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2 years ago

6

A flat sheet of paper of area 0.250 m2 is oriented so that the normal to the sheet is at an angle of 60 to a uniform electric fi

eld of magnitude 14 N>C. (a) Find the magnitude of the electric flux through the sheet. (b) Does the answer to part (a) depend on the shape of the sheet

1 answer:

Harman [31]2 years ago

4 0

Answer:

a. 1.75 Nm²/C

b. Yes.

Explanation:

a. Electric Flux is given as:

Φ = E*A*cosθ

Where E = electric flux

A = Surface area

Φ = 14 * 0.25 * cos60

Φ = 1.75 Nm²/C

b. Yes, the shape of the sheet will affect the Flux through it. This is because flux is dependent on area of the surface and the area is dependent on the shape of the surface.

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hodyreva [135]

20) When light passes from air to glass and then to air

21) When a light ray enters a medium with higher optical density, it bends towards the normal

22) Index of refraction describes the optical density

23) Light travels faster in the material with index 1.1

24) Glass refracts light more than water

25) Index of refraction is $n=\frac{c}{v}$

26) Critical angle: $[tex]sin \theta_c = \frac{n_2}{n_1}$ [/tex]

27) Critical angle is larger for the glass-water interface

Explanation:

20)

It is possible to slow down light and then speed it up again by making light passing from a medium with low optical density (for example, air) into a medium with higher optical density (for example, glass), and then make the light passing again from glass to air.

This phenomenon is known as refraction: when a light wave crosses the interface between two different mediums, it changes speed (and also direction). The speed decreases if the light passes from a medium at lower optical density to a medium with higher optical density, and viceversa.

21)

The change in direction of light when it passes through the boundary between two mediums is given by Snell's law:

$n_1 sin \theta_1 = n_2 sin \theta_2$

with

$n_1, n_2$ are the refractive index of 1st and 2nd medium

$\theta_1, \theta_2$ are the angle of incidence and refraction (the angle between the incident ray (or refracted ray) and the normal to the boundary)

The larger the optical density of the medium, the larger the value of n, the smaller the angle: so, when a light ray enters a medium with higher optical density, it bends towards the normal.

22)

The index of refraction describes the optical density of a medium. More in detail:

A high index of refraction means that the material has a high optical density, which means that light travels more slowly into that medium
A low index of refraction means that the material has a low optical density, which means that light travels faster into that medium

Be careful that optical density is a completely different property from density.

23)

As we said in part 22), the index of refraction describes the optical density of a medium.

In this case, we have:

A material with refractive index of 1.1
A material with refractive index of 2.2

As we said previously, light travels faster in materials with a lower refractive index: therefore in this case, light travels more quickly in material 1, which has a refractive index of only 1.1, than material 2, whose index of refraction is much higher (2.2).

24)

Rewriting Snell's law,

$sin \theta_2 = \frac{n_1}{n_2}sin \theta_1$ (1)

For light moving from air to water:

$n_1 \sim 1.00$ is the index of refraction of air

$n_2 = 1.33$ is the index of refraction ofwater

In this case, $\frac{n_1}{n_2}=\frac{1.00}{1.33}=0.75$

For light moving from air to glass,

$n_2 = 1.51$ is the index of refraction of glass

And so

$\frac{n_1}{n_2}=\frac{1.00}{1.51}=0.66$

From eq.(1), we see that the angle of refraction $\theta_2$ is smaller in the 2nd case: so glass refracts light more than water, because of its higher index of refraction.

25)

The index of refraction of a material is

$n=\frac{c}{v}$

c is the speed of light in a vacuum

v is the speed of light in the material

So, the index of refraction is inversely proportional to the speed of light in the material:

The higher the index of refraction, the slower the light
The lower the index of refraction, the faster the light

26)

From Snell's law,

$sin \theta_2 = \frac{n_1}{n_2}sin \theta_1$

We notice that when light moves from a medium with higher refractive index to a medium with lower refractive index, $n_1 > n_2$ , so $\frac{n_1}{n_2}>1$ , and since $sin \theta_2$ cannot be larger than 1, there exists a maximum value of the angle of incidence $\theta_c$ (called critical angle) above which refraction no longer occurs: in this case, the incident light ray is completely reflected into the original medium 1, and this phenomenon is called total internal reflection.

