Question

An airplane flies east at a speed of 120 m/s the airplane has a mass of 3200

Answer 1

1) The momentum of the airplane is $3.84\cdot 10^5 kg m/s$ east

2) The impulse applied to the airplane is 32,000 kg m/s west

Explanation:

1)

The momentum of an object is defined as

$p=mv$

where

m is the mass of the object

v is its velocity

The airplane in this problem has

m = 3200 kg is its mass

v = 120 m/s east is its velocity

So, its momentum is

$p=(3200)(120)=3.84\cdot 10^5 kg m/s$

And since momentum is a vector, it also has a direction, which is the same as the velocity (east).

2)

The impulse applied to an object is equal to the change in momentum of the object:

$I=\Delta p = m(v-u)$

where

m is the mass

v is the  final velocity

u is the initial velocity

For the airplane in this problem,

m = 3200 kg

u = 120 m/s

v = 110 m/s

So, the impulse is

$I=(3200)(110-120)=-32,000 kg m/s$

And the negative sign means the direction of the impulse is opposite to the airplane motion, so west.

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