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3 years ago

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!! I CANNOT RETAKE THIS!!

Physics

1 answer:

Varvara68 [4.7K]3 years ago

6 0

The answer to your question is,

4 kilometers north

-Mabel <3

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Calculate the electric field at the center of a square

pantera1 [17]

Answer:

$E_y=1175510.2\ N.C^{-1}$

The Magnitude of electric field is in the upward direction as shown directly towards the charge $q_1$ .

Explanation:

Given:

side of a square, $a=52.5\ cm$

charge on one corner of the square, $q_1=+45\times 10^{-6}\ C$

charge on the remaining 3 corners of the square, $q_2=q_3=q_4=-27\times 10^{-6}\ C$

Distance of the center from each corners $=\frac{1}{2} \times diagonals$

$diagonal=\sqrt{52.5^2+52.5^2}$

$diagonal=74.25\cm=0.7425\ m$

∴Distance of center from corners, $b=0.3712\ m$

Now, electric field due to charges is given as:

$E=\frac{1}{4\pi\epsilon_0}\times \frac{q}{b^2}$

For charge $q_1$ we have the field lines emerging out of the charge since it is positively charged:

$E_1=9\times 10^9\times \frac{45\times 10^{-6}}{0.3712^2}$

$E_1=2938775.5\ N.C^{-1}$

Force by each of the charges at the remaining corners:

$E_2=E_3=E_4=9\times 10^9\times \frac{27\times 10^{-6}}{0.3712^2}$

$E_2=E_3=E_4=1763265.3\ N.C^{-1}$

Now, net electric field in the vertical direction:

$E_y=E_1-E_4$

$E_y=1175510.2\ N.C^{-1}$

Now, net electric field in the horizontal direction:

$E_y=E_2-E_3$

$E_y=0\ N.C^{-1}$

So the Magnitude of electric field is in the upward direction as shown directly towards the charge $q_1$ .

8 0

3 years ago

What is the total number of pounds of carbon dioxide that 2 trees remove from the air each year?

Vikentia [17]

Answer:

On average, one acre of new forest can sequester about 2.5 tons of carbon annually. Young trees absorb CO2 at a rate of 13 pounds per tree each year. Trees reach their most productive stage of carbon storage at about 10 years at which point they are estimated to absorb 48 pounds of CO2 per year.

5 0

3 years ago

Read 2 more answers

Plane-polarized light is incident on a single polarizing disk, with the direction of E0 parallel to the direction of the transmi

Vladimir [108]

Answer: $65.9^{\circ}$

Explanation:

Given

Intensity must be reduced by a factor of 6

Intensity is given by $I=I_o\cos ^2\theta\\$

Substitute $I$ by $\frac{I_o}{6}$

$\Rightarrow \dfrac{I_o}{6}=I_o\cos ^2\theta\\\\\Rightarrow \cos^2\theta =\dfrac{1}{6}\\\\\Rightarrow \cos \theta=\dfrac{1}{\sqrt{6}}\\\\\Rightarrow \theta=65.9^{\circ}$

So, the disk must be rotated by an angle of $65.9^{\circ}$ .

7 0

2 years ago

Consider a box weight 2700 newtons resting on the ground. If the area of the box touching the ground is 1.5 square meters, what

Neko [114]

Answer:

The answer to your question is 1800 Pa

Explanation:

Data

Weight = 2700 N

Area = 1.5 m²

Pressure = ?

Formula

Pressure = Force / Area

The Pressure is defined as the force exerted per unit area.

-Substitution

Pressure = 2700 / 1.5

-Result

Pressure = 1800 Pa

The units of pressure are Pascals (Pa)

6 0

3 years ago

A plane electromagnetic wave, with wavelength 4.1 m, travels in vacuum in the positive direction of an x axis. The electric fiel

marusya05 [52]

(a) $7.32\cdot 10^7 Hz$

The frequency of an electromagnetic waves is given by:

$f=\frac{c}{\lambda}$

where

$c=3.0\cdot 10^8 m/s$ is the speed of light

$\lambda=4.1 m$ is the wavelength of the wave in the problem

Substituting into the equation, we find

$f=\frac{3.0\cdot 10^8 m/s}{4.1 m}=7.32\cdot 10^7 Hz$

(b) $4.60\cdot 10^8 rad/s$

The angular frequency of a wave is given by

$\omega = 2\pi f$

where

f is the frequency

For this wave,

$f=7.32\cdot 10^7 Hz$

So the angular frequency is

$\omega=2\pi(7.32\cdot 10^7 Hz)=4.60\cdot 10^8 rad/s$

(c) $1.53 m^{-1}$

The angular wave number of a wave is given by

$k=\frac{2\pi}{\lambda}$

where

$\lambda$ is the wavelength of the wave

For this wave, we have

$\lambda=4.1 m$

so the angular wave number is

$k=\frac{2\pi}{4.1 m}=1.53 m^{-1}$

(d) $1.03\cdot 10^{-6}T$

For an electromagnetic wave,

$E=cB$

where

E is the magnitude of the electric field component

c is the speed of light

B is the magnitude of the magnetic field component

For this wave,

E = 310 V/m

So we can re-arrange the equation to find B:

$B=\frac{E}{c}=\frac{310 V/m}{3\cdot 10^8 m/s}=1.03\cdot 10^{-6}T$

(e) z-axis

In an electromagnetic wave, the electric field and the magnetic field oscillate perpendicular to each other, and they both oscillate perpendicular to the direction of propagation of the wave. Therefore, we have:

- direction of propagation of the wave --> positive x axis

- direction of oscillation of electric field --> y axis

- direction of oscillation of magnetic field --> perpendicular to both, so it must be z-axis

(f) 127.5 W/m^2

The time-averaged rate of energy flow of an electromagnetic wave is given by:

$I=\frac{E^2}{2\mu_0 c}$