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2 years ago

Explain why materials such as plastic foam, feathers, and fur are poor conductors of heat.

1 answer:

8 0

They are poor conductors because they keep the heat trapped in one spot, while other materials like metal let the heat go through itself.

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A disk of radius a has a total charge Q uniformly distributed over its surface. The disk has negligible thickness and lies in th

sleet_krkn [62]

The electric potential V(z) on the z-axis is : V = $(\frac{Q}{a^2} ) [ (a^2 + z^2)^{\frac{1}{2} } -z$

The magnitude of the electric field on the z axis is : E = kб 2 $\pi$ ( 1 - [z / √(z² + a² ) ] )

<u>Given data :</u>

V(z) =2kQ / a²(v(a² + z²) ) -z

<h3>Determine the electric potential V(z) on the z axis and magnitude of the electric field</h3>

Considering a disk with radius R

Charge = dq

Also the distance from the edge to the point on the z-axis = √ [R² + z²].

The surface charge density of the disk ( б ) = dq / dA

Small element charge dq = б( 2πR ) dr

dV $\frac{k.dq}{\sqrt{R^2+z^2} } \\\\= \frac{k(\alpha (2\pi R)dR}{\sqrt{R^2+z^2} }$ ----- ( 1 )

Integrating equation ( 1 ) over for full radius of a

∫dv = $\int\limits^a_o {\frac{k(\alpha (2\pi R)dR)}{\sqrt{R^2+z^2} } } \,$

V = $\pi k\alpha [ (a^2+z^2)^\frac{1}{2} -z ]$

= $\pi k (\frac{Q}{\pi \alpha ^2})[(a^2 +z^2)^{\frac{1}{2} } -z ]$

Therefore the electric potential V(z) = $(\frac{Q}{a^2} ) [ (a^2 + z^2)^{\frac{1}{2} } -z$

Also

The magnitude of the electric field on the z axis is : E = kб 2 $\pi$ ( 1 - [z / √(z² + a² ) ] )

Hence we can conclude that the answers to your question are as listed above.

Learn more about electric potential : brainly.com/question/25923373

7 0

2 years ago

Help me with this question please

vichka [17]

Answer:

Its true i'm pretty sure

Explanation:

7 0

2 years ago

Read 2 more answers

When you catch a baseball, what kind of work do you do, negative or positive? By catching the baseball, what do you change? Does

bagirrra123 [75]

Negative energy by catching it. Changes the force and movement of the baseball. Loses energy. Kinetic energy

3 0

3 years ago

Read 2 more answers

A parallel-plate capacitor has a plate area of 0.2m^2 and a plate separation of 0.1mm. To obtain an electric field of 2.0 × 10^6

Oduvanchick [21]

Answer:

3.536*10^-6 C

Explanation:

The magnitude of the charge is expresses as Q = CV

C is the capacitance of the capacitor

V is the voltage across the capacitor

Get the capacitance

C = ε0A/d

ε0 is the permittivity of the dielectric = 8.84 x 10-12 F/m

A is the area = 0.2m²

d is the plate separation = 0.1mm = 0.0001m

Substitute

C = 8.84 x 10-12 * 0.2/0.0001

C = 1.768 x 10-8 F

Get the potential difference V

Using the formula for Electric field intensity

E = V/d

2.0 × 10^6 = V/0.0001

V = 2.0 × 10^6 * 0.0001

V = 2.0 × 10^2V

Get the charge on each plate.

Q = CV

Q = 1.768 x 10-8 * 2.0 × 10^2

Q = 3.536*10^-6 C

Hence the magnitude of the charge on each plate should be 3.536*10^-6 C

5 0

2 years ago

An object of mass m is hung from a spring and set into oscillation. The period of the oscillation is measured and recorded as T.

sergejj [24]

Answer: $\sqrt{2}T$

Explanation:

Given

object of mass m is suspended from spring and set in oscillation with time Period T

We know Time period of a mass in oscillation is given by

$T=2\pi \sqrt{\frac{m}{k}}$

where k=spring constant

When mass m is replaced by a mass of 2 m time period is given by

$T'=2\pi \sqrt{\frac{2m}{k}}$

$T'=\sqrt{2}\times 2\pi \sqrt{\frac{m}{k}}$

$T'=\sqrt{2}T$

i.e. New time period becomes $\sqrt{2}$ times of previous one

7 0

2 years ago