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Serjik [45]
3 years ago
15

A small charged bead has a mass of 3.6 g. It is held in a uniform electric field E = (200,000 N/C, up). When the bead is release

d, it accelerates upward with an acceleration of 24 m/s^2. What is the charge on the bead?
Physics
1 answer:
DiKsa [7]3 years ago
6 0

Answer:

The charge on the bead is  q = 6.084 *10^{-7}\ C

Explanation:

From the question we are told that

     The  mass of the bead is  m  =  3.6 \ g =  0.0036 \ kg

      The  magnitude of the electric field is  E =  200,000 \ N/C

       The  acceleration of the bead is  a =  24 m/s^2

Generally,  the  electric force on the bead is  mathematically represented as  

         F_ e =  q E

Where q is the charge on the bead

   Now the gravitational force opposing the upward movement of the bead is  mathematically represented as

       F_g  =  mg

Generally the net force on the bead is  mathematically represented as

       F  =  F_e - F_g = m* a

=>    qE -  mg  = ma

Now substituting values

       q *  200000 -   0.0036 *9.8   = 0.0036 * 24

     q = 6.084 *10^{-7}\ C

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What voltage battery would you need to send 2.5A of current through a light bulb of resistance 3.6 ohm
nadezda [96]
<h3><u>Given</u> :</h3>

Current flow light bulb = 2.5\sf{A}

Resistance of light bulb = 3.6Ω

<h3><u>To Find </u>:</h3>

We have to find voltage of battery.

<h3><u>Solution</u> :</h3>

➠ As per ohm's law, current flow through a conductor is directly proportional to the applied potential difference.

➝ V ∝ I

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Where, R is the resistance of conductor.

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A 75 kg bungee jumper leaps from a bridge. When he is 30 meters above the water, and moving at a speed of 20 m/s, the bungee cor
jasenka [17]

Answer:

k = 52.2 N / m

Explanation:

For this exercise we are going to use the conservation of mechanical energy.

Starting point. When it is 30 m high

        Em₀ = K + U = ½ m v² + m g h

Final point. Right when you hit the water

        Em_{f} = K_{e} = ½ k x²

in this case the distance the bungee is stretched is 30 m

        x = h

as they indicate that there are no losses, energy is conserved

        Em₀ = Em_{f}

       ½ m v² + m g h = ½ k h²

       k = \frac{m (v^{2} + 2 g h)}{h^{2} }

let's calculate

       k = \frac{75 \ ( 20^{2}  + 2 \ 9.8 \ 30)}{30^{2} }

       k = 52.2 N / m

3 0
2 years ago
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