1 5kg cat is lifted 2 m into the air.how much gpe does it gain?

S Suppose you wish to fabricate a uniform wire from a mass m of a metal with density rhom and resistivity rho. If the wire is to

denpristay [2]

The diameter of the wire is 2.8 * 10^-3 m.

<h3>What is the length?</h3>

Mass of the wire = 1.0 g or 1 * 10^-3 Kg

Resistance = 0.5 ohm

Resistivity of copper = 1.7 * 10^-8 ohm meter

Density of copper = 8.92 * 10^3 Kg/m^3

V = m/d

But v = Al

Al = m/d

A = m/ld

Resistance = ρl/A

= ρl/m/ld =

l^2 = Rm/ρd

l = √ Rm/ρd

l = √0.5 * 1 * 10^-3 / 1.7 * 10^-8 * 8.92 * 10^3

l = 1.82 m

A = πr^2

Also;

A = m/ld

A = 1 * 10^-3 Kg / 1.82 m * 8.92 * 10^3 Kg/m^3

A = 6.2 * 10^-5 m^2

r^2 = A/ π

r = √A/ π

r = √6.2 * 10^-5 m^2/3.142

r = 1.4 * 10^-3 m

Diameter = 2r = 2( 1.4 * 10^-3 m) = 2.8 * 10^-3 m

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Missing parts;

Suppose you wish to fabricate a uniform wire from 1.00g of copper. If the wire is to have a resistance of R=0.500Ω and all the copper is to be used, what must be (a) the length and (b) the diameter of this wire?

5 0

1 year ago

To reduce inductive reactance, what devices are normally placed on transmission or distribution lines?

UkoKoshka [18]

Answer : Capacitors

Explanation : Capacitors are normally placed on transmission or distribution lines when to reduce inductive reactance.

This is because it enhances electromechanical and voltage stability , limit voltage dips at network nodes and reduces the power loss.

So, we can say that inductive reactance normally replace by the capacitors.

4 0

2 years ago

A small metal ball is suspended from the ceiling by a thread of negligible mass. The ball is then set in motion in a horizontal

Gemiola [76]

Answer:

Time taken, $T=2\pi \sqrt{\dfrac{l\ cos\theta}{g}}$

Explanation:

It is given that, a small metal ball is suspended from the ceiling by a thread of negligible mass. The ball is then set in motion in a horizontal circle so that the thread’s trajectory describes a cone as shown in attached figure.

From the figure,

The sum of forces in y direction is :

$T\ cos\theta-mg=0$

$T=\dfrac{mg}{cos\theta}$

Sum of forces in x direction,

$T\ sin\theta=\dfrac{mv^2}{r}$

$mg\ tan\theta=\dfrac{mv^2}{r}$ .............(1)

Also, $r=l\ sin\theta$

Equation (1) becomes :

$mg\ tan\theta=\dfrac{mv^2}{l\ sin\theta}$

$v=\sqrt{gl\ tan\theta.sin\theta}$ ...............(2)

Let t is the time taken for the ball to rotate once around the axis. It is given by :

$T=\dfrac{2\pi r}{v}$

Put the value of T from equation (2) to the above expression:

$T=\dfrac{2\pi r}{\sqrt{gl\ tan\theta.sin\theta}}$

$T=\dfrac{2\pi l\ sin\theta}{\sqrt{gl\ tan\theta.sin\theta}}$

On solving above equation :

$T=2\pi \sqrt{\dfrac{l\ cos\theta}{g}}$

Hence, this is the required solution.

4 0

2 years ago

Bosons & Fermions A new type of quantum object has been discovered. When a large collection of these are cooled to almost ab

Lena [83]

Answer:

Only 2,3,4 are true

Explanation:

Bosons Particles are particles that condense to the same state. Bosons particle have integral spin like 0 , $\hbar$ , $2$\hbar$ , $3$\hbar$$ , etc. Bosons particles always have asymmetric wave function and there is exchange of particles.