At what speed would a 3.00 x 10^4 kg airplane have to fly and with a momentum of 1.60 x 10^9 kg.m/s

Physics

1 answer:

Ymorist [56]3 years ago

3 0

Answer:

5.3×10⁴ m/s

Explanation:

From the question,

Momentum = mass× velocity

M = mV................ Equation 1

Where M = momentum of the airplane, m = mass of the airplane, V = Velocity of the airplane

make V the subject of the equation

V = M/m.................. Equation 2

Given: M = 1.6×10⁹ Kg.m/s, m = 3.0×10⁴ kg

Substitute these values into equation 2

V = 1.6×10⁹/3.0×10⁴

V = 5.3×10⁴ m/s

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zubka84 [21]

Here's a formula that's simple and useful, and if you're really in
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This one is so simple and useful that I'd suggest memorizing it,
so it's always in your toolbox.

This formula tells how far an object travels in how much time,
when it's accelerating:

   Distance = (1/2 acceleration) x (Time²).

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For your student who dropped an object out of the window,

   Distance = 19.6 m
   Acceleration = gravity = 9.8 m/s²

D = 1/2 G T²

    19.6 =   4.9   T²

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8 0

3 years ago

Read 2 more answers

A 0.2 kg hockey park is sliding along the eyes with an initial velocity of -10 m/s when a player strikes it with his stick, caus

babunello [35]

Answer:

The impulse applied by the stick to the hockey park is approximately 7 kilogram-meters per second.

Explanation:

The Impulse Theorem states that the impulse experimented by the hockey park is equal to the vectorial change in its linear momentum, that is:

$I = m\cdot (\vec{v}_{2} - \vec{v_{1}})$ (1)

Where:

$I$ - Impulse, in kilogram-meters per second.

$m$ - Mass, in kilograms.

$\vec{v_{1}}$ - Initial velocity of the hockey park, in meters per second.

$\vec{v_{2}}$ - Final velocity of the hockey park, in meters per second.

If we know that $m = 0.2\,kg$ , $\vec{v}_{1} = -10\,\hat{i}\,\left[\frac{m}{s}\right]$ and $\vec {v_{2}} = 25\,\hat{i}\,\left[\frac{m}{s} \right]$ , then the impulse applied by the stick to the park is approximately: