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Trava [24]
3 years ago
11

How long does it take electrons to get from a car battery to the starting motor? Assume the current is 300 A and the electrons t

ravel through a copper wire with cross-sectional area 0.21 cm2 and length 0.85 m.The number of charge carriers per unit volume is 8.49 $ 1028 m#
Physics
1 answer:
klasskru [66]3 years ago
7 0

Answer:

808.25 seconds

Explanation:

From the relation, Velocity = \frac{Length}{time}

If t = time

v = velocity

l = length

t = \frac{l}{v} \\

The current density, J, of a charge is given by:

J = nqv..........................(1)\\J = \frac{I}{A} ................(2)

Where number of charge carriers per unit volume n = 8.49 * 10²⁸

Charge of the electron, q = 1.6 * 10⁻¹⁹

Cross sectional area of the wire, A = 0.21 cm² = 0.000021 m²

Equating (1) and (2)

nqv = \frac{I}{A} \\v = \frac{I}{nqA} \\

Since t = l/v.......(3)

Put the formula for v into equation (3)

t = \frac{nqAl}{I} \\t = \frac{8.49*10^{28}*1.6 *10^{-19}* 2.1 * 10^{-5} * 0.85}{300}

t = 808.25 seconds

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A satellite orbits a planet of unknown mass in a circular orbit of radius 2.3 x 104 km. The gravitational force on the satellite
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Answer:

The  kinetic energy is KE  =  7.59  *10^{10} \  J

Explanation:

From the question we are told that

       The  radius of the orbit is  r =  2.3 *10^{4} \ km  = 2.3  *10^{7} \ m

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where v is the speed of the satellite which is mathematically represented as

     v  = \sqrt{\frac{G  M}{r^2} }

=>  v^2  =  \frac{GM }{r}

substituting this into the equation

      KE  =  \frac{ 1}{2} *\frac{GMm}{r}

Now the gravitational force of the planet is mathematically represented as

      F_g  = \frac{GMm}{r^2}

Where M is the mass of the planet and  m is the mass of the satellite

 Now looking at the formula for KE we see that we can represent it as

     KE  =  \frac{ 1}{2} *[\frac{GMm}{r^2}] * r

=>    KE  =  \frac{ 1}{2} *F_g * r

substituting values

       KE  =  \frac{ 1}{2} *6600 * 2.3*10^{7}

         KE  =  7.59  *10^{10} \  J

 

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