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kati45 [8]
3 years ago
11

When a positive charge is released and moves along an electric field line: it moves to a position of lower electric potential an

d lower kinetic energy
true or false
Physics
1 answer:
Tju [1.3M]3 years ago
8 0

Hi there!

<u>False. </u>

Imagine if we released a positive charge between two parallel charged plates (one negative, and the other positive).

Since like charge repels, the positive charge would be repelled by the positive plate and would move towards negative plate, as opposite charge attracts.

The Electric potential of the POSITIVE charge increases as you move closer to the positive plate, with the potential decreasing as you move closer to the negative plate. Since the positive charge will accelerate towards the negative plate, it is LOSING electric potential energy.

As it loses electric potential energy, due to the conservation of energy, it must GAIN kinetic energy. There is work being done on the positive charge due to the potential difference between the plates, which increases its kinetic energy.

Therefore, the given statement is false.

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The feel of weight comes due to the normal reaction force given by the support. Hence, the condition of weightlessness is when the normal reaction force becomes zero. So, during free fall there is no support which can provide the normal reaction. Hence, the bungee jumper feels weightless as she falls towards the earth because of the lack of support force that balances gravity.

Hence, the answer is 3.

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3 years ago
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An electron enters a region between two large parallel plates made of aluminum separated by a distance of 2.0 cm and kept at a p
kotykmax [81]

Answer:

a)v=1.77\times 10^6\ m/s

b)v=3.872\times 10^6\ m/s

c)v=5.5\times 10^6\ m/s

d)v=6.7\times 10^6\ m/s

e)v=7.7\times 10^6\ m/s

Explanation:

Given that

d = 2 cm

V = 200 V

u=4\times 10^5\ m/s

We know that

F = E q

F = m a

E = V/d

So

m a =  q .V/d b            

a=\dfrac{q.V}{m.d}                 ---------1

The mass of electron

m=9.1\times 10^{-31}\ kg

The charge on electron

q=1.6\times 10^{-19}\ C

Now by putting the all values in equation 1

a=\dfrac{1.6\times 10^{-19}\times 200}{9.1\times 10^{-31}\times 0.02}\ m/s^2

a=1.5\times 10^{15}\ m/s^2

We know that

v^2=u^2+2as

a)

s = 0.1 cm

v^2=(4\times 10^5)^2+2\times 1.5\times 10^{15}\times 0.1\times 10^{-2}

v=\sqrt{3.16\times 10^{12}}\ m/s

v=1.77\times 10^6\ m/s

b)

s = 0.5 cm

v^2=(4\times 10^5)^2+2\times 1.5\times 10^{15}\times 0.5\times 10^{-2}

v=\sqrt{1.5\times 10^{13}}\ m/s

v=3.872\times 10^6\ m/s

c)

s = 1 cm

v^2=(4\times 10^5)^2+2\times 1.5\times 10^{15}\times 1\times 10^{-2}

v=\sqrt{3.06\times 10^{13}}\ m/s

v=5.5\times 10^6\ m/s

d)

s = 1.5 cm

v^2=(4\times 10^5)^2+2\times 1.5\times 10^{15}\times 1.5\times 10^{-2}

v=\sqrt{4.5\times 10^{13}}\ m/s

v=6.7\times 10^6\ m/s

e)

s = 2 cm

v^2=(4\times 10^5)^2+2\times 1.5\times 10^{15}\times 2\times 10^{-2}

v=\sqrt{6.06\times 10^{13}}\ m/s

v=7.7\times 10^6\ m/s

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The rain is minor compared to the wind damage.


Tornadoes are not really the strongest storms on Earth.

Welll ... I guess that could depend on what "strong storm" means.

Sure, a tornado is incredibly vicious and powerful. But it's only a few

miles wide, and it only lasts for a few hours and then it's all over.

The tropical cyclones ... hurricanes, monsoons, typhoons ... don't have

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