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gayaneshka [121]
3 years ago
11

A positively charged body exerts an electric field that causes a positively charged particle to move away from it. If work is do

ne to push this particle towards the body, what form of energy will the particle have?
a. mechanical energy
b. electric potential energy
c. heat energy
d. kinetic energy
Physics
1 answer:
ollegr [7]3 years ago
8 0

Answer:

b. electric potential energy.

Explanation:

The energy required to move a charge against the electric field is known as the electric potential energy. As in above case positively charged body is exerting an electric field on the positive charge. As the same charges repel so the charge tend to move away. In order to push it towards the body we need a work done. As it is hard to push the positive charged particle towards the positive electric field. So in the cases like these particle occupies the electric potential energy.

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How dose silicon shape our technological reality?
vichka [17]
Silicon is used in the manufacturing of microchips and is also used in motherboards. We use it in our daily lives. It is a high tech element. 
3 0
3 years ago
An object weighs 32 newtons. What is its mass if a gravitometer indicates that g = 8.25 m/s?
Elan Coil [88]

Explanation:

weight=mass×g

32=mass×8.25

mass=

\frac{128}{33}  = 3.878kg

8 0
2 years ago
A plane electromagnetic wave travels northward. At one instant, its electric field has a magnitude of 9.6 V/m and points eastwar
lara31 [8.8K]

Answer:

The values is  B  = 3.2 *10^{-8} \  T

The  direction is out of the plane

Explanation:

From the question we are told that

  The  magnitude of the electric field is  E  =  9.6 \  V/m

 

The  magnitude of the magnetic field is mathematically represented as

       B  = \frac{E}{c}

where c is the speed of light with value

      B  = \frac{ 9.6}{3.0 *10^{8}}

     B  = 3.2 *10^{-8} \  T

Given that the direction off the electromagnetic wave( c ) is  northward(y-plane ) and  the electric field(E) is eastward(x-plane ) then the magnetic field will be acting in the out of the page  (z-plane  )

     

3 0
3 years ago
A very long uniform line of charge has charge per unit length λ1 = 4.68 μC/m and lies along the x-axis. A second long uniform li
Kitty [74]

Answer:

E_{net} = 6.44 \times 10^5 N/C

Explanation:

As we know that electric field due to infinite line charge distribution at some distance from it is given as

E = \frac{2k \lambda}{r}

now we need to find the electric field at mid point of two wires

So here we need to add the field due to two wires as they are oppositely charged

Now we will have

E_{net} = \frac{2k\lambda_1}{r} + \frac{2k\lambda_2}{r}

now plug in all data

\lambda_1 = 4.68 \muC/m

\lambda_2 = 2.48 \mu C/m

r = 0.200 m

now we have

E_{net} = \frac{2k}{r}(4.68 + 2.48)

E_{net} = \frac{2(9\times 10^9)}{0.200}(7.16 \times 10^{-6})

E_{net} = 6.44 \times 10^5 N/C

8 0
2 years ago
A satellite at a particular point along an elliptical orbit has a gravitational potential energy of 5100 MJ with respect to Eart
serious [3.7K]

To solve this problem we will apply the theorem given in the conservation of energy, by which we have that it is conserved and that in terms of potential and kinetic energy, in their initial moment they must be equal to the final potential and kinetic energy. This is,

E_{initial} = E_{final}

PE_{initial}+KE_{initial} = PE_{final}+KE_{final}

Replacing the 5100MJ for satellite as initial potential energy, 4200MJ for initial kinetic energy and 5700MJ for final potential energy we have that

KE_{final} = (PE_{initial}+KE_{initial} )-PE_{final}

KE_{final} = (5100+4200)-5700

KE_{final} = 3600MJ

Therefore the final kinetic energy is 3600MJ

5 0
3 years ago
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