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KATRIN_1 [288]
4 years ago
14

The diameter of a 12-gauge copper wire is 0.081 in. The maximum safe current it can 17) carry (in order to prevent fire danger i

n building construction) is 20 A. At this current, what is the drift velocity of the electrons?
Physics
1 answer:
Soloha48 [4]4 years ago
6 0

Answer:

0.44m/s

Explanation:

drift velocity=I/nAq

diameter 12 gauge

wire=0.081inches=0.081*2.5=0.2025cm radius=0.10125cm area=pi*R^2 =20/8.5*10^22*3.14*0.10125^2*10^-4*1.6*10^-19*

V = 0.44m/s

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When a balloon is rubbed with human hair, the balloon acquires an excess static charge. This implies that some materials can give up electrons more readily than others.

Answer: Option C

<u>Explanation:</u>

We know that charges can neither be created nor be destroyed by law of conservation of charges. So when we rub two objects, it is natural to have a transfer of charges. But the charges which get transferred may be negligible in most of the cases leading to no significant observations.

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1.60 kg frictionless block is attached to an ideal spring with force constant 315 N/m . Initially the spring is neither stretche
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Answer:

(a). The amplitude of the motion is 0.926 m.

(b). The block’s maximum acceleration is 182.31 m/s².

(c). The maximum force the spring exerts on the block is 291.69 N.

Explanation:

Given that,

Mass of block =1.60 kg

Force constant = 315 N/m

Speed = 13.0 m/s

(a). We need to calculate the amplitude of the motion

Using conservation of energy

\dfrac{1}{2}mv^2=\dfrac{1}{2}kA^2

A^2=\dfrac{mv^2}{k}

Put the value into the relation

A^2=\dfrac{1.60\times13.0^2}{315}

A=\sqrt{0.858}

A=0.926\ m

(b). We need to calculate the block’s maximum acceleration

Using formula of acceleration

a=A\omega^2

a=A\times\dfrac{k}{m}

Put the value into the formula

a=0.926\times\dfrac{315}{1.60}

a=182.31\ m/s^2

(c). We need to calculate the maximum force the spring exerts on the block

Using formula of force

F=ma

Put the value into the formula

F= 1.60\times182.31

F=291.69\ N

Hence, (a). The amplitude of the motion is 0.926 m.

(b). The block’s maximum acceleration is 182.31 m/s².

(c). The maximum force the spring exerts on the block is 291.69 N.

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