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1 year ago

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A source charge generates an electric field of 4286 N/C at a distance of 2.5 m. What is the magnitude of the source charge?

2 answers:

mrs_skeptik [129]1 year ago

8 0

The Answer is 3.0uc. I took the quiz.

choli [55]1 year ago

4 0

Answer : Charge, $q=3.0\ \muC$

Explanation :

It is given that :

Electric field, $E=4286\ N/C$

Distance, $r=2.5\ m$

We know that the electric field at any point is given by :

$E=k\dfrac{q}{r^2}$

k is the electrostatic constant.

q is the electric charge

r is the distance.

$q=\dfrac{E\times r^2}{k}$

$q=\dfrac{4286\ N/C\times (2.5\ m)^2}{9\times 10^9\ Nm^2/C^2}$

$q=0.0000029\ C$

$q=2.9\ \mu C$

or

$q=3.0\ \mu C$

The correct option is (b).

Hence, this is the required solution.

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A beam of electrons moves at right angles to a magnetic field of 4.5 × 10-2 tesla. If the electrons have a velocity of 6.5 × 106

defon

Answer:

$4.7\cdot 10^{-14}N$

Explanation:

For a charge moving perpendicularly to a magnetic field, the force experienced by the charge is given by:

$F=qvB$

where

q is the magnitude of the charge

v is the velocity

B is the magnetic field strength

In this problem,

$q=1.6\cdot 10^{-19} C$

$v=6.5\cdot 10^6 m/s$

$B=4.5\cdot 10^{-2} T$

So the force experienced by the electrons is

$F=(1.6\cdot 10^{-19}C)(6.5\cdot 10^6 m/s)(4.5\cdot 10^{-2} T)=4.7\cdot 10^{-14}N$

3 0

1 year ago

A 1-m-long monopole car radio antenna operates in the AM frequency of 1.5 MHz. How muchcurrent is required to transmit 4 W of po

Zanzabum

Answer:

The current needed to transmit Power of 4 W is 28.47 A

Solution:

As per the question:

Length of the antenna, $L_{a} = 1 m$

Frequency, $\vartheta = 1.5 MHz = 1.5\times 10^{6} Hz$

Power transmitted, $P_{t} = 4 W$

Now,

For a monopole antenna:

$\lambda_{a} = \frac{c}{\vartheta}$

where

$\lambda_{a}$ = wavelength transmitted by the antenna

c = speed of light in vacuum

$\lambda_{a} = \frac{3\times 10^{8}}{1.5\times 10^{6}} = 200 m$

Now,

Since, the value of $\lambda_{a}$ >> $L_{a}$ thus the monopole is a Hertian monopole.

The resistance is calculated as:

$R = \frac{1}{2}(\frac{dL_{a}}{\lambda_{a}})^{2}\times 80\pi^{2}$

$R = \frac{1}{2}(\frac{1}{200)^{2}\times 80\pi^{2} = 9.869\times 10^{- 3} = 9.869 m\Omega$

$P_{radiated} = P_{t}$

$P_{radiated} = \frac{R}{I^{2}}$

Now, the current I is given by:

$I = \sqrt{\frac{2P_{t}}{R}} = \sqrt{\frac{2\times 4}{9.869\times 10^{- 3}}} = 28.47 A$

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2 years ago

Many industries are powered via distant power stations. Calculate the current flowing through a 7,300m long 10. copper power lin

Oliga [24]

Answer:

Current, I = 1000 A

Explanation:

It is given that,

Length of the copper wire, l = 7300 m

Resistance of copper line, R = 10 ohms

Magnetic field, B = 0.1 T

$\mu_o=4\pi \times 10^{-7}\ T-m/A$

Resistivity, $\rho=1.72\times 10^{-8}\ \Omega-m$

We need to find the current flowing the copper wire. Firstly, we need to find the radius of he power line using physical dimensions as :

$R=\rho \dfrac{l}{A}$

$R=\rho \dfrac{l}{\pi r^2}$

$r=\sqrt{\dfrac{\rho l}{R\pi}}$

$r=\sqrt{\dfrac{1.72\times 10^{-8}\times 7300}{10\pi}}$

r = 0.00199 m

or

$r=1.99\times 10^{-3}\ m=2\times 10^{-3}\ m$

The magnetic field on a current carrying wire is given by :

$B=\dfrac{\mu_o I}{2\pi r}$

$I=\dfrac{2\pi rB}{\mu_o}$

$I=\dfrac{2\pi \times 0.1\times 2\times 10^{-3}}{4\pi \times 10^{-7}}$

I = 1000 A

So, the current of 1000 A is flowing through the copper wire. Hence, this is the required solution.

4 0

1 year ago

10. A girl pulls a wagon along a level path for a distance of 44 m. The handle of

Rina8888 [55]

The work done on the wagon is 3549 J

Explanation:

The work done by a force when moving an object is given by

$W=Fd cos \theta$

where :

F is the magnitude of the force

d is the displacement

$\theta$ is the angle between the direction of the force and of the displacement

In this problem we have the following data:

F = 87 N is the magnitude of the force

d = 44 m is the displacement of the wagon

$\theta=22^{\circ}$ is the angle between the direction of the force and the displacement

Substituting, we find the work done

$W=(87)(44)(cos 22^{\circ})=3549 J$

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

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2 years ago

Why doesn't a ball roll on forever after being kicked at a soccer game?

Naddika [18.5K]

Because of gravity and friction.

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2 years ago

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