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Gnesinka [82]
3 years ago
15

A point charge q1 is held stationary at the origin. A second charge q2 is placed at point a, and the electric potential energy o

f the pair of charges is 5.4 x 10^-8J. When the second charge is moved to point b, the electric force on the charge does -1.9x10^-8J of work. What is the electric potential energy of the pair of charges when the second charge is at point b?
Physics
1 answer:
MAVERICK [17]3 years ago
7 0

Answer:

U_{b}=+7.3*10^{-8}J

Explanation:

When a particle moves from a point where the potential energy is Ua to point where it is Ub,the change in potential energy is is equal to work done

So

W_{a-b}=U_{a}-U_{b}\\ U_{b}=U_{a}-W_{a-b}\\

Where Wa-b here is negative this means Ub is greater Ua. Therefore potential energy increases

So

U_{b}=(+5.4*10^{-8}J )-(-1.9*10^{-8}J)\\U_{b}=+7.3*10^{-8}J  

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2 years ago
You are coasting on your 12-kg bicycle at 13 m/s and a 5.0-g bug splatters on your helmet. The bug was initially moving at 1.5 m
Brut [27]

Answer:

a) Pi,c = 1066 kgm/s

b) Pi,b = 0.0075 kgm/s  

c) ΔV = - 0.0007 m/s

d) ΔV = - 0.0008 m/s

Explanation:

Given:-

- The mass of the bicycle, mc = 12 kg

- The mass of passenger, mp = 70 kg

- The mass of the bug, mb = 5.0 g

- The initial speed of the bicycle, vpi = 13 m/s

- The initial speed of the bug, vbi = 1.5 m/s

Find:-

a.What is the initial momentum of you plus your bicycle?

b.What is the initial momentum of the bug?

c.What is your change in velocity due to the collision the bug?

d.What would the change in velocity have been if the bug were traveling in the opposite direction?

Solution:-

- First we will set our one dimensional coordinate system, taking right to be positive in the direction of bicycle.

- The initial linear momentum (Pi,c) of the passenger and the bicycle would be:

                       Pi,c = vpi* ( mc + mp)

                       Pi,c = 13* ( 12+ 70 )

                       Pi,c = 1066 kgm/s  

- The initial linear momentum (Pi,b) of the bug would be:

                       Pi,b = vbi*mb

                       Pi,b = 0.005*1.5

                       Pi,b = 0.0075 kgm/s  

- We will consider the bicycle, the passenger and the bug as a system in isolation on which no external unbalanced forces are acting. This validates the use of linear conservation of momentum.

- The bicycle, passenger and bug all travel in the (+x) direction after the bug splatters on the helmet.

                       Pi = Pf

                       Pi,c + Pi,b = V*(mb + mc + mp)

Where,    V : The velocity of the (bicycle, passenger and bug) after collision.

                      1066 + 0.0075 = V*( 0.005 + 12 + 70 )

                      V = 1066.0075 / 82.005

                      V = 12.9993 m/s

- The change in velocity is Δv = 13 - 12.9993 =  - 0.00070 m/s      

- If the bug travels in the opposite direction then the sign of the initial momentum of the bug changes from (+) to (-).

- We will apply the linear conservation of momentum similarly.

                      Pi = Pf

                      Pi,c + Pi,b = V*(mb + mc + mp)        

                      1066 - 0.0075 = V*( 0.005 + 12 + 70 )

                      V = 1065.9925 / 82.005

                      V = 12.99911 m/s

- The change in velocity is Δv = 13 - 12.99911 =  -0.00088 m/s      

7 0
3 years ago
Read 2 more answers
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