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dybincka [34]
3 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]3 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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A bald eagle is flying to the left with a speed of 34 meters
Shtirlitz [24]

Answer:

the speed after 3 seconds is 10 m/s

Explanation:

The computation of the speed is shown below:

As we know that

V = U  + at

Here,

U = 34 m/s

a =  - 8 m/s²

t = 3 Sec

V = velocity after 3 sec

V  = 34 + (-8)3

 = 34 - 24

 V = 10 m/s

Hence, the speed after 3 seconds is 10 m/s

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3 years ago
What type of nuclear radiation is emitted when carbon-14 decays
dalvyx [7]

Answer:

Beta radiation

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6 0
3 years ago
How many seconds will it take to travel 3,600 meters if your speed is 90 meters per second?
klio [65]
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4 0
3 years ago
Find the magnitude of the sum
shusha [124]

Answer:

\displaystyle |\vec{v_1}+\vec{v_2}|=4.15m

Explanation:

<u>Sum of Vectors in the Plane</u>

Given two vectors

\displaystyle \vec{v_1}\ ,\ \vec{v_2}

They can be expressed in their rectangular components as

\displaystyle \vec{v_1}=

\displaystyle \vec{v_2}=

The sum of both vectors can be done by adding individually its components

\displaystyle \vec{v_1}+\vec{v_2}=

If the vectors are given as a magnitude and an angle (M\ ,\ \theta ), each component can be found as

\displaystyle \vec{v_1}=

\displaystyle \vec{v_2}=

The first vector has a magnitude of 3.14 m and an angle of 30°, so

\displaystyle \vec{v_1}=

\displaystyle \vec{v_1}=

The second vector has a magnitude of 2.71 m and an angle of -60°, so

\displaystyle \vec{v_2}=

\displaystyle \vec{v_2}=

The sum of the vectors is

\displaystyle \vec{v_1}+\vec{v_2}=

\displaystyle \vec{v_1}-\vec{v_2}=

Finally, we compute the magnitude of the sum

\displaystyle |\vec{v_1}+\vec{v_2}|=\sqrt{(4.08)^2+(-0.78)^2}

\displaystyle |\vec{v_1}+\vec{v_2}|=\sqrt{17.25}

\displaystyle |\vec{v_1}+\vec{v_2}|=4.15m

3 0
3 years ago
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