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3 years ago

A kangaroo jumps to a vertical height of 2.8 m. How long was it in the air before returning to earth

Physics

1 answer:

BaLLatris [955]3 years ago

8 0

The answer would be 2.8m height on earth takes
2.8=1/2*9.8*t^2 => <span>s = ut +1/2at^2 </span>

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A 6.72 particle moves through a region of space where an electric field of magnitude 1270 N/C points in the positive direction,

Romashka [77]

Answer:

$v_{y} = -104 m/s$

Explanation:

Using:

Force = electric field * charge

$F=e*q$

Force = magnitude of charge * velocity * magnetic field * sin tither

$F_{x2}= |q|*v*B*sin \alpha$

Force on particle due to electric field:

$F_{x1}= E*q = (1270N/C)*(-6.72*10^{-6} ) = -8.53*10^{-3}$

Force on particle due to magnetic field:

$F_{x2}= |q|*v*B*sin \alpha = (6.72*10^{-6} )*(1.15)*(sin90)*v = (7.728*10^{-6})*(v)$

$F_{x2}$ is in the positive x direction as $F_{x1}$ is in the negative x direction while net force is in the positive x direction.

Magnetic field is in the positive Z direction, net force is in the positive x direction.

According to right hand rule, Force acting on particle is perpendicular to the direction of magnetic field and velocity of particle. This would mean the force is along the y-axis. As this is a negatively charged particle, the direction of the velocity of the particle is reversed. Therefore velocity of particle, v, has to be in the negative y direction.

Now,

$F_{xnet}- F_{x1 } = F_{x2 }$

$(6.13*10^{-3}) - (8.53*10^{-3} ) = (7.728*10^{-6})*(v)$

$v = (F_{xnet} - F_{x1}) / (F_{x2} )$

$=((6.13*10^{-3} ) - (8.53*10^{-3})) / (7.728*10^{-6})$

$= (- 104.25) m/s$

$v_{y} = -104 m/s$

8 0

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