## Hovercraft B/C

Justin72835
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### Re: Hovercraft B/C

What was the original formulation of Newton's Second Law that takes into account changes in mass?
$F=\frac{d(mv)}{dt}$

Interestingly, Newton was referring to momentum when formulating this law and not force  
A large cylindrical tank with height 10 meters and radius 2.5 meters is filled up entirely with water. A circular hole with a radius of 10 cm is punctured underneath the tank. Given that the top of the tank is open to the atmosphere, how long will it take for the entire volume of water to flow out of the tank (assuming no change in flow rate due to changing height?
Last edited by Justin72835 on March 12th, 2018, 8:44 pm, edited 1 time in total.
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### Re: Hovercraft B/C

$F=\frac{d(mv)}{dt}$

Interestingly, Newton was referring to momentum when formulating this law and not force  
A large cylindrical tank with height 10 meters and radius 2.5 meters is filled up entirely with water. A circular hole with a radius of 10 cm is punctured underneath the tank. Given that the top of the tank is open to the atmosphere, how long will it take for the entire volume of water to flow out of the tank?
$g(10 m) = \frac{v^2}{2}$
$v = \sqrt{g(5 m)}$
$\frac{V}{t} = \pi (0.1 m)^2 * \sqrt{g(5 m)}$
$\frac{\pi (2.5 m)^2(10 m)}{t} = \pi (0.1 m)^2 * \sqrt{g(5 m)}$
$t = \frac{\pi (2.5 m)^2(10 m)}{\pi (0.1 m)^2 * \sqrt{g(5 m)}} = \textbf{890 s}$

Justin72835
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### Re: Hovercraft B/C

A large cylindrical tank with height 10 meters and radius 2.5 meters is filled up entirely with water. A circular hole with a radius of 10 cm is punctured underneath the tank. Given that the top of the tank is open to the atmosphere, how long will it take for the entire volume of water to flow out of the tank?
$g(10 m) = \frac{v^2}{2}$
$v = \sqrt{g(5 m)}$
$\frac{V}{t} = \pi (0.1 m)^2 * \sqrt{g(5 m)}$
$\frac{\pi (2.5 m)^2(10 m)}{t} = \pi (0.1 m)^2 * \sqrt{g(5 m)}$
$t = \frac{\pi (2.5 m)^2(10 m)}{\pi (0.1 m)^2 * \sqrt{g(5 m)}} = \textbf{890 s}$
I got [b]446 seconds[/b], which is basically half your answer. With that said, I think you made a mistake in your second line. Since you're multiplying both sides by two, the expression underneath the square root is g(20m) instead of g(5m).

All your other work is correct though, so nice job! You can do the next one!
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### Re: Hovercraft B/C

Oh whoops, haha.
Find the final speed of a 100 kg roller coaster car that has dropped 50 m (from rest), gone back up 30 m, dropped 60 m, gone back up 30 m, and then at that level been pushed for 5 seconds with a force of 50 Newtons.

MattChina
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### Re: Hovercraft B/C

Oh whoops, haha.
Find the final speed of a 100 kg roller coaster car that has dropped 50 m (from rest), gone back up 30 m, dropped 60 m, gone back up 30 m, and then at that level been pushed for 5 seconds with a force of 50 Newtons.
2.5 m/s
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### Re: Hovercraft B/C

Oh whoops, haha.
Find the final speed of a 100 kg roller coaster car that has dropped 50 m (from rest), gone back up 30 m, dropped 60 m, gone back up 30 m, and then at that level been pushed for 5 seconds with a force of 50 Newtons.
2.5 m/s
No, not quite. Remember that the car has dropped and therefore gained speed.

MattChina
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### Re: Hovercraft B/C

Oh whoops, haha.
Find the final speed of a 100 kg roller coaster car that has dropped 50 m (from rest), gone back up 30 m, dropped 60 m, gone back up 30 m, and then at that level been pushed for 5 seconds with a force of 50 Newtons.
2.5 m/s
No, not quite. Remember that the car has dropped and therefore gained speed.
33.8m/s?
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### Re: Hovercraft B/C

2.5 m/s
No, not quite. Remember that the car has dropped and therefore gained speed.
33.8m/s?
Yep (at least that's what I remember). Your turn!

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### Re: Hovercraft B/C

A hovercraft with a mass of 100 kg going at 4m/s has its thrust turned off abruptly. It is traveling across the ice with a friction coefficient of 0.38. How far will it travel across the ice?
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### Re: Hovercraft B/C

A hovercraft with a mass of 100 kg going at 4m/s has its thrust turned off abruptly. It is traveling across the ice with a friction coefficient of 0.38. How far will it travel across the ice?
a = 9.8 m/s^2 * 0.38 = 3.724 m/s^2
0 = (4 m/s)^2 - 2 * 3.724 m/s^2 * d
d = 2.15 m