The value of the critical angle is given by

$sin \theta_c = \frac{n_2}{n_1}$

For angles of incidence above this value, total internal reflection occurs.

27)

Using:

$sin \theta_c = \frac{n_2}{n_1}$

For the interface glass-air,

$n_1 \sim 1.51\\n_2 = 1.00$

The critical angle is

$\theta_c = sin^{-1}(\frac{n_2}{n_1})=sin^{-1}(\frac{1.00}{1.51})=41.5^{\circ}$

For the interface glass-water,

$n_1 \sim 1.51\\n_2 = 1.33$

The critical angle is

$\theta_c = sin^{-1}(\frac{n_2}{n_1})=sin^{-1}(\frac{1.33}{1.51})=61.7^{\circ}$

So, the critical angle is larger for the glass-water interface.

Learn more about refraction:

brainly.com/question/3183125

brainly.com/question/12370040

#LearnwithBrainly

7 0

2 years ago

A magnetic field of 0.080 T is in the y-direction. The velocity of wire segment S has a magnitude of 78 m/s and components of 18

Annette [7]

Answer:

Part a)

Induced EMF when length vector is along Z direction is 0.72 V

Part b)

Induced EMF when length vector is along Y direction is ZERO

Explanation:

As we know that the motional EMF induced in the wire is given as

$E = (v\times B). L$

1)

As we know that

$v = 18\hat i + 24\hat j + 72\hat k$

$B = 0.080 \hat j$

$L = 0.50 \hat k$

now we have

$\vec v \times \vec B = 1.44\hat k - 5.76 \hat i$

so we have

$E = 1.44 (0.50) = 0.72 V$

2)

If the length vector is along Y direction then we have

$L = 0.50 \hat j$

so again we have

$\vec v \times \vec B = 1.44\hat k - 5.76 \hat i$

so we have

EMF = 0

6 0

2 years ago

A strip 1.2 mm wide is moving at a speed of 25 cm/s through a uniform magnetic field of 5.6 t. what is the maximum hall voltage

Alex787 [66]

The equation for Hall voltage Vh is:

Vh=v*B*w, where v is the velocity of the strip, B is the magnitude of the magnetic field, and w is the width of the strip.

v=25 cm/s = 0.25 m/s
B=5.6 T
w= 1.2 mm = 0.0012 m

We input the numbers into the equation and get:

Vh= 0.25*5.6*0.0012 = 0.00168 V

The maximum Hall voltage is Vh= 0.00168 V.

4 0

2 years ago

Hoosier Manufacturing operates a production shop that is designed to have the lowest unit production cost at an output rate of 1

abruzzese [7]

Answer:

90.77%

its capacity utilization rate for the month is 90.77%

Explanation:

The capacity utilisation rate can be expressed mathematically as;

Capacity utilisation rate = capacity used/Best operating level × 100%

Given;

Total Number of production time = 205hours

Production output/capacity used = 21400 units

Best operation rate = 115units/hour

Best operation output for the month of July( at best operation level )

=115units/hour × 205 hours = 23575 units

Capacity utilisation rate = 21400/23575 × 100%

= 90.77%

3 0

2 years ago

A person weighing 0.70 kn rides in an elevator that has an upward acceleration of 1.5 m/s2. what is the magnitude of the normal

creativ13 [48]

First of all, we can find the mass of the person, since we know his weight W:
$W=mg=0.70 kN=700 N$
And so
$m= \frac{W}{g}= \frac{700 N}{9.81 m/s^2}=71.4 kg$

We know for Newton's second law that the resultant of the forces acting on the person must be equal to the product between the mass and the acceleration a of the person itself:
$\sum F = ma$
There are only two forces acting on the person: his weight W (downward) and the vincular reaction Rv of the floor against the body (upward). So we can rewrite the previous equation as
$R_v -W = ma$
We know the acceleration of the system, $a=1.5 m/s^2$ (upward, so with same sign of Rv), so we can solve to find the value of Rv, the normal force exerted by the elevator's floor on the person:
$R_v = ma+W=(71.4 kg)(1.5 m/s^2)+700 N =807N$

8 0

2 years ago