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2 years ago

If a neutral metal comb is held near an object with a negative charge, the Como become charged by

Physics

2 answers:

stich3 [128]2 years ago

8 0

Answer:

the answer is A. induction.

Explanation:

LenKa [72]2 years ago

4 0

The local charging of the comb is due to induction. The negative object's charge pushes electrons in the comb away from the side close to the charged object causing that part to be positively charged. Note that the comb's net charge is still zero provided it doesn't touch the object. When you move the comb away its electrons will redistribute.

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A waitperson carrying a tray with a platter on it tips the tray at an angle of 12° below the horizontal. If the gravitational fo

Nadusha1986 [10]

Answer:

The answer is 1.0 N

Explanation:

inclination of tray=12^{\circ}

gravitational Force=5 N

Now this gravitational force has two component i.e.

5\sin \theta is parallel to the tray =1.039 N

5\cos \theta is perpendicular to the tray =4.890 N

7 0

2 years ago

What quantities are related by Ohm's law? Check all that apply. voltage conductivity current resistance insulation

Free_Kalibri [48]

Answer: Current, resistance and voltage are the quantities which are related by Ohm's law.

Explanation:

A law which states that electric current is directly proportional to voltage and inversely proportional to resistance is called Ohm's law.

Mathematically, it is represented as follows.

$I = \frac{V}{R}$

where,

I = current

V = voltage

R = resistance

This means that the quantities related by Ohm's law include current, voltage and resistance.

Thus, we can conclude that current, resistance and voltage are the quantities which are related by Ohm's law.

7 0

3 years ago

Read 2 more answers

A ball is dropped from the top of a building.

son4ous [18]

Kinetic energy is greatest at the lowest point of a roller coaster and least at the highest point

8 0

3 years ago

Read 2 more answers

A small plastic bead has been charged to -15nC.

marishachu [46]

Answer:

Explanation:

q = - 15 n C = - 15 x 10^-9 c

(a) d = 0.9 cm = 0.009 m

electric field on proton, E = Kq/d²

E = 9 x 10^9 x 15 x 10^-9 / (0.009)²

E = 1.67 x 10^6 N/C

Force on proton, F = charge on proton x Electric field

F = 1.6 x 10^-19 x 1.67 x 10^6 = 2.67 x 10^-13 N

acceleration of proton, a = F / m

a = (2.67 x 10^-13) / (1.67 x 10^-27)

a = 1.6 x 10^14 m/s²

(B) the direction of acceleration is towards the bead, as the force is attractive.

electric field on proton, E = Kq/d²

E = 9 x 10^9 x 15 x 10^-9 / (0.009)²

E = 1.67 x 10^6 N/C

Force on electron, F = charge on electron x Electric field

F = 1.6 x 10^-19 x 1.67 x 10^6 = 2.67 x 10^-13 N

acceleration of proton, a = F / m

a = (2.67 x 10^-13) / (9.1 x 10^-31)

a = 3 x 10^17 m/s²

(D) the direction of acceleration is away from the bead, as the force is repulsive.

8 0

3 years ago

could anyone determinate the period knowing that it performs 4000 vibrations in 0.5 minutes, Sorry my english is bad

dexar [7]

Answer:

f = 4000 / 30 sec = 133.3 vibrations/sec

P = 1 / f = .0075 sec period of 1 vibration

4 0

2 years ago