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ehidna [41]
3 years ago
6

A small ball of mass 2.00 kilograms is moving at a velocity 1.50 meters/second. It hits a larger, stationary ball of mass 5.00 k

ilograms. What is the kinetic energy of the system after the collision if the collision is elastic?
Physics
1 answer:
rewona [7]3 years ago
3 0

The kinetic energy of the small ball before the collision is

                             KE  =  (1/2) (mass) (speed)²

                                     = (1/2) (2 kg) (1.5 m/s)

                                     =    (1 kg)  (2.25 m²/s²)

                                     =        2.25 joules.

Now is a good time to review the Law of Conservation of Energy:

                     Energy is never created or destroyed. 
                     If it seems that some energy disappeared,
                     it actually had to go somewhere.
                     And if it seems like some energy magically appeared,
                     it actually had to come from somewhere.

The small ball has 2.25 joules of kinetic energy before the collision.
If the small ball doesn't have a jet engine on it or a hamster inside,
and does not stop briefly to eat spinach, then there won't be any
more kinetic energy than that after the collision.  The large ball
and the small ball will just have to share the same 2.25 joules.

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4 0
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to 10 Hz. Superimposed on this signal is 60-Hz noise with an amplitude of 0.1 V. It is desired to attenuate the 60-Hz signal to
givi [52]

Answer:

G \sqrt{1 +(\frac{f}{f_c})^{2n}} = 1

If we square both sides we got:

G^2 (1+\frac{f}{f_c})^{2n}= 1

We divide both sides by G^2 and we got:

(1+\frac{f}{f_c})^{2n} = \frac{1}{G^2}

Now we can apply log on both sides and we got:

2n ln(1+\frac{f}{f_c}) = ln (\frac{1}{G^2})

And solving for n we got:

n = \frac{ ln (\frac{1}{G^2})}{2ln(1+\frac{f}{f_c})}

And replacing we got:

n = \frac{ln (\frac{1}{0.1^2})}{2ln(1+\frac{60}{10})}

n = \frac{4.60517}{3.8918}=1.18

And since n needs to be an integer the correct answer would be n=2 for the filter order.

Explanation:

For this case we can use the formula for the Butterworth filter gain given by:

[tec] G = \frac{1}{\sqrt{1 +(\frac{f}{f_c})^{2n}}}[/tex]

Where:

G represent the transfer function and we want that G =0.1 since the desired signal is less than 10% of it's value

f_c = 10 Hz represent the corner frequency

f= 60 Hz represent the original frequency

n represent the filter order and that's the variable that we need to find

G \sqrt{1 +(\frac{f}{f_c})^{2n}} = 1

If we square both sides we got:

G^2 (1+\frac{f}{f_c})^{2n}= 1

We divide both sides by G^2 and we got:

(1+\frac{f}{f_c})^{2n} = \frac{1}{G^2}

Now we can apply log on both sides and we got:

2n ln(1+\frac{f}{f_c}) = ln (\frac{1}{G^2})

And solving for n we got:

n = \frac{ ln (\frac{1}{G^2})}{2ln(1+\frac{f}{f_c})}

And replacing we got:

n = \frac{ln (\frac{1}{0.1^2})}{2ln(1+\frac{60}{10})}

n = \frac{4.60517}{3.8918}=1.18

And since n needs to be an integer the correct answer would be n=2 for the filter order.

7 0
3 years ago
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