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4 years ago

What happens when a resultant electric field exist in conductor

Physics

1 answer:

klio [65]3 years ago

7 0

Answer:

When an electric field exists in a conductor a current will flow.

This implies a voltage difference between two points on the conductor.

Electrostatics pertains to static charge distributions.

That means that an object such as a charged spherical conductor will be at the same potential (voltage) on both its outer and inner surfaces.

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What travels by vibrating particles? Mechincal Waves or ElecrtoMagnetic Waves.

Naddik [55]

Answer:mechanical waves.

Explanation:

Mechanical waves require the particles of the medium to vibrate in order for energy to be transferred. For example, water waves, earthquake/seismic waves, sound waves, and the waves that travel down a rope or spring are also mechanical waves.

5 0

3 years ago

A bottle lying on the windowsill falls off and takes 4.95 seconds to reach the ground. The distance from the windowsill to the g

mr Goodwill [35]

The distance an object falls from rest through gravity is

   D = (1/2) (g) (t²)

Distance = (1/2 acceleration of gravity) x (square of the falling time)

We want to see how the time will be affected
if ' D ' doesn't change but ' g ' does.
So I'm going to start by rearranging the equation
to solve for ' t '.

D = (1/2) (g) (t²)

Multiply each side by 2 :    2 D =      g t²

Divide each side by ' g ' : 2 D/g = t²

Square root each side: t = √ (2D/g)

Looking at the equation now, we can see what happens
to ' t ' when only ' g ' changes:

-- ' g ' is in the denominator; so bigger 'g' ==> shorter 't'
         and smaller 'g' ==> longer 't' .

-- They don't change by the same factor, because 1/g is inside
the square root. So 't' changes the same amount as √1/g does.

Gravity on the surface of the moon is roughly 1/6 the value
of gravity on the surface of the Earth.

So we expect ' t ' to increase by √6 = 2.45 times.

It would take the same bottle (2.45 x 4.95) = 12.12 seconds
to roll off the same window sill and fall 120 meters down to the
surface of the Moon.

4 0

3 years ago

Read 2 more answers

Suppose that a public address system emits sound uniformly in all directions and that there are no reflections . The intensity a

Marina86 [1]

Answer:

$I_{2}=2.39*10^{-5} W/m^{2}$

Explanation:

The definition of the intensity in terms of power is given by:

$I=\frac{P}{A}$

Where:

P is the power
A is the area

If the sound emits uniformly in all directions and that there are no reflections, we can assume the geometry of the wave sound is spherical.

Let's recall the area of a sphere is $A = 4\pi R^{2}$

To the first location we have:

$I_{1}=3*10^{-4} W/m^{2}=\frac{P}{4\pi 22^{2}}$

and to the second location we have:

$I_{2}=\frac{P}{4\pi 78^{2}}$

Now, we can divide each intensity to find the second intensity.

$\frac{I_{2}}{I_{1}}=\frac{\frac{P}{4\pi 78^{2}}}{\frac{P}{4\pi 22^{2}}}$

$I_{2}=I_{1}* \frac{22^{2}}{78^{2}}$

$I_{2}=3*10^{-4}\frac{22^{2}}{78^{2}}$

$I_{2}=2.39*10^{-5} W/m^{2}$

I hope it helps you!

5 0

3 years ago

Read 2 more answers

A transformer is to be used to provide power for a computer disk drive that needs 6.0 V (rms) instead of the 120 V (rms) from th

Ipatiy [6.2K]

Answer:

$N_{2}$ =20 turns

Explanation:

The given case is a step down transformer as we need to reduce 120 V to 6 V.

number of turns on primary coil N_{P}= 400

current delivered by secondary coil I_{S}= 500 mA

output voltage = 6 V (rms)

we know that

$I_{p}=\frac{V_{out}}{V_{in}\times I_{s}}$

putting values we get

$I_{p}=\frac{6}{120\times 0.5}$

$I_{p}= 0.1 A$

to calculate number of turns in secondary

$\frac{N_{2}}{400} =\frac{6}{120}$

therefore, $N_{2}$ =20 turns

5 0

3 years ago

One end of a thin rod is attached to a pivot, about which it can rotate without friction. Air resistance is absent. The rod has

Mars2501 [29]

Answer:

6.86 m/s

Explanation:

This problem can be solved by doing the total energy balance, i.e:

initial (KE + PE) = final (KE + PE). { KE = Kinetic Energy and PE = Potential Energy}

Since the rod comes to a halt at the topmost position, the KE final is 0. Therefore, all the KE initial is changed to PE, i.e, ΔKE = ΔPE.

Now, at the initial position (the rod hanging vertically down), the bottom-most end is given a velocity of v0. The initial angular velocity(ω) of the rod is given by ω = v/r , where v is the velocity of a particle on the rod and r is the distance of this particle from the axis.

Now, taking v = v0 and r = length of the rod(L), we get ω = v0/ 0.8 rad/s

The rotational KE of the rod is given by KE = 0.5Iω², where I is the moment of inertia of the rod about the axis of rotation and this is given by I = 1/3mL², where L is the length of the rod. Therefore, KE = 1/2ω²1/3mL² = 1/6ω²mL². Also, ω = v0/L, hence KE = 1/6m(v0)²

This KE is equal to the change in PE of the rod. Since the rod is uniform, the center of mass of the rod is at its center and is therefore at a distane of L/2 from the axis of rotation in the downward direction and at the final position, it is at a distance of L/2 in the upward direction. Hence ΔPE = mgL/2 + mgL/2 = mgL. (g = 9.8 m/s²)

Now, 1/6m(v0)² = mgL ⇒ v0 = $\sqrt{6gL}$

Hence, v0 = 6.86 m/s

4 0

3 years ago

What happens when a resultant electric field exist in conductor​

What happens when a resultant electric field exist in conductor