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3 years ago

7

A cube has one corner at the origin and the opposite cornerat the point (L,L,L). The sides of the cube are parallel to thecoordi

nate planes. The electric field in and around the cube isgiven by E=(a+bx)x+cy.Part A Find the total electric flux phi E through the surface of the cube. Express your answer in terms of a,b,c and L. Phi E=Part B This part will be visible after youcomplete previous item(s). Part C What is the net charge q inside the cube? Express your answer in terms of a,b,c ,L,and epsilon. q=

1 answer:

ella [17]3 years ago

3 0

Answer:

Explanation:

in this case, flux and area vectors are parallel . therefore, flux is just the dot product of electric field and area vector.

(flux=EAcos0 =EA)

flux through the face x face

Physics homework question answer, step 1, image 1

the flux through -x face

Physics homework question answer, step 1, image1

Step 2

the flux through y face,

Physics homework question answer, step 2, image 2

the flux through -y face,

Physics homework question answer, step 2, image 2

the flux through the z faces are zero,

therefore, the net flux is,

Physics homework question answer, step 2, image 3

...

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The Mistuned Piano Strings Two identical piano strings of length 0.800 m are each tuned exactly to 480 Hz. The tension in one of

IRINA_888 [86]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The answer is

$T_2 = 1.008$ % higher than $T_1$

$T_2 = 0.99$ % lower than $T_1$

Explanation:

From the question we are told that

The first string has a frequency of $f_1 = 230 Hz$

The period of the beat is $t_{beat} = 0.99s$

Generally the frequency of the beat is

$f_{beat} = \frac{1}{t_{beat}}$

Substituting values

$f_{beat} = \frac{1}{0.99}$

$= 1.01 Hz$

From the question

$f_2 - f_1 = f_{beat}$ for $f_2$ having a higher tension

So

$f_2 - 230 = 1.01$

$f_2 = 231.01Hz$

From the question

$\frac{f_2}{f_1} = \sqrt{\frac{T_2}{T_1} }$

$\frac{T_2}{T_1} = \frac{f_2^2}{f_1^2}$

Substituting values

$\frac{T_2}{T_1} = \frac{(231.01)^2}{(230)^2}$

$T_2 = 1.008$ % higher than $T_1$

For $f_2$ having a lower tension

$f_1 - f_2 = f_{beat}$

So

$230 - f_2 = 1.01$

$f_2 = 230 -1.01$

$= 228.99$

From the question

$\frac{f_2}{f_1} = \sqrt{\frac{T_2}{T_1} }$

$\frac{T_2}{T_1} = \frac{f_2^2}{f_1^2}$

Substituting values

$\frac{T_2}{T_1} = \frac{(228.99)^2}{(230)^2}$

$T_2 = 0.99$ % lower than $T_1$

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4 years ago

At which position would a person on Earth be able to view a solar eclipse?<br> A<br> B<br> C<br> D

sweet-ann [11.9K]

The answer would be A, since at that position the Earth is fully/partially being engulfed in a shadow cast by the moon from the sun’s rays.

Hope this helps!

5 0

3 years ago

Which statement describes static equilibrium?

scoray [572]

Answer:

the forces on an object are balanced and there is no motion..(c)

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4 years ago

A light string is wrapped around the outer rim of a solid uniform cylinder of diameter 75.0 cm that can rotate without friction

anyanavicka [17]

Answer:18.5 kg

Explanation:

Given

diameter $d=75 cm$

mass of stone $m=3 kg$

velocity of stone $v=3.4 m/s$

height fallen $h=2.40 m$

using

$v^2-u^2=2as$

$(3.4)^2=2\times a\times 2.4$

$a=2.4 m/s^2$

Tension of string will Provide Torque to cylinder

$T\times r=I\times \alpha$

where $I=moment\ of\ inertia$

$\alpha =angular\ acceleration$

$T=\frac{Mr^2}{2}\times \frac{a}{r}$

$T=\frac{Ma}{2}$

and $mg-T=ma$

$mg-ma=T$

Put value of T

$mg-ma=\frac{Ma}{2}$

$mg=a(m+\frac{M}{2})$

$3\times 9.8=2.4\cdot (m+\frac{M}{2})$

$12.25=3+\frac{M}{2}$

$9.25=\frac{M}{2}$

$M=18.5 kg$

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4 years ago

*R3= 9 ohms <br><br> What is the voltage drop running through resistor five?

joja [24]

The correct answer is: The voltage drop across $R_5$ is 45V.

Explanation:
First you need to find the total current of the circuit, which is:
According to Ohm's law:

V = I * R

I = V/R --- (1)

Where
R = Total Equivalent resistance of the circuit
V = Total Voltage = 90V
I -= Total Current

First let us find the equivalent resistance:
$R = R_1 + R_2 + R_5 + \frac{R_3*R_4}{R_3+R_4} \\
R = 3 + 6 + 15 + \frac{9*18}{9+18} \\
R = 30$

Now plug in the values in (1):
(1) => I = 90/30 = 3A

Now that we have current of the circuit, we can now find the voltage drop across $R_5$ = 15 ohms using:
$V_5 = I * R_5$

Where $V_5$ is the voltage drop across resistor 5;
$V_5 = 3 * 15 = 45V$

5 0

3 years ago