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lesya692 [45]
3 years ago
5

An insulating rod having linear charge density λ = 40.0 mC/m and linear mass density μ = 0.100 kg/m is released from rest in a u

niform electric field E = 100 V/m directed perpendicular to the rod.
(a) Determine the speed of the rod after it has traveled 2.00 m.
(b) What If

Physics
1 answer:
Nezavi [6.7K]3 years ago
5 0

Answer:

a) Speed of the rod = 0.4m/s

b) The part b says what if ? how does your answer to part (a) change if the electric field is not perpendicular to the rod?

If the electric field is not perpendicular to the rod, then a rrsolution has to be don eon the x -axis by resolving with respect tp Cos theta, the value pf the speed will still remains the same.

Explanation:

The detailed and mathematical derivation is shown in the attachment.

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describe the journey that light takes from the sun to your eye when you are looking at fish in a pond
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Rays of the light fast moves to the water and to the fish and then they are reflected

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7 0
2 years ago
An engine draws energy from a hot reservoir with a temperature of 1250 K and exhausts energy into a cold reservoir with a temper
dimulka [17.4K]

Answer:

The power output of this engine is  P =  17.5 W

The  the maximum (Carnot) efficiency is  \eta_c  = 0.7424

The  actual efficiency of this engine is  \eta _a  = 0.46

Explanation:

From the question we are told that

    The temperature of the hot reservoir is  T_h = 1250 \ K

      The temperature of the cold reservoir  is  T_c  =  322 \ K

     The energy absorbed from the hot reservoir is E_h  = 1.37 *10^{5} \ J

       The energy exhausts into  cold reservoir is  E_c  = 7.4 *10^{4} J

The power output is mathematically represented as

      P  =  \frac{W}{t}

Where t is the time taken which we will assume to be 1 hour =  3600 s  

W is the workdone which is mathematically represented as

      W =  E_h  -E_c

substituting values

       W = 63000 J

So

    P =  \frac{63000}{3600}

    P =  17.5 W

The Carnot efficiency is mathematically represented as

          \eta_c  =  1 - \frac{T_c}{T_h}

         \eta_c  =  1 - \frac{322}{1250}

         \eta_c  = 0.7424

The actual efficiency is mathematically represented as

        \eta _a  =   \frac{W}{E_h}

substituting values

         \eta _a  =  \frac{63000}{1.37*10^{5}}

         \eta _a  = 0.46

     

7 0
4 years ago
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