Ask question

Ask question

All categories

3 years ago

9

A 20-cm-diameter disk emits light uniformly from its surface. 40 cm from this disk, along its axis, is a 16.0-cm-diameter opaque

black disk; the faces of the two disks are parallel. 40 cm beyond the black disk is a white viewing screen. The lighted disk illuminates the screen, but there's a shadow in the center due to the black disk. What is the diameter of the completely dark part of this shadow

1 answer:

Marianna [84]3 years ago

7 0

Answer:

132

Explanation:

if you round it is correct

You might be interested in

An ___ is a deep valley on land that forms along a divergent boundary

Gennadij [26K]

This is a Rift Valley.

Hope this helped!

7 0

3 years ago

Read 2 more answers

Two fully charged cylindrical capacitors are connected to two identical batteries. The capacitors are identical except that the

Leni [432]

Answer:

Part(a): The relative capacitance is $\dfrac{C_{A}}{C_{B}} = 0.33$

Part(b): The relative energy stored is $\dfrac{U_{A}}{U_{B}} = 0.33$

Part(c): The relative charge stored is $\dfrac{Q_{A}}{Q_{B}} = 0.33$

Explanation:

We know the capacitance ( $C$ ) of a capacitor having charge ( $Q$ ) and subjected to a potential difference of ( $V$ ) is given by

$C = \dfrac{Q}{V}$

Also, the energy ( $U$ ) stored by a capacitor can be written as

$U = \dfrac{1}{2}C~V^{2}$

Let us assume that the inner radius of the Capacitor B, as shown in the figure, be $\textbf{r_{i}^{B}}$ $\bf{r_{i}^{B}}$ , the outer radius be $\bf{r_{o}^{B}}$ , the inner radius of Capacitor A be $\bf{r_{i}^{A}}$ and the outer radius be $\bf{r_{o}^{B}}$ .

Given in the problem,

$&& r_{o}^{B} = 2~r_{B}^{i}\\&& r_{o}^{A} = 4~r_{B}^{i}\\&& and~r_{i}^{B} = 4~r_{o}^{B} = 8~r_{B}^{i}$

Now, the capacitance ( $C$ ) of a cylindrical capacitor is given by,

$\bf{C = \dfrac{2~\pi~\epsilon_{0}~L}{ln(\dfrac{r_{o}}{r_{i}})}}$

where $\epsilon_{o}$ is the permittivity of the free space, $L$ is the length of the cylindrical capacitor.

Part(a):

The capacitance of capacitor A,

$C_{A} = \dfrac{2~\pi~\epsilon_{0}L}{ln(\dfrac{r_{o}^{A}}{r_{i}^{A}})} = \dfrac{2~\pi~\epsilon_{0}L}{ln(\dfrac{8~r_{i}^{B}}{r_{i}^{B}})} = \dfrac{2~\pi~\epsilon_{0}L}{ln(8)}$

and the capacitance of capacitor B,

$C_{B} = \dfrac{2~\pi~\epsilon_{0}L}{ln(\dfrac{r_{o}^{B}}{r_{i}^{B}})} = \dfrac{2~\pi~\epsilon_{0}L}{ln(\dfrac{2~r_{i}^{B}}{r_{i}^{B}})} = \dfrac{2~\pi~\epsilon_{0}L}{ln(2)}$

giving the relative capacitance of each capacitor to be

$\dfrac{C_{A}}{C_{B}} = \dfrac{ln(2)}{ln(8)} = \dfrac{ln(2)}{3~\ln(2)} = \dfrac{1}{3} = 0.33$

Part(b):

Energy stored by capacitor A,

$U_{A} = \dfrac{1}{2}~C_{A}~V^{2}$

Energy stored by capacitor B,

$U_{B} = \dfrac{1}{2}~C_{B}~V^{2}$

giving the relative energy stored by each capacitor to be

$\dfrac{U_{A}}{U_{B}} = \dfrac{C_{A}}{C_{B}} = 0.33$

Part(c):

The charge stored by capacitor A,

$Q_{A} = C_{A}~V$

The charge stored by capacitor B,

$Q_{B} = C_{B}~V$

giving the relative charge stored by each capacitor to be

$\dfrac{Q_{A}}{Q_{B}} = \dfrac{C_{A}}{C_{B}} = 0.33$

8 0

3 years ago

A water pipe tapers down from an initial radius of R1 = 0.21 m to a final radius of R2 = 0.11 m. The water flows at a velocity v

Aleks [24]

Answer:

$0.116 m^3/s$

Explanation:

The volume flow rate of a fluid in a pipe is given by:

$Q=Av$

where

A is the cross-sectional area of the pipe

v is the speed of the fluid

In this problem, at the initial point we have

v = 0.84 m/s is the speed of the water

r = 0.21 m is the radius of the pipe, so the cross-sectional area is

$A=\pi r^2 = \pi (0.21 m)^2 =0.138 m^2$

So, the volume flow rate is

$Q=(0.138 m^2)(0.84 m/s)=0.116 m^3/s$

8 0

3 years ago

The ability to do work or cause change is called

ipn [44]

Answer:

Energy

Explanation:

PLAESE MARK AS BRAINLIEST RATE 5 STARS AND THANK ME.

HOPE THIS HELPS

3 0

3 years ago

A desk was moved 12 m using 280 j of work. how much force was used to move the desk?

Oksana_A [137]

Let F = required force, N

Given:
d = 12 m, distance
W = 280 J, work done

By definition,
W = F*d,
therefore
(F N)*(12 m) = (280 J)
F = 280/12 = 23.33 N

Answer: The force is 23.3 N (nearest tenth)

4 0

3 years ago