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2 years ago

A copper wire of resistivity 2.6 × 10-8 Ω m, has a cross sectional area of 35 × 10-4 cm2

Physics

1 answer:

KengaRu [80]2 years ago

6 0

Answer:

the length of the wire is 134.62 m.

Explanation:

Given;

resistivity of the copper wire, ρ = 2.6 x 10⁻⁸ Ωm

cross-sectional area of the wire, A = 35 x 10⁻⁴ cm² = ( 35 x 10⁻⁴) x 10⁻⁴ m²

resistance of the wire, R = 10Ω

The length of the wire is calculated as follows;

$R = \frac{\rho L}{A} \\\\L = \frac{RA}{\rho} \\\\L= \frac{10 \times (35\times 10^{-4}) \times 10^{-4}}{2.6 \times 10^{-8}} \\\\L = 134.62 \ m$

Therefore, the length of the wire is 134.62 m.

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The Sun has a mass of 1.99x10^30 kg and a radius of 6.96x10^8 m. Calculate the acceleration due to gravity, in meters per second

just olya [345]

Answer:

$g=274\ m/s^2$

Explanation:

Mass of the Sun, $M=1.99\times 10^{30}\ kg$

The radius of the Sun, $r=6.96\times 10^8\ m$

We need to find the acceleration due to gravity on the surface of the Sun. It is given by the formula as follows :

$g=\dfrac{GM}{r^2}\\\\g=\dfrac{6.67\times 10^{-11}\times 1.99\times 10^{30}}{(6.96\times 10^8)^2}\\\\g=274\ m/s^2$

So, the value of acceleration due to gravity on the Sun is $274\ m/s^2$ .

8 0

2 years ago

What are organelles? Describe the role of two organelles within the cell.

zloy xaker [14]

Organelles are small structures found in cells that carry out certain tasks. Two examples of organelles are the Nucleus and the Mitochondria. Think of the nucleus as the brain of its cell, it controls activities and it contains a majority of the cells genetic material. The mitochondria is the part of the cells tasked with cellular respiration, which is the act of taking nutrients from a cell and turning it into energy.

4 0

3 years ago

An electron is confined to a one dimensional region, bounded by an infinite potential. If the energy of the electron in its firs

OLga [1]

Answer:

The energy in its ground state is 10 meV.

Explanation:

It is given that,

The energy of the electron in its first excited state is 40 meV.

Energy of the electron in any state is given by :

$E=\dfrac{n^2\pi^2h^2}{8mL^2}$

For ground state, n = 1

$E_1=\dfrac{\pi^2h^2}{8mL^2}$ .............(1)

For first excited state, n = 2

$40=\dfrac{2^2\pi^2h^2}{8mL^2}$ .............(2)

Dividing equation (1) and (2), we get :

$\dfrac{E_1}{40}=\dfrac{1}{4}$

$E_1=10\ meV$

So, the energy in its ground state is 10 meV. Hence, this is the required solution.

4 0

3 years ago

Question 20

oksian1 [2.3K]

Answer:

The image distance is 17.56 cm

Explanation:

We have,

Height of light bulb is 3 cm.

The light bulb is placed at a distance of 50 cm. It means object distance is, u =-50 cm

Focal length of the lens, f = +13 cm

Let v is distance between image and the lens. Using lens formula :

$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{13}+\dfrac{1}{(-50)}\\\\v=17.56\ cm$

So, the image distance is 17.56 cm.

5 0

2 years ago

A bullet fired vertically at a velocity of 36m/s .after 45 the bullet hit the top of a bulid how height is a bulid?

zzz [600]

Answer:

The height of the building is 8,302.5 m

Explanation:

Given;

velocity of the projectile, u = 36 m/s

time of motion, t = 45 s

Let the upward direction of the bullet be negative,

The height of the building is calculated as;

$h = ut - \frac{1}{2} gt^2\\\\h = (36\times 45) - (\frac{1}{2} \times 9.8 \times 45^2)\\\\h = 1620 - 9922.5\\\\h = -8,302.5 \ m\\\\The \ height \ of \ the \ building \ is \ 8,302.5 \ m$

3 0

2 years ago