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3 years ago

8

A +15 nC point charge is placed on the x axis at x = 1.5 m, and a -20 nC charge is placed on the y axis at y = -2.0m. What is th

e magnitude of the electric field at the origin?

1 answer:

IceJOKER [234]3 years ago

8 0

Answer: $E=75\ N/m$

Explanation:

Given

First charge of $q_1=15\ nC$ is placed at $x=1.5\ m$

Second charge $q_2=-20\ nC$ is placed at $y=-2\ m$

Electric field is given by

$E=\frac{kq}{r^2}$

Electric field due to $q_1$ is away from it

$E_1=\frac{9\times 10^9\times 15\times 10^{-9}}{(1.5)^2}$

$E_1=60\ N/m$

Electric field due to $q_2$

$E_2=\frac{9\times 10^9\times 20\times 10^{-9}}{2^2}$

$E_2=45\ N/m$

Net electric field will be vector addition of two

$\vec{E_{net}}=\vec{E_1}+\vec{E_2}$

$\vec{E_{net}}=-60\hat{i}-45\hat{j}$

Magnitude of Electric field is

$E=\sqrt{60^2+45^2}$

$E=75\ N/m$

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rewona [7]

Answer: 75V

Explanation:

Given that,

total resistance (Rtotal) = 150Ω

Current (I) = 0.5A

Change in electric potential (V) = ?

Recall that potential difference is the product of amount of current and the amount of resistance in the circuit. And its unit is volts.

So, apply the formula V = I x Rtotal

V = 0.5A x 150Ω

V = 75V

Thus, the change in electric potential across the circuit is 75 Volts

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3 years ago

The bright-line spectrum of an element in the gaseous phase is produced as 1. protons move from lower energy states to higher en

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Answer:

4. Electrons move from higher energy states to lower energy states.

Explanation:

When electrons fall from a higher (excited) energy state to a lower energy state, it loses/gives out energy.

This energy is given out by the emission of photons (quanta of light) by the electron.

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3 years ago

What speed would a proton need to achieve in order to circle Earth 1790 km above the magnetic equator, where the Earth's mag- ne

irinina [24]

Answer:

The velocity is 31.25 m/s and direction is toward west.

Explanation:

Given that,

Distance $h= 1790 km = 1.790\times10^{6}\ m$

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Mass of proton $m=1.673\times10^{-21}\ Kg$

Radius of earth $R =6.38\times10^{6}\ m$

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We need to calculate the speed

Using formula of magnetic field

$Bvq=\dfrac{mv^2}{r}$

$v=\dfrac{Bqr}{m}$

Put the value into the formula

$v=\dfrac{4\times10^{-8}\times1.6\times10^{-19}\times8170000}{1.673\times10^{-21}}$

$v=31.25\ m/s$

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3 years ago

While entering a freeway, a car accelerates from rest at a rate of 2.40 m/s2 for 12.0 s. (a) Draw a sketch of the situation. (b)

ArbitrLikvidat [17]

Answer:

a) See attached picture, b) We know the initial velocity = 0, initial position=0, time=12.0s, acceleration= $2.40m/s^{2}$ , c) the car travels 172.8m in those 12 seconds, d) The car's final velocity is 28.8m/s

Explanation:

a) In order to draw a sketch of the situation, I must include the data I know, the data I would like to know and a drawing of the car including the direction of the movement and its acceleration, just like in the attached picture.

b) From the information given by the problem I know:

initial velocity =0

acceleration = $2.40m/s^{2}$

time = 12.0 s

initial position = 0

c)

unknown:

displacement.

in order to choose the appropriate equation, I must take the knowns and the unknown and look for a formula I can use to solve for the unknown. I know the initial velocity, initial position, time, acceleration and I want to find out the displacement. The formula that contains all this data is the following:

$x=x_{0}+V_{x0}t+\frac{1}{2}a_{x}t^{2}$

Once I got the equation I need to find the displacement, I can plug the known values in, like this:

$x=0+0(12s)+\frac{1}{2}(2.40\frac{m}{s^{2}} )(12s)^{2}$

after cancelling the pertinent units, I get that my answer will be given in meters. So I get:

$x=\frac{1}{2} (2.40\frac{m}{s^{2}} )(12s)^{2}$

which solves to:

$x=172.8m$

So the displacement of the car in 12 seconds is 172.8m, which makes sense taking into account that it will be accelerating for 12 seconds and each second its velocity will increase by 2.4m/s.

d) So, like the previous part of the problem, I know the initial position of the car, the time it travels, the initial velocity and its acceleration. Now I also know what its final position is, so we have more than enough information to find this answer out.

I need to find the final velocity, so I need to use an equation that will use some or all of the known data and the unknown. In order to solve this problem, I can use the following equation:

$a=\frac{V_{f}-V_{0} }{t}$

Next, since I need to find the final velocity, I can solve the equation just for that, I can start by multiplying both sides by t so I get:

$at=V_{f}-V_{0}$

and finally I can add $V_{0}$ to both sides so I get:

$V_{f}=at+V_{0}$

and now I can proceed and substitute the known values:

$V_{f}=at+V_{0}$

$V_{f}=(2.40\frac{m}{s^{2}}} (12s)+0$

which solves to:

$V_{f}=28.8m/s$

8 0

3 years ago

Read 2 more answers

You stand on a straight desert road at night and observe a vehicle approaching. This vehicle is equipped with two small headligh

choli [55]

Answer:

They can be seen from a distance of 4.372 kilometers.

Explanation:

Using the Reyligh creterion for diffraction through a circular aperture we have

$\frac{x}{D}=\frac{1.22\lambda }{d}$

where symbol's have their usual meaning

thus applying values we get

$D=\frac{dx}{1.22\lambda }$

$\therefore D=\frac{0.633\times 4.61\times 10^{-3}}{1.22\times 547\times 10^{-9}}\\\\D=4372.77m\\=4.372km$

5 0

3 years ago