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Step2247 [10]
3 years ago
12

A coin is dropped in a 15.0 m deep well.

Physics
1 answer:
g100num [7]3 years ago
5 0

Answer:

About 1.75 seconds

Explanation:

We use the kinematic equation for free fall under the action of gravity:

h(t)=h_i -\frac{1}{2} g\,t^2\\0 = 15-\frac{1}{2} 9.8\,t^2\\t^2=\frac{15*2}{9.8} \\t=\sqrt{\frac{15*2}{9.8}} \\t\approx 1.75\,\,sec

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Answer:

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6 0
3 years ago
Solve for v<br> when<br> d = 10 and<br> t=5
VashaNatasha [74]

Answer:

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Explanation:

7 0
3 years ago
Read 2 more answers
Please help me
qaws [65]

-- We know that the y-component of acceleration is the derivative of the
y-component of velocity.

-- We know that the y-component of velocity is the derivative of the
y-component of position.

-- We're given the y-component of position as a function of time.

So, finding the velocity and acceleration is simply a matter of differentiating
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Now, the position function may look big and ugly in the picture.  But with the
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to be trying to write it out, given the text-formatting capabilities of the wonderful
envelope-pushing website we're working on here.

From the picture . . . . . y (t) = (1/2) (a₀ - g) t² - (a₀ / 30t₀⁴ ) t⁶

First derivative . . . y' (t) = (a₀ - g) t  -  6 (a₀ / 30t₀⁴ ) t⁵  =  (a₀ - g) t  -  (a₀ / 5t₀⁴ ) t⁵

There's your velocity . . . /\ .

Second derivative . . . y'' (t) = (a₀ - g) -  5 (a₀ / 5t₀⁴ ) t⁴ = (a₀ - g) -  (a₀ /t₀⁴ ) t⁴

and there's your acceleration . . . /\ .
That's the one you're supposed to graph.

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Pick, or look up, some reasonable figures for a₀ and t₀
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The big name in model rocketry is Estes.  Their website will give you
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6 0
3 years ago
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Ilia_Sergeevich [38]


i think true 
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5 0
3 years ago
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Korvikt [17]
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8 0
4 years ago
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