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2 years 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? (Us

e k=) 1. 2 µC 3. 0 µC 1. 2 C 3. 0 C.

1 answer:

svp [43]2 years ago

8 0

The magnitude of the source charge is 3 μC which generates 4286 N/C of the electric field. Option B is correct.

What does Gauss Law state?

It states that the electric flux across any closed surface is directly proportional to the net electric charge enclosed by the surface.

$Q = \dfrac {ER^2}k$

Where,

$E$ = electric force = 4286 N/C

$k$ = Coulomb constant = $8.99 \times 10^9 \rm\ N m ^2 /C ^2$

$Q\\
$ = charges = ?

$r$ = distance of separation = 2.5 m

Put the values in the formula,

$Q = \dfrac {4286\times 2.5 ^2}{8.99 \times 10^9 }\\\\
Q = 3\rm \ \mu C$

Therefore, the magnitude of the source charge is 3 μC.

Learn more about Gauss's law:

brainly.com/question/1249602

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A seagull flies at a velocity of 9.00 m/s straight into the wind.

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c)Compared to the 133.33 seconds it would take without the wind, the overall round-trip time is affected by the wind.

<h3>What is velocity?</h3>

The change of distance with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.

The given data in the problem is

A seagull flies at a velocity, $\rm v_{SA} = 9 \ m/sec$

The time the bird takes,t=18.0 min

The distance traveled relative to the earth = 6.00 km

a)

The seagull's relative velocity with reference to the ground as;

$\rm v_{sg} = \frac{6.00 \times 10^3 \ m }{(20 min) \times \frac{60 s }{1 \ min}} \\\\ v_{sg}= 5.00 \ m/sec$

Air velocity with reference to the ground is;

$\rm v_{AG}= v_{SG}-v_{SA} \\\\ v_{AG} = 5.00 \ m/sec - 9.00 \ m/sec \\\\ v_{AG} = -4.00 \ m/sec$

b)

If the bird turns around and flies with the wind, The time will he take to return 6.00 km is;

$\rm v_{SG}=v_{SA}+v_{AG} \\\\ v_{SG}=-900 \ m/sec +(-4.00 \ m/sec) \\\\ v_{SG}= -13.00 \ m/sec$

The time the bird takes;

$\rm t = \frac{x_{SG}}{v_{SG}} \\\\ t = \frac{6.00 \times 10^3 \ m }{13.00 \ m/sec } \\\\ t = 462 m/sec \\\\ t = 7 \ min \ and \ 42 \ sec$

c)\

The total round-trip time compared to what it would be with no wind. is;

$\rm t = 20 \ min( \frac{60 \ sec }{1 \ min} )+ 462 \ sec \\\\ t = 1200 \ sec +6 462 \ ec \\\\ t= 1662 \ sec$

The time for the round trip is;

$\rm t = \frac{12 \times 10^ 3 }{ 9 \ m/sec } \\\\ t = 1333.33 \ sec$

Hence the wind's speed, the time bird would need to fly back the total round-trip time will be 4 m/s, 7 minutes and 42 seconds and 1333.33 sec.

To learn more about the velocity, refer to the link: brainly.com/question/862972.

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2 years ago

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Answer:

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Explanation:

Applying the law of conservation of momentum, we have:

Δ $p_{x = 0}$

$p_{x} = p"_{x}$

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Δ $p_{y} = 0$

$p_{y} = p"_{y}$

$0 = m_{1} v"_{1} - m_{2} v"_{2} sin$ Θ (Equation 2)

From Equation 1:

$v"_{2} = \frac{m_{1}v_{1}}{m_{2}cos}$ Θ

From Equation 2:

$m_{2} v"_{2}$ sinΘ = $m_{1} v_{1}$

$v"_{1} = \frac{m_{2} v"_{2}sinΘ}{m_{1} }$

Replacing Equation 3 in Equation 4:

$v"_{1}=\frac{m_{2}\frac{m_{1}v_{1}}{m_{2}cosΘ}sinΘ}{m_{1}}$

$v"_{1}=v_{1}\frac{sinΘ}{cosΘ}$

$v"_{1}=v_{1}tan$ Θ (Equation 5)

And we found Θ from the Equation 5:

tanΘ= $\frac{v"_{1}}{v_{1}}$

Θ= $tan^{-1}(\frac{v"_{1}}{v_{1}})$

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3 years ago