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vovangra [49]
3 years ago
13

Two stationary positive point charges, charge 1 of magnitude 3.90 nC and charge 2 of magnitude 1.80 nC, are separated by a dista

nce of 46.0 cm. An electron is released from rest at the point midway between the two charges, and it moves along the line connecting the two charges.
What is the speed vfinal of the electron when it is 10.0 cm from charge 1?
Physics
1 answer:
soldi70 [24.7K]3 years ago
6 0

Answer:

v = 7793150 m/s

Explanation:

First, we are going to calculate the electrical potential in the point middle between the two charges

Remember that the electrical potential can be calculated as:

v = \frac{kQ}{r}

                 Where     k = 8.9874 x 10^{9} \frac{Nm^{2} }{C^{2} }

and it is satisfy the superposition principle, thus

v = \frac{8.9874x10^{9}(3.90x10^{-9} ) }{0.23} +  \frac{8.9874x10^{9}(1.80x10^{-9} ) }{0.23}

v = 222.73v

The electrical potential at 10 cm from charge 1 is:

v = \frac{8.9874x10^{9}(3.90x10^{-9} ) }{0.1} +  \frac{8.9874x10^{9}(1.80x10^{-9} ) }{0.36}

v = 395.44 v

Since the work - energy theorem, we have:

q\Delta v = \frac{mv^{2} }{2}

                     where q is the electron's charge and m is the electron's mass

Therefore:

v = \sqrt{\frac{2q\Delta v}{m} }

v = 7793150 m/s

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san4es73 [151]
Nine times more (squared speed)
8 0
3 years ago
Please help Need good grade
ivann1987 [24]

Answer:

The other angle is 120°.

Explanation:

Given that,

Angle = 60

Speed = 5.0

We need to calculate the  range

Using formula of range

R=\dfrac{v^2\sin(2\theta)}{g}...(I)

The range for the other angle is

R=\dfrac{v^2\sin(2(\alpha-\theta))}{g}....(II)

Here, distance and speed are same

On comparing both range

\dfrac{v^2\sin(2\theta)}{g}=\dfrac{v^2\sin(2(\alpha-\theta))}{g}

\sin(2\theta)=\sin(2\times(\alpha-\theta))

\sin120=\sin2(\alpha-60)

120=2\alpha-120

\alpha=\dfrac{120+120}{2}

\alpha=120^{\circ}

Hence, The other angle is 120°

6 0
4 years ago
The amplitude of a standing sound wave in a long pipe closed at the left end is sketched below.
ollegr [7]

(1) The harmonic number for the mode of oscillation is 3.

(2) The pitch (frequency) of the sound is 579.55 Hz

(3) The level of the water inside the vertical pipe is 0.1 m.

<h3>The harmonic number</h3>

The harmonic number for the mode of oscillation illustrated for the closed pipe is 3.

<h3>Frequency of the wave</h3>

The pitch (frequency) of the sound is calculated from third harmonic formula;

f = 3v/4L

where;

  • v is speed of sound
  • L is length of the pipe

f = (3 x 340) / (4 x 0.44)

f = 579.55 Hz

<h3>level of the water</h3>

wave equation for first harmonic of a closed pipe is given as

f  = v/(4L)

251.1  = 340/(4L)

4L = 340/251.1

4L = 1.35

L = 1.35/4

L = 0.34 m

level of water = 0.44 m - 0.34 m = 0.1 m

Thus, the level of the water inside the vertical pipe is 0.1 m.

Learn more about harmonics of closed pipes here: brainly.com/question/27248821

#SPJ1

3 0
1 year ago
Really need help! ;(<br> Could you please explain it? :) <br> GIVING 10 POINTS
ale4655 [162]

Answer:b

Explanation: If you look at the line on the graph, you can see that it is going downward, meaning it has a negative slope, and choice b is the only one that has a negative slope

4 0
2 years ago
Read 2 more answers
A tin can collapses if all air inside it is taken out why
Veseljchak [2.6K]

That only happens when the tin can is IN air.

In the familiar, comfy part of Earth's atmosphere where we live, the normal pressure of air is around 14.6 pounds on every square inch of everything. That's a big part of the reason why we're built with bodies that generate that same amount of pressure on the INSIDE pressing OUT. That way, we always have the same pressure pushing in both directions, so we know that we won't get crushed or blow up like balloons.

But we have to be careful with our bodies or other things when they're in places where the atmospheric pressure on the outside is NOT normal.

-- When a deep-sea diver goes hundreds of feet down in the ocean, and the pressure of the water is much GREATER than normal air.

-- When an astronaut has to go outside ... where there's NO air ... and fix something on the International Space Station.

When the pressure on the outside becomes very unusual, we have to wear special suits to protect our bodies from the unusual conditions.

The tin can in the story is a lot like our bodies. As long as it has air inside and air outside, the pressure is the same in both directions, so there's no particular force trying to deform the can. But ...

-- If you seal the can with the air inside it, take the can into a vacuum chamber, and pump the air out of the vacuum chamber, then the can only has pressure inside. It'll expand, and eventually spring a little hole in the metal, and all the air inside will blow out.

-- If you take all the air OUT of the can (so the can is REALLY 'empty'), then the pressure on it is all from the outside. In that situation, the can simply collapses, because there's nothing inside to provide pressure in the outward direction.

One more little thing to think about:

When you want some toothpaste to come drizzling out of the tube onto your brush, what do you do ? Do you perhaps squeeze the tube, and increase the pressure on the outside ?

4 0
3 years ago
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