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kvasek [131]
3 years ago
12

To understand the behavior ofthe electric field at the surface of a conductor, and itsrelationship to surface charge on the cond

uctor.
A conductor is placed in an external electrostatic field. Theexternal field is uniform before the conductor is placed within it.The conductor is completely isolated from any source of current orcharge.

PART A)
Which of the following describes the electricfield inside this conductor?

It is in thesame direction as the original external field.
It is in theopposite direction from that of the original externalfield.
It has adirection determined entirely by the charge on itssurface.
It is alwayszero.
PART B)
The charge density inside theconductor is:

0
non-zero;but uniform
non-zero;non-uniform
infinite
Physics
1 answer:
valentinak56 [21]3 years ago
7 0

Answer:

PART A

It is always zero

PART B

Answer is : 0

Explanation:

PART A :It is always zero because according to my research, The net electric field inside a conductor is always zero. If the net electric field were not zero, a current would flow inside the conductor. This would build up charge on the exterior of the conductor. This charge would oppose the field, ultimately (in a few nanoseconds for a metal) canceling the field to zero.

PART B: You already know that there is a zero net electric field inside a conductor; therefore, if you surround any internal point with a Gaussian surface, there will be no flux at any point on this surface, and hence the surface will enclose zero net charge. This surface can be imagined around any point inside the conductor with the same result, so the charge density must be zero everywhere inside the conductor. This argument breaks down at the surface of the conductor, because in that case, part of the Gaussian surface must lie outside the conducting object, where there is an electric field.

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Sun warms the earth with Electromagnetic Radiations.

Sun emits out short range radiation in the form of Electromagnetic (EM) Radiation. Our Atmosphere is transparent to EM radiation. It does not heats up the atmosphere but when it collide with the solid objects or particles it is absorbed, reflected or refracted and the object heats up and start emitting radiation in the form of long wave radiation. Our Earth does the same. The long wave radiation emitted by the earth surface is absorbed by our atmosphere and it heats up. Therefore, we can say our atmosphere is heated from below.
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3 years ago
Like the alto in a choir the octavina plays the lower melody . what musical element does this tell?
Katena32 [7]

<em>L</em><em>ike the alto in a choir the octavina plays the lower melody . what musical element does this tell?</em>

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3 years ago
Water at 20°C flows by gravity through a smooth pipe from one reservoir to a lower one. The elevation difference is 60 m. The pi
Serga [27]

Answer:

Flow Rate = 80 m^3 /hours  (Rounded to the nearest whole number)

Explanation:

Given

  • Hf = head loss
  • f = friction factor
  • L = Length of the pipe = 360 m
  • V = Flow velocity, m/s
  • D = Pipe diameter = 0.12 m
  • g = Gravitational acceleration, m/s^2
  • Re = Reynolds's Number
  • rho = Density =998 kg/m^3
  • μ = Viscosity = 0.001 kg/m-s
  • Z = Elevation Difference = 60 m

Calculations

Moody friction loss in the pipe = Hf = (f*L*V^2)/(2*D*g)

The energy equation for this system will be,

Hp = Z + Hf

The other three equations to solve the above equations are:

Re = (rho*V*D)/ μ

Flow Rate, Q = V*(pi/4)*D^2

Power = 15000 W = rho*g*Q*Hp

1/f^0.5 = 2*log ((Re*f^0.5)/2.51)

We can iterate the 5 equations to find f and solve them to find the values of:

Re = 235000

f = 0.015

V = 1.97 m/s

And use them to find the flow rate,

Q = V*(pi/4)*D^2

Q = (1.97)*(pi/4)*(0.12)^2 = 0.022 m^3/s = 80 m^3 /hours

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3 years ago
When you jump upward, your hang time is the time your feet are off the ground. Does hang time depend on the vertical component o
hjlf

Answer:

It only depends on the vertical component

Explanation:

Hello!

The horizontal component will tell you how much you travel in that direction.

You could have a large horizontal velocity, but if the vertical velocity is zero, you will never be out of the ground. Similarly, you could have a zero horizontal velocity, but if you have a non-zero vertical velocity you will be some time off the ground. This time can be calculated by two means, one is using the equation of motion (position as a function of time) and the other using the velocity as a fucntion of time.

For the former you must find the time when the position is zero.

Lets consider the origin of teh coordinate system at your feet

y(t) = vt - (1/2)gt^2

We are looking for a time t' for which y(t')=0

0 = vt' - (1/2)gt'^2

vt' = (1/2)gt'^2

The trivial solution is when t'=0 which is the initial position, however we are looking for t'≠0, therefore we can divide teh last equation by t'

v = (1/2)gt'

Solving for t'

t' = (2v/g)

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3 years ago
I need help pelase ASAP!
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