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2 years ago

All known frequencies of the visible spectrum are..

Physics

1 answer:

crimeas [40]2 years ago

7 0

Explanation:

$b. \: light$

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What value must q2 have if the electric potential at point a is to be zero?

Ulleksa [173]

<span>let r be side of square
V (B) = k q2/r + k Q / sqrt[r^2 + r^2] = 0 given
q2/r + Q / r root2 = 0
q2 = - Q / root2
q2/Q = - 1/root2
q2/Q = - root2 /2 = - 0.707

q2 equals negative 0.707 times Q</span>

5 0

3 years ago

An electric current in a conductor varies with time according to the expression i(t) = 110 sin (120πt), where i is in amperes an

Dafna1 [17]

As we know that current is defined as rate of flow of charge

$i = \frac{dq}{dt}$

so by rearranging the equation we can say

$q = \int i dt$

here we know that

$i(t) = 110 sin(120\pi t)$

here we will substitute it in the above equation

$q = \int 110 sin(120\pi t) dt$

$q = 110 [- \frac{cos(120\pi t)}{120\pi}]$

now here limits of time is from t = 0 to t = 1/180s

so here it will be given as

$q = \frac{110}{120\pi}( -cos0 + cos(\frac{2\pi}{3}))$

$q = 0.44 C$

so total charge flow will be 0.44 C

6 0

3 years ago

Read 2 more answers

A positive charge of 0.00047 C is 15 m from a negative charge of 0.00089 C. What is the force of one of the charges due to the o

Lady_Fox [76]

Answer:

16.732 N

Explanation:

Given:

q1 = 0.00047 C = 4.7 x 10^-4 C

q2 = 0.00089 C = 8.9 x 10^-4 C

d = 15 m

k = 9 x 10^9 N m^2 / C^2

To Find:

F = ?

Solution:

F = k x q1 x q2/d^2

F = 9 x 10^9 x 4.7 x 10^-4 x 8.9 x 10^-4 / 15 x 15

F = 9 x 4.7 x 8.9 x 10^9 x 10^-4 x 10^-4 / 225

F = 9 x 4.7 x 8.9 x 10^9 x 10^-8 / 225

F = 9 x 4.7 x 8.9 x 10 / 225

F = 418.3/25

F = 1673.2/100

Therefore, F = 16.732 N

PLZ MARK ME AS BRAINLIEST!!!

8 0

3 years ago

15. Sandra decided to talk a walk at her neighborhood park. She walks 20 meters

FinnZ [79.3K]

Answer:

Long question good luck:)..............

Explanation:

8 0

2 years ago

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,

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