I did all the word but my answers are not matching my books?(adsbygoogle = window.adsbygoogle || []).push({});

1) Can a 2.5 mm-diameter copper wire have the same resistence as a tungsten wire of the same length? Give numerical detials.Answer, yes, 4.6 mm

Here is what I did:

( p = resistivity )

R = p(L/A) so I set R(Cu) = R(tug) and since L is the same it cancels out.

Area = 1/2pie2^2 ---> Cu area = (.5)(3.14)(1.25mm) ---> A = .393 mm

So p(A) = p(A) ---> (1.68 x 10^-8)(.393 mm) = (5.6 x 10^-8)(A)

A = .1179 mm

therefore diameter would be... .1179 mm = 2(pie)r^2

r = .2729 mm which mean diameter =.546 mm... so what did I do incorrect?

2) A small immersion heater can be used in a car to heat a cup of water. If the heater can heat 120 mL of water from 25C to 95 C in 8.0 minutes, (a) How much current does it draw from the car's 12 V battery, (b) What is the resistence?Assume the manufacturer's clain of 60% effciency.Answers: 10 A, 1.2 Olhms

What I did...

Used specific heat equation ---> Q = mcT where I changed 120 mL of water into .120 kg ---> Q = (.120 kg)(4185 J/kg C)(95C-25C)

Q = 35154 J

Then I used the equation PE = 1/2Q(where Q is the charger)V

35151 J = (.5)(Q)(12 V)

Q = 5859 C

Finally I used I(current) = Q/T(time) ---> 5859C/480s

I = 12 A

And for Resistence I used V/I = R ---> 12 V/12 A =1 Olhm

I am off my two units for some odd reason in my current and off by .2 in my resistence. Did I make a mistake?

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# Electric Current problem.

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