Question

Hello. Can anyone help me with this 2 questions. I need to submit it tonight. Please provide the answer with the working.Don't answer it if you don't know! Thanks.​

Answer 1

Explanation:

you know that you can change and reform any equation by doing the same operation to everything on the left and on the right side of the "=" sign ?

that way you can change the appearance of terms and the locations of variables or constants without changing the expressed balance.

this is like in a 2-bowl scale, where you can add or remove the same weight on both sides to keep the scale in balance.

and that is the principle, when we transform equations.

4.

V = RI

we want to have "I" alone on one side. so, what do we need to do ? a division by "R", of course.

again, it has to happen on both sides of the "scale" ...

V/R = I

I = 50/2.5 = 20 A (or amps)

5.

the area of a circle is

A = pi × r²

with r being the radius.

now, r should stand alone (and is the result of the other values). so, what do we need to do ?

well, first we divide both sides by pi :

A/pi = r²

and now ? how do we get r from r² ? we pull the square root (sqrt()) ! again, as always, both sides :

r = sqrt(A/pi)

r = sqrt(100/pi) = sqrt(31.83098862...) =

= 5.641895835... ≈ 5.64 mm

Answer 2

Question 4

Ohm's Law

`V=RI`

where:

• V = potential difference in volts, V
• R = resistance in ohms,
  `\sf \Omega`
• I = current in amperes, A

Part (a)

Given equation:

`\implies V=RI`

Divide both sides by R:

`\implies (V)/(R)=(RI)/(R)`

Cancel the common factor:

`\implies (V)/(R)=( \diagup\!\!\!\!\!R\:I)/(\diagup\!\!\!\!\!R)`

Therefore:

`\implies I=(V)/(R)`

Part (b)

Given:

• V = 50 V
• R = 2.5
  `\sf \Omega`

Substitute the given values into the formula and solve for I:

`\implies I=(V)/(R)`

`\implies I=(50)/(2.5)`

`\implies I=20\:\:\sf A`

Question 5

`\textsf{Area of a circle}=\pi r^2 \quad \textsf{(where r is the radius)}`

Part (a)

Given formula:

`\implies A=\pi r^2`

Divide both sides by
`\pi`:

`\implies (A)/(\pi)=(\pi r^2)/(\pi)`

Cancel the common factor:

`\implies (A)/(\pi)=(\diagup\!\!\!\!\!\pi r^2)/(\diagup\!\!\!\!\!\pi)`

`\implies r^2=(A)/(\pi)`

Square root both sides:

`\implies √(r^2)=\sqrt{(A)/(\pi)}`

`\implies r=\sqrt{(A)/(\pi)}`

Part (b)

Given:

• A = 100 mm²

Substitute the given value into the formula and solve for r:

`\implies r=\sqrt{(A)/(\pi)}`

`\implies r=\sqrt{(100)/(\pi)}`

`\implies r=5.64\:\:\sf mm \:(2 \:dp)`