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Semenov [28]
2 years ago
11

A silver wire 2.6 mm in diameter transfers a charge of 420 Cin 80 min. Silver contains 5.8 x 10^{28} free electrons per cubic me

ter.
What is the current in the wire?
What is the magnitude of the drift velocity of the electronsin the wire?
Physics
1 answer:
never [62]2 years ago
4 0

1) Current in the wire: 0.0875 A

The current in the wire is given by:

I=\frac{Q}{t}

where

Q is the charge passing a given point in the conductor

t is the time elapsed

In this problem, we have

Q = 420 C is the total charge passing through a given point in a time of

t = 80 min = 4800 s

So, the current is

I=\frac{420 C}{4800 s}=0.0875 A

2) Drift velocity of the electrons: 1.78\cdot 10^{-6} m/s

The drift velocity of the electrons in the wire is given by:

u = \frac{I}{nAq}

where

I = 0.0875 A is the current

n=5.8\cdot 10^{28} is the number of free electrons per cubic meter

A is the cross-sectional area

q=1.6\cdot 10^{-19} C is the charge of one electron

The radius of the wire is

r=\frac{d}{2}=\frac{2.6 mm}{2}=1.3 mm=0.0013 m

So the cross-sectional area is

A=\pi r^2=\pi (0.0013 m)^2=5.31\cdot 10^{-6} m^2

So, the drift velocity is

u = \frac{(0.0875 A)}{(5.8\cdot 10^{28})(5.31\cdot 10^{-6})(1.6\cdot 10^{-19}C)}=1.78\cdot 10^{-6} m/s

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PolarNik [594]

Answer:

a. 150 N

Explanation:

Gravitational Force: This is the force that act on a body under gravity.

The gravitational force always attract every object on or near the earth's surface. The earth therefore, exerts an attractive force on every object on or near it.

The S.I unit of gravitational force is Newton(N).

Mathematically, gravitational force of attraction is expressed as

(i) F = GmM/r² ........................ Equation 1 ( when it involves two object of different masses on the earth)

(ii) F = mg ............................... Equation 2 ( when it involves one mass and the gravitational field).

Given: m = 17 kg, g = 8.8 m/s²

Substituting into equation 2,

F = 17(8.8)

F = 149.6 N

F ≈ 150 N.

Thus the gravitational force = 150 N

The correct option is a. 150 N

5 0
2 years ago
A 2 kg stone is tied to a 0.5 m string and swung around a circle at a constant angular velocity of 12 rad/s. the angular momentu
Illusion [34]
Starting from the angular velocity, we can calculate the tangential velocity of the stone:
v=\omega r= (12 rad/s)(0.5 m)= 6 m/s

Then we can calculate the angular momentum of the stone about the center of the circle, given by
L=mvr
where
m is the stone mass
v its tangential velocity
r is the radius of the circle, that corresponds to the length of the string.

Substituting the data of the problem, we find
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3 0
2 years ago
9ma electric current is flowing through a conducting wire , then the number of electron passing through it in 3 min is
Leviafan [203]

Answer:

1.0125 x 10^19

Explanation:

current flowing through conductive wire= 9mA = 9 x 10^ -3 A

charge passing per 3 min

Q = It

= 9 x 10^ -3 x (3 x 60)

= 1.620 C

no of electrons in charge

Q = ne

1.620 = n x 1.6 x 10 ^ -19

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2 years ago
10 kg cart and a 5 kg cart are placed on identical surfaces. The 10 kg cart experiences a net force of 12 N to the left, while t
SashulF [63]
F=ma

For the first (10kg) cart,
12=10a
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For the second (5kg) cart,
8=5a
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Therefore, the lighter (5kg) cart experiences a greater acceleration.
7 0
2 years ago
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hram777 [196]

Answer:

22.7 meters

Explanation:

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In this case, the problem asks to find the distance covered by the ball. This will be the sum of the distances covered by the ball in each part of its motion, therefore:

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(instead, the displacement will be the difference between the final and initial position of the ball, therefore:

d=13.2 m-9.5 m=3.7 m)

4 0
2 years ago
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