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11111nata11111 [884]
3 years ago
5

Positive electric charge Q is distributed uniformly along a thin rod of length 2a. The rod lies along the x-axis between x = -a

and x = +a (fig. 23.27). Calculate how much work you must do to bring a positive point charge q from infinity to the point x = +L on the x-axis, where:L> a.
Physics
1 answer:
Semmy [17]3 years ago
4 0

Answer:

kQq(ln(L+2a/L+a))/2a

Explanation:

This question is a example of potential difference calculation.

firstly we have to calculate the potential at point +L which we can find by integration.

steps for finding it

Firstly take a small section of dx length on road of charge (Qdx/2a) which is at x distance from the point +L.

Then find dv due to that using v=kq1/r formula

So required function will be dv=kQdx/(2a*x) we will integrate on both side with limit x=L+2a to x=L+a.

Then multiply v by charge which we will be carrying from infinity to that point.

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If two lines in a system of linear equations have the same slope and same y-intercept, how many solutions will the system have?
arsen [322]

Answer:

We will have <u>infinite solutions </u>to the system of linear equations.

Explanation:

Well, when we have two lines with the <u>same slopes and the same y-interception</u>, both of them <u>are overlapped, </u>so we will have <u>infinite solutions </u>to the system of linear equations.

This kind of system is called <u>dependent system.</u>

I hope it helps you!

8 0
3 years ago
3. A model rocket is launched straight upward at 58.8 m/s.
SIZIF [17.4K]

Let us assume that rocket only runs in initial energy and not using its own to flying.

Also , let upward direction is +ve and downward direction is -ve .

Initial velocity , u = 58.8 m/s .

Acceleration due to gravity , g=-9.8\ m/s^2 .

Final velocity , v - = 0 m/s .

We know , by equation of motion .

v^2-u^2=2gh\\\\2gh_{max}=0^2-58.8^2\\\\h_{max}=\dfrac{0^2-58.8^2}{-2\times 9.8}\\\\h_{max}= 176.4\ m

Hence, this is the required solution .

8 0
3 years ago
Need help with question 1) and 2) please ASAP ??? 15 points ?
Vaselesa [24]
1- interaction between 2 objects
2- action- reaction force pairs
3 0
3 years ago
a steel sphere and brass ring have diameter 25cm and 24.9cm at 15°C.If the sphere and the ring are heated together.what is the t
Oksi-84 [34.3K]

Answer:

Explanation:

Due to heat energy , metal expands . Formula for linear expansion is as follows .

L = l ( 1 + α Δt )

where L is expanded length , l is original length , α is coefficient of linear expansion and Δt is increase in temperature .

To pass the sphere through the ring , the diameter of both ring and sphere should be same after heating . Let after increase of temperature Δt , their diameter becomes same as L  . The linear coefficient of brass and steel are

20 x 10⁻⁶ and 12 x 10⁻⁶ respectively .

For steel sphere ,

L = 25 ( 1 + 12 x 10⁻⁶ Δt )

For brass ring

L = 24.9 ( 1 + 20 x 10⁻⁶ Δt )

25 ( 1 + 12 x 10⁻⁶ Δt ) = 24.9 ( 1 + 20 x 10⁻⁶ Δt )

1.004( 1 + 12 x 10⁻⁶ Δt ) =  ( 1 + 20 x 10⁻⁶ Δt )

1.004 + 12.0482 x 10⁻⁶ Δt  =   1 + 20 x 10⁻⁶ Δt

.004 = 7.9518 x 10⁻⁶ Δt

Δt  = 4000 / 7.9518

= 503⁰C.

final temp = 503 + 15 = 518⁰C  .

6 0
3 years ago
A diver jumps from a high platform and stays in the air for 1 second. How high was the platform?
ivolga24 [154]

Answer: Depends

Explanation:

Depends on how much the diver weighs.

6 0
2 years ago
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