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SIZIF [17.4K]
3 years ago
14

An electron is released from rest at the negative plate of a parallel plate capacitor. The charge per unit area on each plate is

σ = 1.69E-7 C/m2, and the plates are separated by a distance of 1.75E-2 m. How fast is the electron moving just before it reaches the positive plate?
Physics
1 answer:
Andreas93 [3]3 years ago
3 0

Answer:

v=1.08\times 10^7\ m/s

Explanation:

Initial speed of the electron, u = 0

The charge per unit area of each plate, \dfrac{Q}{A}=1.69\times 10^{-7}\ C/m^2

Separation between the plates, d=1.75\times 10^{-2}\ m

An electron is released from rest, u = 0

Using equation of kinematics,

v^2-u^2=2ad..........(1)

Acceleration of the electron in electric field, a=\dfrac{qE}{m}............(2)

Electric field, E=\dfrac{\sigma}{\epsilon_o}............(3)

From equation (1), (2) and (3) :

v=\sqrt{\dfrac{2q\sigma d}{m\epsilon_o}}

v=\sqrt{\dfrac{2\times 1.6\times 10^{-19}\times 1.69\times 10^{-7}\times 1.75\times 10^{-2}}{9.1\times 10^{-31}\times 8.85\times 10^{-12}}}

v = 10840393.1799 m/s

or

v=1.08\times 10^7\ m/s

So, the electron is moving with a speed of 1.08\times 10^7\ m/s before it reaches the positive plate. Hence, this is the required solution.

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A boat moves through the water of a river at 10m/s relative to the water, regardless of the boat ‘s direction . If the water in
katen-ka-za [31]

Answer:

The appropriate solution is "61.37 s".

Explanation:

The given values are:

Boat moves,

= 10 m/s

Water flowing,

= 1.50 m/s

Displacement,

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Now,

The boat is travelling,

= 10+1.50

= 11.5 \ m/s

Travelling such distance for 300 m will be:

⇒ v = \frac{d}{t} \ sot \ t

      =\frac{d}{v}

On putting the values, we get

      =\frac{300}{11.5}

      =26.08 \ s

Throughout the opposite direction, when the boat seems to be travelling then,

= 10-1.50

= 8.5 \ m/s

Travelling such distance for 300 m will be:

⇒ v=\frac{v}{t} \ sot \ t

      =\frac{d}{v}

On putting the values, we get

      =\frac{300}{8.5}

      =35.29 \ s

hence,

The time taken by the boat will be:

= 26.08+35.29

= 61.37 \ s

8 0
3 years ago
Which major planet has the largest . . . A. semimajor axis? B. average orbital speed around the Sun? C. orbital period around th
Yuliya22 [10]
<h2>Mercury, Neptune, and Jupiter </h2>

Explanation:

  • Mercury has the largest semimajor axis that is 5.791 x 107 in km.
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  • Neptune has the longest orbital speed around the sun of any planet in the Solar System which is equivalent to 164.8 years (or 60,182 Earth days)
  • Jupiter has the largest eccentricity.

Hence, the answer is Mercury, Neptune, and Jupiter respectively.

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Answer:

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Explanation:

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∑F = Fr

F_{engine}-F_{brake} =F_{r}\\F_{r}=79-31\\F_{r}=48[N]

The movement remains forward, since the force produced by the movement is greater than the braking force.

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