-cm-long uniformly charged plastic rod is sealed inside a plastic bag. The net electric flux through the bag is 7.50 × 10

Rina8888 [55]

Answer:

66.375 x 10⁻⁶ C/m

Explanation:

Using Gauss's law which states that the net electric flux (∅) through a closed surface is the ratio of the enclosed charge (Q) to the permittivity (ε₀) of the medium. This can be represented as ;

∅ = Q / ε₀ -----------------(i)

Where;

∅ = 7.5 x 10⁵ Nm²/C

ε₀ = permittivity of free space (which is air, since it is enclosed in a bag) = 8.85 x 10⁻¹² Nm²/C²

Now, let's first get the charge (Q) by substituting the values above into equation (i) as follows;

7.5 x 10⁵ = Q / (8.85 x 10⁻¹²)

Solve for Q;

Q = 7.5 x 10⁵ x 8.85 x 10⁻¹²

Q = 66.375 x 10⁻⁷ C

Now, we can find the linear charge density (L) which is the ratio of the charge(Q) to the length (l) of the rod. i.e

L = Q / l ----------------------(ii)

Where;

Q = 66.375 x 10⁻⁷ C

l = length of the rod = 10.0cm = 0.1m

Substitute these values into equation (ii) as follows;

L = 66.375 x 10⁻⁷C / 0.1m

L = 66.375 x 10⁻⁶ C/m

Therefore, the linear charge density (charge per unit length) on the rod is 66.375 x 10⁻⁶ C/m.

3 0

3 years ago

After hitting the spring, the block is bounced back up the ramp. The maximum compression of the spring is Δx=0.03m, and the spri

Lyrx [107]

Answer:

v = 1.28 m/s

Explanation:

Given that,

Maximum compression of the spring, $\Delta x=0.03\ m$

Spring constant, k = 800 N/m

Mass of the block, m = 0.2 kg

To find,

The velocity of the block when it first reaches a height of 0.1 m above the ground on the ramp.

Solution,

When the block is bounced back up the ramp, the total energy of the system remains conserved. Let v is the velocity of the block such that,

Initial energy = Final energy

$\dfrac{1}{2}kx^2=mgh+\dfrac{1}{2}mv^2$

Substituting all the values in above equation,

$\dfrac{1}{2}\times 800\times 0.03^2=0.2\times 9.8\times 0.1+\dfrac{1}{2}\times 0.2\times v^2$

v = 1.28 m/s

Therefore the velocity of block when it first reaches a height of 0.1 m above the ground on the ramp is 1.28 m/s.

5 0

2 years ago

PLEASE HELP 15 POINTS AND BRAINIEST!!!<br> Reporting fake answers

Zolol [24]

Answer:

(3) The period of the satellite is independent of its mass, an increase in the mass of the satellite will not affect its period around the Earth.

(4) he gravitational force between the Sun and Neptune is 6.75 x 10²⁰ N

Explanation:

(3) The period of a satellite is given as;

$T = 2\pi \sqrt{\frac{r^3}{GM} }$

where;

T is the period of the satellite

M is mass of Earth

r is the radius of the orbit

Thus, the period of the satellite is independent of its mass, an increase in the mass of the satellite will not affect its period around the Earth.

(4)

Given;

mass of the ball, m₁ = 1.99 x 10⁴⁰ kg

mass of Neptune, m₂ = 1.03 x 10²⁶ kg

mass of Sun, m₃ = 1.99 x 10³⁰ kg

distance between the Sun and Neptune, r = 4.5 x 10¹² m

The gravitational force between the Sun and Neptune is calculated as;

$F_g = \frac{Gm_2m_3}{r^2} \\\\F_g = \frac{6.67\times 10^{-11} \times 1.03 \times 10^{26}\times 1.99\times 10^{30}}{(4.5\times 10^{12})^2} \\\\F_g = 6.751 \times 10^{20} \ N$

5 0

3 years ago

A 26-g steel-jacketed bullet is fired with a velocity of 630 m/s toward a steel plate and ricochets along path CD with a velocit

Alecsey [184]

Answer:

F = - 3.53 10⁵ N

Explanation:

This problem must be solved using the relationship between momentum and the amount of movement.

I = F t = Δp

To find the time we use that the average speed in the contact is constant (v = 600m / s), let's use the uniform movement ratio

v = d / t

t = d / v

Reduce SI system

m = 26 g ( 1 kg/1000g) = 26 10⁻³ kg

d = 50 mm ( 1m/ 1000 mm) = 50 10⁻³ m

Let's calculate

t = 50 10⁻³ / 600

t = 8.33 10⁻⁵ s

With this value we use the momentum and momentum relationship

F t = m v - m v₀

As the bullet bounces the speed sign after the crash is negative

F = m (v-vo) / t

F = 26 10⁻³ (-500 - 630) / 8.33 10⁻⁵

F = - 3.53 10⁵ N

The negative sign indicates that the force is exerted against the bullet

5 0

3 years ago

Two marbles, one twice as heavy as the other, are dropped to the ground from the roof of a building. Just before hitting the gro

Andreyy89

Answer:

B. twice as much kinetic energy

Explanation:

Lets take the mass of the first marble =2 m

the mass of the second marble = m

We know that velocity of particle does not depends on their mass that is the velocity of both mass will be same after dropping from the roof.

We know that kinetic energy of a mass is given as

$KE=\dfrac{1}{2}Mv^2$

Kinetic energy for heavier mass

$KE=\dfrac{1}{2}\times 2m\times V^2$

Kinetic energy for light mass

$KE'=\dfrac{1}{2}\times m\times V^2$

KE=2 KE '

Form above two equation we can say that ,the kinetic energy for the heavier mass is twice the lighter mass.

Therefore the answer will be B.

7 0

3 years ago