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olga2289 [7]
3 years ago
7

Un avión de rescate en Alaska deja caer un paquete de provisiones a un grupo de exploradores extraviados. Si el avión viaja hori

zontalmente a 40 m/s y una altura de 100 m sobre el suelo, ¿Donde cae el paquete en relación con el punto que se soltó?
Physics
1 answer:
posledela3 years ago
4 0

Answer:

180.4 m

Explanation:

The package in relation to the point where it was released falls a certain distance that is calculated by applying the horizontal motion formulas , as the horizontal speed of the plane and the height above the ground are known, the time that It takes the package to reach its destination and then the horizontal distance (x) is calculated from where it was dropped, as follows:    

$V_{ox}=v_x = 40 \ m/s$

   h = 100 m  

    x =?

     Height formula h:

     $h=g \times \frac{t^2}{2}$

      Time t is cleared:

     $t = \sqrt{\frac{2h}{g}}$

      $t = \sqrt{\frac{2 \times 100}{9.8}}$

      t = 4.51 sec

 Horizontal distance formula x:

       $x=V_x \times t$

        x = 40 m / sec x 4.51 sec

        x = 180.4 m

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Coulomb's law for the magnitude of the force FFF between two particles with charges QQQ and Q′Q′Q^\prime separated by a distance
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Answer:

{F_{tot} = -1.092*10^{-2}N

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q_1=-13.5nC=-13.5*10^{-9}C

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The separation between charges q_3 and q_1 is

d_{31}= (-1.240mm)-(-1.735mm)=0.495mm=4.95*10^{-4} m

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F_{31} = (9*10^9NmC^{-2})\dfrac{47*10^{-9}*(-13.5*10^{-9})}{4.95*10^{-4}} =-0.01152N

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The separation between charges q_2  and q_3 is

d_{23} =-1.240mm-0=-1.240*10^{-3}m

therefore, the force between them is

F_{23} =(9*10^9Nm^2C^{-2})\dfrac{47*10^{-9}C*(-1.735*10^{-9}C)}{-1.240*10^{-3}m}=5.94*10^{-4}N

Therefore the total force on charge q_3 is

F_{tot} =5.94*10^{-4}-0.01152N = -0.010926N\\\\\boxed{F_{tot} = -1.092*10^{-2}N}

8 0
3 years ago
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