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earnstyle [38]
3 years ago
6

I really need help with this, i dont know what to do first, PLEASE I NEED HELP DUE TOMORROW!!!

Physics
1 answer:
enot [183]3 years ago
5 0

Can I tell you what makes this problem so hard ?

It's having all the data WITHOUT HAVING THE STORY !

We first have to figure out what all those things are.  I mean, we don't even know what  F  is, what  d  is, what  Kef  or  Vi  is, or how  W  figures in to the whole thing.  You really have no mercy !

If my hunch is correct, the story goes like this:

-- There's an object sailing along, minding its own business, not bothering anybody, and its speed is 7.2 meters per second.

-- Somebody jumps out in front of the object and begins to push back on it with 215 Newtons of force, trying to slow it down and stop it.

-- The object is only able to go another 13 meters, pushing the guy backwards but slowing down, and then it stops.

-- The question is:  What is the mass of the object ?

Now I'll go ahead and solve the problem that I just invented:

-- Kinetic energy = (1/2) (mass) (speed²)

Before anybody touched it, the object's kinetic energy was

KE = (1/2) (mass) (7.2 m/s)²

KE = (25.92) x (mass)

-- Since that's the energy the object had, THAT's how much work the guy has to do in order to make the object stop.

Work = (force) x (distance)

Work = (215 N) x (13 meters)

Work = 2,795 N-m

-- And there you go.  The work the guy did to stop the object is the amount of energy the object had before he came along.

(25.92) x (mass of the object) = 2,795 N-m

Divide each side by 25.92:

Mass of the object = (2,795 N-m) / (25.92)

<em>Mass = 107.83 kilograms</em>

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Explanation:

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\dfrac{120(50 \ Hz)}{P_1}= \dfrac{120( 60 \Hz)}{P_2} \\ \\ \dfrac{6000 \ Hz}{P_1}= \dfrac{7200 \ Hz}{P_2} \\ \\ \dfrac{P_2}{P_1}=\dfrac{7200}{6000} \\ \\ \\ \dfrac{P_2}{P_1}= \dfrac{12}{10}

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3 years ago
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Answer:

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Q = \frac{2\pi L(t_{hf} - t_{cf}) }{\frac{1}{h_{hf}r_1}+\frac{ln(r_2/r_1)}{k_A} + \frac{ln(r_3/r_2)}{k_B} +\frac{1}{h_{cf}r_3}}

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t_{f} = Temperature at the outside of the pipe = 20 °C

r₁ =internal  radius of pipe = 4.0 cm

r₂ = Outer radius of pipe = 4.5 cm

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