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Talja [164]
3 years ago
13

Find an expression for the minimum frictional coefficient needed to keep a car with speed v on a banked turn of radius R designe

d for speed v0.
Physics
1 answer:
solniwko [45]3 years ago
4 0
"v0" means that there are no friction forces at that speed
<span>mgsinΘ = (mv0²/r)cosΘ → the variable m cancels </span>
<span>sinΘ/cosΘ = tanΘ = v0² / gr 
</span><span>Θ = arctan(v0² / gr) </span>

<span>When v > v0, friction points downslope: </span>
<span>mgsinΘ + µ(mgcosΘ + (mv²/r)sinΘ) = (mv²/r)cosΘ → m cancels: </span>
<span>gsinΘ + µ(gcosΘ + (v²/r)sinΘ) = (v²/r)cosΘ </span>
<span>µ = ((v²/r)cosΘ - gsinΘ) / (gcosΘ + (v²/r)sinΘ) </span>
<span>where Θ is defined above. </span>

<span>When v > v0, friction points upslope: </span>
<span>mgsinΘ - µ(mgcosΘ + (mv²/r)sinΘ) = (mv²/r)cosΘ → m cancels: </span>
<span>gsinΘ - µ(gcosΘ + (v²/r)sinΘ) = (v²/r)cosΘ </span>
<span>µ = (gsinΘ - (v²/r)cosΘ) / (gcosΘ + (v²/r)sinΘ) </span>
<span>where Θ is defined above. </span>
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A particle with charge 7.76×10^(−8)C is moving in a region where there is a uniform 0.700 T magnetic field in the +x-direction.
kodGreya [7K]

Answer:

The  z-component of the force is  \= F_z  =  0.00141 \ N    

Explanation:

From the question we are told that

          The charge on the particle is q =  7.76 *0^{-8} \  C    

           The magnitude of the magnetic field is  B =  0.700\r i \ T

            The  velocity of the particle toward the x-direction is  v_x  =  -1.68*10^{4}\r  i  \ m/s

           The  velocity of the particle toward the y-direction is

v_y  =  -2.61*10^{4}\ \r j  \ m/s

           The  velocity of the particle toward the z-direction is

v_y  =  -5.85*10^{4}\ \r k  \ m/s

Generally the force on this particle is mathematically represented as

          \= F  =  q (\= v   X  \= B )

So  we have    

          \= F  =  q ( v_x \r  i + v_y \r  j  +  v_z \r k  )  \ \ X \ (  \= B i)

         \= F  = q (v_y B(-\r  k) + v_z B\r j)      

  substituting values

       \= F  = (7.7 *10^{-8})([ (-2.61*10^{4}) (0.700)](-\r  z) + [(5.58*10^{4}) (0.700)]\r y)    

      \= F=  0.00303\ \r j +0.00141\ \r k                  

So the z-component of the force is  \= F_z  =  0.00141 \ N    

Note :  The  cross-multiplication template of unit vectors is  shown on the first uploaded image  ( From Wikibooks ).

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