Ask question

Ask question

All categories

3 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?

2 answers:

ipn [44]3 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]3 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.

You might be interested in

Find the are of a parallelogram that has a base of 7 1/2 and a height of 5 2/7

Amiraneli [1.4K]

Use the formula for the area of a parallelogram:

$\sf A=bh$

Plug in what we know:

$\sf A=(7\dfrac{1}{2})(5\dfrac{2}{7})$

Convert both of them into improper fractions. We do this by multiplying the denominator to the whole number, adding it to the numerator, which becomes our new numerator, and we keep the denominator the same:

$\sf A=(\dfrac{15}{2})(\dfrac{37}{7})$

Now multiply the numerators and denominators together:

$\sf A=\dfrac{15}{2}\cdot\dfrac{37}{7}\rightarrow\dfrac{15\cdot 37}{2\cdot 7}\rightarrow\dfrac{555}{14}$

Convert it back into a mixed number. 14 goes into 555 thirty-nine times. 14 * 39 = 546. 555 - 546 = 9. So we have 39 wholes and 9 left over, or:

$\sf 39\dfrac{9}{14}$

So this is the area of the parallelogram.

6 0

3 years ago

Suppose that each child born is equally likely to be a boy or a girl. Consider a family with exactly three children. Let BBG ind

Gemiola [76]

Answer:

(a)

$S = \{GGG, GGB, GBG, GBB, BBG, BGB, BGG, BBB\}$

(b)

i.

$1\ girl = \{GBB, BBG, BGB\}$

$P(1\ girl) = 0.375$

ii.

$Atleast\ 2 \ girls = \{GGG, GGB, GBG, BGG\}$

$P(Atleast\ 2 \ girls) = 0.5$

iii.

$No\ girl = \{BBB\}$

$P(No\ girl) = 0.125$

Step-by-step explanation:

Given

$Children = 3$

$B = Boys$

$G = Girls$

Solving (a): List all possible elements using set-roster notation.

The possible elements are:

$S = \{GGG, GGB, GBG, GBB, BBG, BGB, BGG, BBB\}$

And the number of elements are:

$n(S) = 8$

Solving (bi) Exactly 1 girl

From the list of possible elements, we have:

$1\ girl = \{GBB, BBG, BGB\}$

And the number of the list is;

$n(1\ girl) = 3$

The probability is calculated as;

$P(1\ girl) = \frac{n(1\ girl)}{n(S)}$

$P(1\ girl) = \frac{3}{8}$

$P(1\ girl) = 0.375$

Solving (bi) At least 2 are girls

From the list of possible elements, we have:

$Atleast\ 2 \ girls = \{GGG, GGB, GBG, BGG\}$

And the number of the list is;

$n(Atleast\ 2 \ girls) = 4$

The probability is calculated as;

$P(Atleast\ 2 \ girls) = \frac{n(Atleast\ 2 \ girls)}{n(S)}$

$P(Atleast\ 2 \ girls) = \frac{4}{8}$

$P(Atleast\ 2 \ girls) = 0.5$

Solving (biii) No girl

From the list of possible elements, we have:

$No\ girl = \{BBB\}$

And the number of the list is;

$n(No\ girl) = 1$

The probability is calculated as;

$P(No\ girl) = \frac{n(No\ girl)}{n(S)}$

$P(No\ girl) = \frac{1}{8}$

$P(No\ girl) = 0.125$

7 0

3 years ago

3. Would it be more efficient to use an inequality vo

myrzilka [38]

Answer:

Inequality

Step-by-step explanation:

There are an infinite number of quantities less than 6 (5, 4, 3, 2, 1, 0, -1, -2, -3...) so it is much easier to write an inequality

4 0

3 years ago

Help meeee pleaseeeeeeeee

QveST [7]

Answer:

y54678

Step-by-step explanation:

vggndndhhcbcbfhfnnf

8 0

3 years ago

The data given for a football kicked from the field follows a quadratic model, h(x) = - 3x2 + 18x + 7. Where h(t) is the height

OLga [1]

Answer:

C) 34

Step-by-step explanation:

use -b/2a to find the x value of the vertex (3)

then plug 3 into equation

- 3(3)² + 18(3) + 7

the answer is 34

4 0

3 years ago