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Burka [1]
4 years ago
13

When a positively charged conductor touches a neutral conductor, the neutral conductor will:

Physics
1 answer:
frutty [35]4 years ago
6 0

Answer:

Lose electrons

Explanation:

When a positively charged conductor touches a neutral conductor, the neutral conductor will lose electrons. Only electrons can move from one conductor to another, so if the neutral conductor ended up with a positive charge it means it lost electrons. The conductor touching and the neutral conductor both end up being charged positively.

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What type of charge is used as the test charge to determine the direction of electric fields?
Dmitriy789 [7]

Answer:

B small positive charge

Explanation:

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3 years ago
What is the change in potential energy of an energetic 72 kg hiker who makes it from the floor of death valley to the top of mt.
storchak [24]
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4 years ago
Two particles each have the same mass but particle #1 has four times the charge of particle #2. Particle #1 is accelerated from
marin [14]

Answer:

 v_2 = 2*v  

Explanation:

Given:

- Mass of both charges = m

- Charge 1 = Q_1

- Speed of particle 1 = v

- Charge 2 = 4*Q_1

- Potential difference p.d = 10 V

Find:

What speed does particle #2 attain?

Solution:

- The force on a charged particle in an electric field is given by:

                                       F = Q*V / r

Where, r is the distance from one end to another.

- The Net force acting on a charge accelerates it according to the Newton's second equation of motion:

                                      F_net = m*a

- Equate the two expressions:

                                      a = Q*V / m*r

- The speed of the particle in an electric field is given by third kinetic equation of motion.

                                      v_f^2 - v_i^2 = 2*a*r

Where, v_f is the final velocity,

            v_i is the initial velocity = 0

                                      v_f^2 - 0 = 2*a*r

Substitute the expression for acceleration in equation of motion:

                                       v_f^2 = 2*(Q*V / m*r)*r

                                       v_f^2 = 2*Q*V / m

                                       v_f = sqrt (2*Q*V / m)

- The velocity of first particle is v:

                                       v = sqrt (20*Q / m)

- The velocity of second particle Q = 4Q

                                       v_2 = sqrt (20*4*Q / m)

                                       v_2 = 2*sqrt (20*Q / m)

                                       v_2 = 2*v  

3 0
3 years ago
Why do raindrops fall with a constant speed during the later stages of their descents?
BartSMP [9]
Firstly they have a acceleration downwards due the force downwards due they gravitational field acting on it's mass.
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Are you wondering that why is the raindrop still moving given that the forces are balanced? If so according to Newton's 1st law an object will keep moving or Remain at rest until a RESULTANT force acts on it.
6 0
4 years ago
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Liula [17]

Answer:D.Refractive Indez

Explanation:

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7 0
3 years ago
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