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DerKrebs [107]
4 years ago
15

A 1.5m wire carries a 7 A current when a potential difference of 54 V is applied. What is the resistance of the wire?

Mathematics
2 answers:
ipn [44]4 years ago
7 0
<h2>Answer:</h2>

The resistance of wire will be 7.71 Ohms

<h2>Step-by-step explanation: </h2>

According to Ohm's Law

V = IR

R= V/ I

Putting the value  

R = 54 / 7

R = 7.71 Ohms

borishaifa [10]4 years ago
6 0

Answer:

R = 7.71 ohms

Step-by-step explanation:

By ohm's law: R= V/I

where,  R = resistance of wire, V = potential difference, I = electric current

Given: V = 54 and I = 7 A

Plug in the above values in the ohm's law, we get

R = 54 V/7 A

R = 7.71 ohms

Hope this will helpful.

Thank you.

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Rly rly need help on this
jeka94

Answer:

D    73º

Step-by-step explanation:

Total angles

a = (sides - 2) * 180

a = (5 - 2) * 180

a = 540

-------------------

Missing angle

x = 540 - (104 + 118 + 96 + 115)

x = 107

a is supplementary to x

m∠A = 180 - 107

m∠A = 73º

3 0
3 years ago
Read 2 more answers
3+5i/-2+3i perform operation
melisa1 [442]

Answer:

\dfrac{9}{13}-\dfrac{19}{13}i

Step-by-step explanation:

Remember i^2=-1

You are given the fraction \dfrac{3+5i}{-2+3i}

First, multiply the numerator and the denominator by -2-3i:

\dfrac{3+5i}{-2+3i}=\dfrac{(3+5i)(-2-3i)}{(-2+3i)(-2-3i)}=\dfrac{(3+5i)(-2-3i)}{(-2)^2-(3i)^2}=\dfrac{(3+5i)(-2-3i)}{4-9i^2}=\dfrac{(3+5i)(-2-3i)}{4+9}

This gives you 13 in denominator, now multiply two complex numbers in numerator:

(3+5i)(-2-3i)=-6-9i-10i-15i^2=-6-19i+15=9-19i

Thus, the initial fraction is

\dfrac{9-19i}{13}=\dfrac{9}{13}-\dfrac{19}{13}i

4 0
3 years ago
0.75,1/2,0.4, and 1/5 from least to greatest
avanturin [10]
0.4, 0.75, 1/2, 1/5
6 0
3 years ago
Read 2 more answers
3 of 8
expeople1 [14]

Answer:

Step-by-step explanation:

i) (4*4)^2-(2(-8))^2=256-(16)^2=256-256=0

ii) (4*16)-2*(-8)^2=64- 2(16)= 64-32=32

4 0
3 years ago
Using the digits 1 to 9, at most one time each, fill in the blanks to make two different pairs of two-digit numbers that have a
agasfer [191]
Do you mean no repeating numbers within the two sets? Because if so, I don't think it's possible.

I started by trying to figure out what the numbers in the tenths place should be. I used subtraction: 71 - ___ = ____. If you try it out, you can't subtract anything with a 6, 7, 8, or 9 in the tenths place because it will leave you with 11 (a repeating digit number), 10 (has a 0), or less (1-digit numbers). Also, a 3 cannot go into the tenths place because when you do 71 - 3_ , your answer will always begin with a 3 (problem because the 3 repeats), or it will contain a 0. 

Therefore, the numbers left for the tenths place are: 1, 2, 4, and 5. 1 and 5 pair up, leaving 4 and 2.
71 - 5_ = 1_   and 71 - 4_ = 2_.

Then, I tried to figure out what numbers go in the ones place. That lead me to realize they act in pairs. The pairs possible in the ones place are 2 and 9, 3 and 8, 4 and 7, 5 and 6. These numbers always go together to result with the final "1" in the "71". Using this information, I looked at the numbers I already used: 1, 2, 4, and 5. Now, looking at the pairs, I eliminated the pairs containing a number already used. This leaves me with only one pair: 3 and 8. Obviously, you need two more pairs to solve the problem, which leads me to my point of saying: This problem is impossible to solve.

I really hope someone can prove me wrong! But this is the solution I have reached for now. :)
4 0
3 years ago
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