If you given volume. for example 200 cm³ how can you change it to area m²

So I'm struggling with rearranging kinematic formulas. Does anyone have any steps or something to help.

bekas [8.4K]

Rearranging formulas is all about simple algebra rules. Just like when solving for x in an equation, you're just isolating whichever variable you want. I'll work this one out for you and hopefully it'll help, but if you need more explanation, then feel free to comment!

D = ViT + 0.5at² Subtract ViT from both sides

D - ViT = 0.5at² Divide both sides by 0.5t²

$\frac{D - ViT}{0.5t^{2} } = \frac{0.5at^{2} }{0.5t^{2} }$ I wrote this step out a little more to show how your fraction will cancel

$\frac{D - ViT}{0.5t^{2} }$ = a I like to flip these around so the single variable is on the right

a = $\frac{D - ViT}{0.5t^{2} }$

7 0

3 years ago

A battery-operated car moves forward as a result of which device? Electromagnet, generator, motor, or a transformer?

Romashka [77]

A battery-operated car moves forward as a result of which device?

A) Electromagnet

B) Generator

<u>C) Motor </u>

D) Transformer

7 0

3 years ago

Read 2 more answers

What is the magnitude of the magnetic field at a point midway between them if the top one carries a current of 19.5 A and the bo

Phantasy [73]

Answer:

The magnetic field will be $\large{\dfrac{1.4 \times 10^{-4}}{d}} T$ , '2d' being the distance the wires.

Explanation:

From Biot-Savart's law, the magnetic field ( $\large{\overrightarrow{B}}$ ) at a distance ' $r$ ' due to a current carrying conductor carrying current ' $I$ ' is given by

$\large{\overrightarrow{B} = \dfrac{\mu_{0}I}{4 \pi}} \int \dfrac{\overrightarrow{dl} \times \hat{r}}{r^{2}}}$

where ' $\overrightarrow{dl}$ ' is an elemental length along the direction of the current flow through the conductor.

Using this law, the magnetic field due to straight current carrying conductor having current ' $I$ ', at a distance ' $d$ ' is given by

$\large{\overrightarrow{B}} = \dfrac{\mu_{0}I}{2 \pi d}$

According to the figure if ' $I_{t}$ ' be the current carried by the top wire, ' $I_{b}$ ' be the current carried by the bottom wire and ' $2d$ ' be the distance between them, then the direction of the magnetic field at 'P', which is midway between them, will be perpendicular towards the plane of the screen, shown by the $\bigotimes$ symbol and that due to the bottom wire at 'P' will be perpendicular away from the plane of the screen, shown by $\bigodot$ symbol.