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Physics Examples

V = I R

$V = I R$

$V$ $V$	$=$ $=$	$?$ $?$
$I$ $I$	$=$ $=$	$0.55 A$ $0.55 A$
$R$ $R$	$=$ $=$	$2 \cdot 10^{2} Ω$ $2 \cdot 10^{2} Ω$

Step 1

Substitute the given values into

V = I R

$V = I R$ .

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Step 1.1

Substitute

0.55 A

$0.55 A$ for

I

$I$ .

V = 0.55 A R

$V = 0.55 A R$

Step 1.2

Substitute

2 \cdot 10^{2} Ω

$2 \cdot 10^{2} Ω$ for

R

$R$ .

V = 0.55 A (2 \cdot 10^{2} Ω)

$V = 0.55 A (2 \cdot 10^{2} Ω)$

V = 0.550 A (2.0 \cdot 10^{2} Ω)

$V = 0.550 A (2.0 \cdot 10^{2} Ω)$

Step 2

Simplify the right side.

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Step 2.1

Multiply

2.0

$2.0$ by

0.550

$0.550$ .

V = 1.1000 A (10^{2} Ω)

$V = 1.1000 A (10^{2} Ω)$

Step 2.2

Raise

10

$10$ to the power of

2

$2$ .

V = 1.1000 A (100 Ω)

$V = 1.1000 A (100 Ω)$

Step 2.3

Multiply

100

$100$ by

1.1000

$1.1000$ .

V = 110.0000 (A Ω)

$V = 110.0000 (A Ω)$

Step 2.4

Rewrite

A Ω

$A Ω$ as

V

$V$ .

V = 110.0000 V

$V = 110.0000 V$

V = 110.0000 V

$V = 110.0000 V$

Step 3

Round to

2

$2$ significant figures.

V = 1.1 \cdot 10^{2} V

$V = 1.1 \cdot 10^{2} V$

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⎡ ⎢ ⎣ \begin{matrix} x^{2} & \frac{1}{2} \sqrt{π} & \int x d x \end{matrix} ⎤ ⎥ ⎦

$[\begin{array}{c} x^{2} & \frac{1}{2} \\ \sqrt{π} & \int x d x \end{array}]$