All categories

4 years ago

If a negatively charged rod touches a conductor, the conductor will be charged by what method?

Physics

2 answers:

musickatia [10]4 years ago

6 0

b friction

thanks for the points

mylen [45]4 years ago

5 0

Answer: D. conduction

Explanation:

Charge is transferred from one object together by mainly three methods:

1. Induction: when a charged object is brought close to an uncharged object, the charge is induced.

2. Friction: when two bodies having different electro-negativity are rubbed against each other, both of them get charged. For example, balloon when rubbed with fur.

3. Conduction: when a charged object is touched to an uncharged object, the uncharged object gets charged.

Thus, when a negatively charged rod touches a conductor, the conductor would be charged by conduction.

You might be interested in

An object at rest on a flat, horizontal surface explodes into two fragments, one seven times as massive as the other. The heavie

IgorLugansk [536]

Answer:

The distance the lighter fragments slides is 343 m

Explanation:

Here, we have

Let the mass of the heavier fragments be m₁

Let the distance the heavy object slide be d₁

Let the mass of the lighter fragments be m₂

Let the final velocity of the heavier fragments be v₁

Let the final velocity of the lighter fragments be v₂

Let the distance the light object slide be d₂

The distance traveled by the heavy fragment = 7.00 m

Therefore since m₁ = 7 × m₂ we have

Initial total momentum = final total momentum

Since the initial total momentum = 0 we have

m₁·v₁ + m₂·v₂ = 0 or

7·m₂·v₁ = -m₂·v₂

∴ v₂ = 7·v₁

The net work done by the heavier block is

W $_{net, 7m}$ = μk × m₁ × g × d₁ = 1/2×m₂×v₂²

Also v₁² = 2μk·g·d₁ and

v₂² = 2μk·g·d₂ so that since v₂ = 7·v₁ or v₁ = v₂/7 we get

(v₂/7)² = 2μk·g·d₁

So v₂²/49 = 2μk·g·d₁

or (2μk·g·d₂)/49 = 2μk·g·d₁

∴ d₂/49 = d₁

d₂ = d₁×49 = 7 × 49 = 343 m

The distance the lighter fragments slides = 343 m.

7 0

3 years ago

Robin fired a bullet of mass 100 gm from a gun of mass 5 kg. The bullet leaves the gun with a speed of 400 m/s. After penetratin

Marina CMI [18]

(a) the recoil or backward velocity of the gun is 8 m/s.

(b) the bullet cannot completely penetrate the plank.

The given parameters include;

mass of the bullet, m₁ = 100 g = 0.1 kg

mass of the gun, m₂ = 5 kg

initial velocity of the bullet, u₁ = 400 m/s

thickness of the plank, x = 10 cm

To calculate the following:

(a) the backward velocity of the gun

let the backward velocity of the gun = u₂

Apply the principle of conservation of linear momentum.

m₁u₁ + m₂u₂ = 0

m₂u₂ = -m₁u₁

$u_2 = -\frac{m_1u_1}{m_2} \\\\u_2 = - \frac{0.1 \times 400}{5} \\\\u_2 = -8 \ m/s$

Thus, the recoil or backward velocity of the gun is 8 m/s.

(b) Can the bullet penetrate the plank of the wood completely ?

the bullet traveled 4 cm and lost ¹/₃ of u₁

the remaining distance to completely penetrate the plank = 6 cm

the final velocity of the bullet at 4 cm, v = 400 - ¹/₃ x 400 m/s = 266.67 m/s

the acceleration of the bullet is calculated as;

v² = u₁² + 2as

2as = v² - u²

$a = \frac{v^2 -u_1^2}{2s} \\\\a = \frac{(266.67)^2 -(400)^2}{2\times 0.04} = -1.111 \times 10^6 \ m/s^2$

Finally, determine the distance traveled by the bullet when it comes to a complete stop, that is the final velocity = 0

$v_f^2 = v^2 + 2ad\\\\2ad = v_f^2 - v^2\\\\d = \frac{v_f^2 - v^2}{2a} \\\\d = \frac{(0) - (266.67)^2}{2(-1.111\times 10^6)} \\\\d = 0.032 \ m$

d = 3.2 cm

The total distance traveled by the bullet inside the plank = 4 cm + 3.2 cm = 7.2 cm

Therefore, the bullet cannot completely penetrate the plank.

To learn more about linear momentum visit: brainly.com/question/15869303

6 0

3 years ago

A light ray transfers from more to less dense medium at certain condition , if we repeat this experiment by increasing the angle

evablogger [386]

Answer:

The correct option is;

Still constant

Explanation:

The relative refractive index ₁n₂ between the two medium can be as follows;

$_1n_2 = \dfrac{The \ speed \ of \ light \ in \ medium \ 1}{The \ speed \ of \ light \ in \ medium \ 2} = \dfrac{c_1}{c_2}$

Therefore, given that the speed of light in medium 1 is constant and the speed of light on medium 2 is also constant, the relative refractive index ₁n₂ = c₁/c₂ is always constant.

4 0

3 years ago

Please explain what is couloms law?

svp [43]

Answer:

Coulomb's law states that the electrical force between two charged objects is directly proportional to the product of the quantity of charge on the objects and inversely proportional to the square of the separation distance between the two objects.

Explanation:

4 0

3 years ago

The angular speed of an automobile engine is increased at a constant rate from 1260 rev/min to 3460 rev/min in 9.90 s. (a) What

larisa86 [58]

Answer:

13333.33 rev/min²

Explanation:

Given:

Initial angular speed of the automobile (ω₁) = 1260 rev/min

Final angular speed of the automobile (ω₂) = 3460 rev/min

Time interval for the change in speed (t) = 9.90 s