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777dan777 [17]
3 years ago
14

A wire is 1.1 m long and 0.91 mm in cross-sectional area. It carries a current of 4.4 A when a 1.8 V potential difference is app

lied between its ends. Calculate the conductivity of the material of which this wire is made. Number Units The number of significant digits is set to 2; the tolerance is +/-2%
Physics
1 answer:
Vlad [161]3 years ago
3 0

Answer:

The conductivity of the material from which the conductor is made is 2.95 x10⁶  [S. m⁻¹]  .

Explanation:

L= 1.1m

S= 0.91 mm²= 9.1 x10⁻⁷ m²

I= 4.4 A

V= 1.8V

\frac{V}{I} =R\\\\\frac{1.8V}{4.4A} =R= 9/22 ohms

R=9/22Ω

R=\frac{L}{σ . S} \\\\Conductivity=sigma=\frac{L}{R.S} \\\\Conductivity= \frac{1.1m}{9/22 ohms . 9.1 E10 (-7) square meters}

Conductivity=2.95 x10⁶  [S. m⁻¹]

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Define resistance and describe what would happen to a light bulb if the voltage increased but the resistance stayed the same. (h
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6 0
1 year ago
A 645 g block is released from rest at height h0 above a vertical spring with spring constant k = 530 N/m and negligible mass. T
Nina [5.8K]

Answer:

a)5.88J

b)-5.88J

c)0.78m

d)0.24m

Explanation:

a) W by the block on spring is given by

W= \frac{1}{2}kx² = \frac{1}{2}(530)(0.149)² =  5.88 J

b)  Workdone by the spring = - Workdone by the block = -5.88J

c) Taking x = 0 at the contact point we have U top = U bottom

So, mgh_o = \frac{1}{2}kx² - mgx

And, h_o= ( \frac{1}{2}kx² - mgx )/(mg) = [\frac{1}{2} (530)(0.149^2)-(0.645)(9.8)(0.149)]/(0.645x9.8)    

   h_o=   0.78m            

d) Now, if the initial initial height of block is 3h_o

h_o = 3 x 0.78 = 2.34m

then, \frac{1}{2}kx² - mgx - mgh_o =0

 

\frac{1}{2}(530)x²  - [(0.645)(9.8)x] - [(0.645)(9.8)(2.34) = 0

265x² - 6.321x - 14.8 = 0  

a=265

b=-6.321

c=-14.8

By using quadratic eq. formula, we'll have the roots

x= 0.24 or x=-0.225

Considering only positive root:

x= 0.24m (maximum  compression of the spring)

4 0
3 years ago
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