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An object is tracked by a radar station and found to have a position vector given by r = (4570-160 t) i + 2930j + 130k , with r in meters and t in seconds. The radar station's x axis points east, its y axis north, and its z axis vertically up. If the object is a 300 kg meteorological missile, what are (a) its linear momentum (in unit vector notation) and (b) the magnitude of the net force on it?

Respuesta :

Answer:

a)P= -4800 i

b)F= 0

Explanation:

Given that

r = (4570-160 t) i + 2930 j + 130 k

We know that velocity is rate of change of the space vector.

V= dr/dt

r = (4570-160 t) i + 2930 j + 130 k

dr/dt= -160 i + 0 + 0

dr/dt= -160 i

V= -160 i

It means that velocity is in only x-direction

We also know that acceleration is the rate of change of velocity .

a= dV/dt

V= -160 i

dV/dt=0

So we can say that acceleration is zero.

 a= 0

From Second law of Newton'

Force =  Mass x acceleration

F= 300 x 0

F= 0

We know that linear momentum P

P = m V

Given that m= 300 kg

P = 300 x (-160 i)

P= -4800 i

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