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3 months ago

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A solid cylinder is radiating power. It has a length that is ten times its radius. It is cut into a number of smaller cylinders,

each of which has the same radius. Each small cylinder has the same temperature as the original cylinder. The total radiant power emitted by the pieces is twice that emitted by the original cylinder. How many smaller cylinders are there? Give your answer as a number with no units.

1 answer:

Ostrovityanka [3.2K]3 months ago

3 0

Answer:

The total count of small cylinders equals 7.

Explanation:

Let

the radius of the large cylinder equal R

and the length as L

L = 10 R

The overall area A = 2 π R² + π R L

The length of individual small cylinders is denoted as l

The number of small cylinders is n

L = n l

The total surface area of the smaller cylinders

A'=n (2 π R² + π R l)

Given that emissions power can be expressed as

P = A ε σ T⁴

For the large cylinder

P = A ε σ T⁴ -----------1

For the smaller cylinders

P'=A' ε σ T⁴ ------2

From 1 and 2

Given that

P'= 2 P

A' ε σ T⁴ =2 A ε σ T⁴

A'=2 A (Other factors remain constant)

n (2 π R² + π R l) =(2 2 π R² + π R L)

n (2 R² + R l) = (2 R² + R L)

$n(2R^2+R\times \dfrac{L}{n}) = 2(2R^2+RL)$

L = 10 R

$n(2R^2+R\times \dfrac{10R}{n}) =2 (2R^2+R\times 10R)$

$n(2+\dfrac{10}{n}) =2( 2+ 10)$

2 n +10 = 2 x 12

2 n +10 = 24

2 n = 24 -10

2 n = 14

n = 7

The total number of small cylinders equals 7.

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b) The electric field has a strength of 100 N/C, directed from point B toward point A, where the charge is negative.

Clarification:

Given:

An object with a charge of q = -6.00 x 10^-9 C starts from rest at point A, making its kinetic energy zero ( $K_{A}$ = 0) and moving to point B at a distance l = 0.500m where its kinetic energy is ( $K_{B}$ = 5.00 x 10^-7J). Additionally, the electric potential of q at point A is VA = +30.0 v.

Required:

(a) We seek to find the electric potential VB

(b) We need to compute the magnitude and orientation of the electric field E.

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(a) Utilizing the given values for VA, $K_{B}$ and q, we derive a relationship among the three parameters and VB to compute VB.

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$K_{A} +U_{A} =K_{B} +U_{B}$ .........................................(1)

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VA-VB = $\int\limits^1_0 {E} \, dl$ ...................................(a)

$=E\int\limits^1_0 {} \, dl$

VA-VB=E $l$ (Restructuring for E)

E= VA-VB/ $l$ ..................................(3)

Now, substituting our values for VA, Vs, and l into equation (3) allows us to compute the electric field E

E= VA-VB/ $l$

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