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dybincka [34]
2 years ago
8

An outside force, Fo, brings two small metal spheres, A and B, at rest from a long distance away to a point where they are 1 met

ers apart. They are then connected by a non-conducting string of negligible mass. The spheres have mass mA = 5.8 g and mB = 9.5 g and both have equal positive charges of 5 μC. (You may assume the length of the string is much greater than the radii of the spheres.)
1)What was the total work done by the outside force to bring the spheres to the point described?
2)What was the total work done by the electric field when the two sphere were brought together as described?
3)What is the potential energy of the two sphere system after they have been brought together as described? (You may assume the spheres had zero potential energy when they were a long distance apart.)
4)Suppose you cut the string. At that instant, what is the magnitude of the acceleration of sphere A?
5)At that instant, what is the magnitude of the acceleration of sphere B?
6)After a very long time, what is the magnitude of the velocity of sphere A?
7)After a very long time, what is the magnitude of the velocity of sphere B?

Physics
1 answer:
Ksenya-84 [330]2 years ago
4 0

Answer:

1)

The total work done by outside force is W_{o}=0.294J

2)

The total work done by the electric field is -W_{o} =-0.294J

3)

The potential energy of the two sphere system is  PE = W_{o} = 0.294J

4)

The magnitude of the acceleration of sphere A is a_{A} = 66.868m/s^2

5)

The magnitude of the acceleration of sphere B is a_{B} = 26.747 \ m/s^2

6)

The magnitude of velocity sphere A after a very long time is  v_{A} = 10.251 \ m/s\

7) The magnitude of velocity sphere B after a very long time is  v_{B} = 4.1 m/s

 

Explanation:

The explanation is shown on the first and second uploaded image

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Ksenya-84 [330]

The concept of power is given by the relationship between intensity and area, that is to say that power is defined as

P = A*I

Our values are given under the condition of,

r_1 = 18m

r_2 = m

The power is proportional to the Area, and in turn, we know that the Area of a circle is the product between \pi  times the radius squared, therefore the power is proportional to the radius squared.

\text{Power} \propto r^2

For both panels we would have to

\frac{\text{Power by panel 1}}{\text{Power by panel 2}} = \frac{r_1^2}{r_2^2}

\frac{P_1}{P_2} = (\frac{18}{6})^2

\frac{P_1}{P_2} = 9

Therefore the correct option is option C.9

5 0
2 years ago
Read 2 more answers
Plz answer brainliest if right!
AleksandrR [38]

Answer: it is D. it is the only possible answer. use the process of elimination. which answers make sense?

4 0
3 years ago
3. Si usted duplicara la amplitud de un M.A.S. ¿cómo cambiaría la frecuencia, velocidad máxima, la aceleración máxima y la energ
WITCHER [35]

Answer:

Explanation:

la frecuencia = ω/2π, nada cambio

v(max) = ωA → ω2Α = 2ωA  duplicara velocidad máxima

a(max) = ω²Α → ω²2Α = 2ω²Α duplicara la aceleración máxima

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8 0
2 years ago
A diver jumps off a diving platform that is 20 meters long. Describe the transfer of energy that occurs during the fall.
kobusy [5.1K]

Answer: gravitational potential energy is converted into kinetic energy

Explanation:

When the diver stands on the platform, at 20 m above the surface of the water, he has some gravitational potential energy, which is given by

E=mgh

where m is the man's mass, g is the gravitational acceleration and h is the height above the water. As he jumps, the gravitational potential energy starts decreasing, because its height h above the water decreases, and he acquires kinetic energy, which is given by

K=\frac{1}{2}mv^2

where v is the speed of the diver, which is increasing. When he touches the water, all the initial gravitational potential energy has been converted into kinetic energy.

8 0
3 years ago
a man can row about 4kmperhr in a still water.he rows the boat 2km up the stream and 2km back to his starting point in 2hr.how f
Nat2105 [25]

<u>Answer</u>:

The stream flowing  at a speed of 2.828 \mathrm{km} / \mathrm{hr}

<u>Explanation</u>:

Given:  

Distance = 2km (both in upstream and downstream)  

The speed in still water be x km/hr.  

The speed in upstream = 4-x  

Speed in downstream = 4+x  

Solution:

We know that, Speed = distance/time  

So, Time = distance/speed

Therefore,  

2=\left(\frac{2}{4-x}\right)+\left(\frac{2}{4+x}\right)

2=\frac{2(4+x)+2(4-x)}{(4-x)(4+x)}

2(4-x)(4+x)=2(4+x)+2(4-x)

2(4-x)(4+x)=2(4+x+4-x)

By cancelling 2 on both sides,

16-x^{2}=8

x^{2}=16-8=8

x=\sqrt{8}

x=2.828 \mathrm{km} / \mathrm{hr}

Result:

Thus the speed of the stream is 2.828 \mathrm{km} / \mathrm{hr}

7 0
3 years ago
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