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3 years ago

15

A gold wire has a cross-sectional area of 1.0 cm^2 and a resistivity of 2.8 × 10^-8 Ω ∙ m at 20°C. How long would it have to be

to have a resistance of 0.001 Ω?

1 answer:

Karo-lina-s [1.5K]3 years ago

4 0

Answer:

Length, l = 3.57 meters

Explanation:

It is given that,

Area of cross-section of gold wire, $A=1\ cm^2=0.0001\ m^2$

Resistivity of gold wire, $\rho=2.8\times 10^{-8}\ \Omega-m$

Resistance, R = 0.001 ohms

Resistance in terms of length and area is given by :

$R=\rho \dfrac{l}{A}$

$l=\dfrac{RA}{\rho}$

$l=\dfrac{0.001\times 0.0001}{2.8\times 10^{-8}}$

l = 3.57 meters

So, the length of the wire is 3.57 meters. Hence, this is the required solution.

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ANTONII [103]

Answer:

1.35m

Explanation:

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The vertical speed at the jumping point can be computed:

$v_v^2 + v_h^2 = v^2$

$v_v^2 + 13^2 = 14^2$

$v_v^2 = 196 - 169 = 27$

$v_v = \sqrt{27} = 5.2 m/s$

When the skier jumps to the its potential energy is converted to kinetic energy:

$E_p = E_k$

$mgh = mv_v^2/2$

where m is the skier mass and h is the vertical distance traveled, $v_v$ is the vertical velocity at jumping point, and h is the highest point.

Let g = 10m/s2

We can divide both sides of the equation by m:

$gh = v_v^2/2$

$h = \frac{v_v^2}{2g} = \frac{27}{2*10} = 1.35 m$

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3 years ago

Calculate the potential energy stored in an object of mass 50 kg at a height of 20 m from the ground.

bearhunter [10]

Answer:

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2 years ago

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Ce unitate de masura are indicele de REFRACTIE?

ivolga24 [154]

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Explanation:

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2 years ago

A satellite of mass M = 270kg is in circular orbit around the Earth at an altitude equal to the earth's mean radius (6370 km). A

zubka84 [21]

To solve this problem we will apply the concepts related to Orbital Speed as a function of the universal gravitational constant, the mass of the planet and the orbital distance of the satellite. From finding the velocity it will be possible to calculate the period of the body and finally the gravitational force acting on the satellite.

PART A)

$V_{orbital} = \sqrt{\frac{GM_E}{R}}$

Here,

M = Mass of Earth

R = Distance from center to the satellite

Replacing with our values we have,

$V_{orbital} = \sqrt{\frac{(6.67*10^{-11})(5.972*10^{24})}{(6370*10^3)+(6370*10^3)}}$

$V_{orbital} = 5591.62m/s$

$V_{orbital} = 5.591*10^3m/s$

PART B) The period of satellite is given as,

$T = 2\pi \sqrt{\frac{r^3}{Gm_E}}$

$T = \frac{2\pi r}{V_{orbital}}$

$T = \frac{2\pi (2*6370*10^3)}{5.591*10^3}$

$T = 238.61min$

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mamaluj [8]

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