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7 days ago

A ray of light passes from air into carbon disulfide (n = 1.63) at an angle of 28.0 degrees to the normal. what is the refracted

angle?

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If the potential difference across the bulb in a certain flashlight is 3.0 V, What is the potential difference across the combin

Yuliya22 [3333]

Answer:

The voltage across the bulb measures 3.0 V,

Explanation:

The bulb's voltage aligns with the voltage of the batteries, as they are the only power source for the bulb. Therefore, the voltage across the batteries is 3.0 V.

3 0

2 months ago

At a location where the acceleration due to gravity is 9.807 m/s2, the atmospheric pressure is 9.891 × 104 Pa. A barometer at th

serg [3582]

Answer:

$8616.7468 \ kg/m^3$

Explanation:

The measurement of pressure is indicated as $p=\rho gh$ where p denotes the pressure, $\rho$ signifies density, and h represents height

Given values include pressure $p=9.891\times 10^4\ Pa$ , gravity's acceleration $g=9.9870\ m/sec^2$ , and height =1.163 m

$\rho =\frac{p}{gh}=\frac{9.891\times 10^4}{9.870\times 1.163}=8616.7468 \ kg/m^3$

5 0

2 months ago

Read 2 more answers

A child's toy is suspended from the ceiling by means of a string. The Earth pulls downward on the toy with its weight force of 8

serg [3582]

The reaction force is F = -8 N. It is noted that the Earth's gravitational force pulls the toy down with a weight of 8.0 N. This problem utilizes Newton's third law of motion, which states that "for every action, there is an equal and opposite reaction." In a mathematical context, it is illustrated as follows: where the force exerted on one object by another is indicated. In this scenario, the force of Earth acting on the toy is the action force. Consequently, the reaction force is characterized as F = -8 N (the negative indicates that the reaction force is directed upward). In essence, the toy exerts an upward force of 8 N on the Earth, thus comprising the required solution.

3 0

1 month ago

An infinite sheet of charge, oriented perpendicular to the x-axis, passes through x = 0. It has a surface charge density σ1 = -2

Maru [3345]

1) For x = 6.6 cm, $E_x=3.47\cdot 10^6 N/C$

2) For x = 6.6 cm, $E_y=0$

3) For x = 1.45 cm, $E_x=-3.76\cdot 10^6N/C$

4) For x = 1.45 cm, $E_y=0$

5) Surface charge density at b = 4 cm: $+62.75 \mu C/m^2$

6) At x = 3.34 cm, the x-component of the electric field equals zero

7) Surface charge density at a = 2.9 cm: $+65.25 \mu C/m^2$

8) None of these regions

Explanation:

The electric field from an infinite charge sheet is perpendicular to it:

$E=\frac{\sigma}{2\epsilon_0}$

where

$\sigma$ is the surface charge density

$\epsilon_0=8.85\cdot 10^{-12}F/m$ represents vacuum permittivity

Outside the slab, the electric field behaves like that of an infinite sheet.

Consequently, the electric field at x = 6.6 cm (situated to the right of both the slab and sheet) results from the combination of the fields from both:

$E=E_1+E_2=\frac{\sigma_1}{2\epsilon_0}+\frac{\sigma_2}{2\epsilon_0}$

where

$\sigma_1=-2.5\mu C/m^2 = -2.5\cdot 10^{-6}C/m^2\\\sigma_2=64 \muC/m^2 = 64\cdot 10^{-6}C/m^2$

The field from the sheet points left (negative, inward), and the slab’s field points right (positive, outward).

Thus,

$E=\frac{1}{2\epsilon_0}(\sigma_1+\sigma_2)=\frac{1}{2(8.85\cdot 10^{-12})}(-2.5\cdot 10^{-6}+64\cdot 10^{-6})=3.47\cdot 10^6 N/C$

and the negative sign indicates a rightward direction.

Both the sheet’s and slab’s fields are perpendicular to their surfaces, directing along the x-axis, hence there's no y-component for the total field.

<pThus, the y-component totals zero.

This happens because both the sheet and slab stretch infinitely along the y-axis. Choosing any x-axis point reveals that the y-component of the field, generated by a surface element dS of either the sheet or slab, $dE_y$ , will be equal and opposite to the corresponding component from the opposite side, $-dE_y$ . Thus, the combined y-direction field is always zero.

This scenario resembles part 1), but the point here is

x = 1.45 cm

which lies between the sheet and the slab. The fields from both contribute leftward as the slab has a negative charge (resulting in an outward field). Thus, the total field computes to

$E=E_1-E_2$

Replacing with expressions from part 1), we get

$E=\frac{1}{2\epsilon_0}(\sigma_1-\sigma_2)=\frac{1}{2(8.85\cdot 10^{-12})}(-2.5\cdot 10^{-6}-64\cdot 10^{-6})=-3.76\cdot 10^6N/C$

where the negative illustrates a leftward direction.

This portion parallels part 2). Since both fields remain perpendicular to the slab and sheet, no component exists along the y-axis, thus the electric field's y-component is zero.

Notably, the slab behaves as a conductor, signifying charge mobility within it.

The net charge on the slab is positive, indicating a surplus of positive charge. With the negatively charged sheet on the left of the slab, positive charges shift towards the left slab edge (at a = 2.9 cm), while negative charges move to the right edge (at b = 4 cm).

The surface charge density per unit area of the slab is

$\sigma=+64\mu C/m^2$

This average denotes the surface charge density on both slab sides at points a and b:

$\sigma=\frac{\sigma_a+\sigma_b}{2}$ (1)

Additionally, the infinite sheet at x = 0 negatively charged $\sigma_1=-2.5\mu C/m^2$ , induces an opposite net charge on the slab's left surface, thus

$\sigma_a-\sigma_b = +2.5 \mu C/m^2$ (2)

Having equations (1) and (2) allows for solving the surface charge densities at a and b, yielding:

$\sigma_a = +65.25 \mu C/m^2\\\sigma_b = +62.75 \mu C/m^2$

We aim to compute the x-component of the electric field at

x = 3.34 cm

This point lies inside the slab, bounded at

a = 2.9 cm

b = 4.0 cm

In a conducting slab, the electric field remains at zero owing to charge equilibrium; thus, the x-component thereof in the slab is zero

From part 5), we determined the surface charge density at x = a = 2.9 cm is $\sigma_a = +65.25 \mu C/m^2$

As mentioned in part 6), conductors have zero electric fields internally. Since the slab is conductive, the electric field inside remains zero; therefore, the regions where the electric field is null are

2.9 cm < x < 4 cm

Thus, the suitable answer is

"none of these regions"

Learn more about electric fields:

8 0

3 months ago

In a demonstration, a 4.00 cm2 square coil with 10 000 turns enters a larger square region with a uniform 1.50 T magnetic field

serg [3582]

Explanation:

It is stated that,

Area of square coil, $A=4\ cm^2=0.0004\ m^2$