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Travka [436]
3 years ago
13

The resistivity of gold is 2.44 x 10-8 ohms.m at room temperature. A gold wire that is 0.9 mm in diameter and 14 cm long carries

a current of 940 mA. What is the electric field in the wire?
Answer

0.090 V/m

0.028 V/m

0.046 V/m

0.0090 V/m

0.036 V/m
Physics
1 answer:
harkovskaia [24]3 years ago
6 0

Answer:

E=0.036 V/m

Explanation:

Given that

Resistivity ,ρ=2.44 x 10⁻⁸ ohms.m

d= 0.9 mm

L= 14 cm

I = 940 m A = 0.94 A

We know that electric field E

E= V/L

V= I R

R=ρL/A

So we can say that

E= ρI/A

Now by putting the values

E=\dfrac{ 2.44\times 10^{-8}\times 0.94}{\dfrac{\pi}{4}(0.9\times 10^{-3})^2}

E=0.036 V/m

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4 0
1 year ago
A 6 kilogram block in outer space is moving at -100 m/s (to the left). It suddenly experiences three forces as shown below (in t
Blizzard [7]

Answer:

x = 1474.9 [m]

Explanation:

To solve this problem we must use Newton's second law, which tells us that the sum of forces must be equal to the product of mass by acceleration.

We must understand that when forces are applied on the body, they tend to slow the body down to stop it.

So as the body continues to move to the left, it is slowing down. Therefore we must calculate this deceleration value using Newton's second law. We must perform a sum of forces on the x-axis equal to the product of mass by acceleration. With leftward movement as negative and rightward forces as positive.

ΣF = m*a

10 +12*sin(60)= - 6*a\\a = - 3.39[m/s^{2}]

Now using the following equation of kinematics, we can calculate the distance of the block, before stopping completely. The initial speed must be 100 [m/s].

v_{f}^{2} =v_{o}^{2}-2*a*x

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Vf = final velocity = 0 (the block stops)

Vo = initial velocity = 100 [m/s]

a = - 3.39 [m/s²]

x = displacement [m]

0 = 100^{2}-2*3.39*x\\x=\frac{10000}{2*3.39}\\x=1474.9[m]

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Answer:

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