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vovangra [49]
3 years ago
11

A wire 3.22 m long and 7.32 mm in diameter has a resistance of 11.9 mΩ. A potential difference of 33.7 V is applied between the

ends. (a) What is the current in amperes in the wire? (b) What is the magnitude of the current density? (c) Calculate the resistivity of the material of which the wire is made.
Physics
1 answer:
Scorpion4ik [409]3 years ago
4 0

Answer:

(a) Current is 2831.93 A

(b) 8.40A/m^2

(c) \rho =15.52\times 10^{-9}ohm-m

Explanation:

Length of wire l = 3.22 m

Diameter of wire d = 7.32 mm = 0.00732 m

Cross sectional area of wire

A=\pi r^2=3.14\times 0.00366^2=4.20\times 10^{-5}m^2

Resistance R=11.9mohm=11.9\times 10^{-3}ohm

Potential difference V = 33.7 volt

(A) current is equal to

i=\frac{V}{R}=\frac{33.7}{11.9\times 10^{-3}}=2831.93A

(B) Current density is equal to

J=\frac{i}{A}

J=\frac{2831.93}{4.20\times 10^{-5}}=8.40A/m^2

(c) Resistance is equal to

R=\frac{\rho l}{A}

11.9\times 10^{-3}=\frac{\rho \times 3.22}{4.20\times 10^{-5}}

\rho =15.52\times 10^{-9}ohm-m

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The question is incomplete. Here is the complete question.

Three crtaes with various contents are pulled by a force Fpull=3615N across a horizontal, frictionless roller-conveyor system.The group pf boxes accelerates at 1.516m/s2 to the right. Between each adjacent pair of boxes is a force meter that measures the magnitude of the tension in the connecting rope. Between the box of mass m1 and the box of mass m2, the force meter reads F12=1387N. Between the box of mass m2 and box of mass m3, the force meter reads F23=2304N. Assume that the ropes and force meters are massless.

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(b) What is the mass of each box?

Answer: (a) Total mass = 2384.5kg;

               (b) m1 = 915kg;

                   m2 = 605kg;

                   m3 = 864.5kg;

Explanation: The image of the boxes is described in the picture below.

(a) The system is moving at a constant acceleration and with a force Fpull. Using Newton's 2nd Law:

F_{pull}=m_{T}.a

m_{T}=\frac{F_{pull}}{a}

m_{T}=\frac{3615}{1.516}

m_{T}=2384.5

Total mass of the system of boxes is 2384.5kg.

(b) For each mass, analyse each box and make them each a free-body diagram.

<u>For </u>m_{1}<u>:</u>

The only force acting On the m_{1} box is force of tension between 1 and 2 and as all the system is moving at a same acceleration.

m_{1} = \frac{F_{12}}{a}

m_{1} = \frac{1387}{1.516}

m_{1} = 915kg

<u>For </u>m_{2}<u>:</u>

There are two forces acting on m_{2}: tension caused by box 1 and tension caused by box 3. Positive referential is to the right (because it's the movement's direction), so force caused by 1 is opposing force caused by 3:

m_{2} = \frac{F_{23}-F_{12}}{a}

m_{2} = \frac{2304-1387}{1.516}

m_{2} = 605kg

<u>For </u>m_{3}<u>:</u>

m_{3} = m_{T} - (m_{1}+m_{2})

m_{3} = 2384.5-1520.0

m_{3} = 864.5kg

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