Bernoulli's Principle is a consequence of the conservation of

a cylinder has a radius of 2.1cm and a length of 8.8cm .total charge 6.1 x 10^-7C is distributed uniformly throughout. the volum

7nadin3 [17]

The answer for the following problem is explained below.

Therefore the volume charge density of a substance (ρ) is 0.04 × $10^{-1}$ C.

Explanation:

Given:

radius (r) =2.1 cm = 2.1 × $10^{-2}$ m

height (h) =8.8 cm = 8.8 × $10^{-2}$ m

total charge (q) =6.1× $10^{-7}$ C

To solve:

volume charge density (ρ)

We know;

ρ =q ÷ v

volume of cylinder = π ×r × r × h

volume of cylinder =3.14 × 2.1 × 2.1 × $10^{-4}$ × 8.8 × $10^{-2}$

volume of cylinder (v) = 122.23 × $10^{-6}$

ρ =q ÷ v

ρ = 6.1× $10^{-7}$ ÷ 122.23 × $10^{-6}$

ρ = 0.04 × $10^{-1}$ C

Therefore the volume charge density of a substance (ρ) is 0.04 × $10^{-1}$ C.

3 0

3 years ago

1. Which cell organelle is popularly known as power house of the cell. Why?

MrRissso [65]

The mitochondria is the powerhouse of the cell because it takes nutrients and breaks it down to provide energy for the cell.

5 0

3 years ago

Responsible for reflex, maintain muscles tone and dopamine. 8 letter word

Lady bird [3.3K]

Exercise is the answer hope i helped you

3 0

3 years ago

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.