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3 years ago

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a 2.00 kg object is moving east at 4.00 m/s when it collides with a 6 kg object that is initially at rest. after the collision t

he larger object moves east at 1 m/s. what is the final velocity of the smaller object after the collision

1 answer:

baherus [9]3 years ago

6 0

Answer:

v = 1.00 m/s east

Explanation:

Conservation of momentum

Let east be the positive direction

2.00(4.00) + 6.00(0.00) = 2.00(v) + 6.00(1.00)

v = 1.00 m/s east

The two items have stuck together.

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A tuning fork labeled 392 Hz has the tip of each of its two prongsvibrating with an amplitude of 0.600mma) What is the maximum s

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a) 1.48 m/s

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$v_{max}=\omega A$

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$v_{max}$ is the maximum speed

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b) $3.0\cdot 10^{-5} J$

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$K=\frac{1}{2}mv_{max}^2$

where

$m=0.0270 g=2.7\cdot 10^{-5} kg$ is the mass of the fly

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After flying for 15 min in a wind blowing 42 km/h at an angle of 19° south of east, an airplane pilot is over a town that is 48

masha68 [24]

Answer:

The speed of the airplane relative to the air is 209.47km/hr

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Whenever we are solving a physics problem, it's really useful to start by drawing a diagram of the problem (See picture attached). It will help us visualize the problem better.

Now, we know that the plane flew for an amount of time of 15 minutes. For our dimensions to be the same, we need to turn those 15min to hours, like this:

$15min*\frac{1hr}{60min}=0.25hr$

Once our time is rewritten as hours, we can now calculate the velocity towards north of the plane.

$V=\frac{distance}{time}$

the plane traveled a distance to the north of 48km so the velocity is:

$V=\frac{48km}{0.25hr}$

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V=192km/hr j

Now, we can calculate the x and y-components of the velocity of the wind. The problem states that the wind is blowing at 42km/hr at an angle of 19° south of east, so the x and y-components of the velocity of the wind are:

$V_{x}=42km/hr*cos(-19^{o} )=39.71 i$

and

$V_{y}=42km/hr*sin(-19^{o} )=-13.67 j$

So the velocity of the wind can be expressed as a vector as:

$V_{wind}=(39.71i - 13.67j)km/hr$

Once we know this, we can find the velocity of the plane with respect of the wind on x and on y:

$V_{plane x}=V_{plane/wind x}+V_{wind x}$

$V_{plane/wind x}=V_{plane x}-V_{wind x}$

$V_{plane/wind x}=(0-39.71 i)km/hr$

$V_{plane/wind x}= -39.71 i km/hr$

and

$V_{plane y}=V_{plane/wind y}+V_{wind y}$

$V_{plane/wind y}=V_{plane y}-V_{wind y}$

$V_{plane/wind y}=192km/hr j - (- 13.67j)km/hr$

$V_{plane/wind x}= 205.67 j km/hr$

So the velocity of the plane with respect to the wind can be rewritten as:

$V_{plane/wind x}= (-39.71i + 205.67 j) km/hr$

Since the problem asks us to find the speed of the plane with respect to the wind, this means that we need to find the magnitude of the velocity, since the speed is a scalar defined to be the magnitude of the velocity.

so:

$speed=\sqrt{(-39.71)^{2}+(205.67)^{2} }$

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Therefore, the speed of the airplane relative to the air is 209.47km/hr

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