princessnavya30
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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.​

Hello Can anyone help me with this 2 questions I need to submit it tonight Please provide the answer with the workingDont answer it if you dont know Thanks class=

Respuesta :

reinhard10158

Step-by-step 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

semsee45

Answer:

Question 4

Ohm's Law

[tex]V=RI[/tex]

where:

  • V = potential difference in volts, V
  • R = resistance in ohms, [tex]\sf \Omega[/tex]
  • I = current in amperes, A

Part (a)

Given equation:

[tex]\implies V=RI[/tex]

Divide both sides by R:

[tex]\implies \dfrac{V}{R}=\dfrac{RI}{R}[/tex]

Cancel the common factor:

[tex]\implies \dfrac{V}{R}=\dfrac{ \diagup\!\!\!\!\!R\:I}{\diagup\!\!\!\!\!R}[/tex]

Therefore:

[tex]\implies I=\dfrac{V}{R}[/tex]

Part (b)  

Given:

  • V = 50 V
  • R = 2.5  [tex]\sf \Omega[/tex]

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

[tex]\implies I=\dfrac{V}{R}[/tex]

[tex]\implies I=\dfrac{50}{2.5}[/tex]

[tex]\implies I=20\:\:\sf A[/tex]

Question 5

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

Part (a)

Given formula:

[tex]\implies A=\pi r^2[/tex]

Divide both sides by [tex]\pi[/tex]:

[tex]\implies \dfrac{A}{\pi}=\dfrac{\pi r^2}{\pi}[/tex]

Cancel the common factor:

[tex]\implies \dfrac{A}{\pi}=\dfrac{\diagup\!\!\!\!\!\pi r^2}{\diagup\!\!\!\!\!\pi}[/tex]

[tex]\implies r^2=\dfrac{A}{\pi}[/tex]

Square root both sides:

[tex]\implies \sqrt{r^2}=\sqrt{\dfrac{A}{\pi}}[/tex]

[tex]\implies r=\sqrt{\dfrac{A}{\pi}}[/tex]

Part (b)

Given:

  • A = 100 mm²

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

[tex]\implies r=\sqrt{\dfrac{A}{\pi}}[/tex]

[tex]\implies r=\sqrt{\dfrac{100}{\pi}}[/tex]

[tex]\implies r=5.64\:\:\sf mm \:(2 \:dp)[/tex]

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