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Snezhnost [94]
3 years ago
14

Madeline fires a bullet horizontally. The rifle is 1.9 meters above the ground. The bullet travels 200 meters horizontally befor

e it hits the ground. What speed did Madeline's bullet have when it exited the rifle?
Physics
1 answer:
Dennis_Churaev [7]3 years ago
7 0

Answer:

322.6 m/s

Explanation:

Given that there are two components of position;

x= vot

y component is;

y= 1.9 - gt^2/2

When the bullet hits the ground, y=0

1.9 -gt^2/2 =0

Where g = 10ms^-2

t= √2 × 1.9/10

t= 0.62 secs

Therefore,

x= vot

vo = x/t

Where, x= 200 m

vo= 200/0.62 =

vo= 322.6 m/s

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A particular coaxial cable is comprised of inner and outer conductors having radii 1 mm and 3 mm respectively, separated by air.
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Answer:

The value is  \rho_s  =  4.026 *10^{-6} \  C/m^2

Explanation:

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    The radius of the outer conductor is  r_2 = 3 \ mm = 0.003 \  m

    The potential at the outer conductor is  V = 1.5 kV  =  1.5 *10^{3} \  V

Generally the capacitance per length of the capacitor like set up of the two conductors is

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Here \epsilon_o is the permitivity of free space with value  \epsilon_o =  8.85*10^{-12} C/(V \cdot m)

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=>   C= 50.6 *10^{-12} \  F/m

Generally given that the potential  of the outer conductor with respect to the inner conductor is positive it then mean that the outer conductor is positively charge

Generally the line  charge density of the outer  conductor is mathematically represented as

      \rho_l  =  C *  V

=>   \rho_d  =  50.6*10^{-12} *  1.5*10^{3}

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        \rho_s  =  \frac{\rho_l }{2 \pi * r_2 }    here 2 \pi r = (circumference \ of \ outer \  conductor  )

=>    \rho_s  =  \frac{7.59 *10^{-8} }{2* 3.142 * 0.003 }

=>    \rho_s  =  4.026 *10^{-6} \  C/m^2

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