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3 years ago

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It took a crew 9 h 36 min to row 8 km upstream and back again. If the rate of flow of the stream was 2 km/h, what was the rowing

speed of the crew in still water?

1 answer:

babunello [35]3 years ago

7 0

Answer:

3 km/h

Explanation:

Let's call the rowing speed in still water x, in km/h.

Rowing speed in upstream is: x - 2 km/h

Rowing speed in downstream is: x + 2 km/h

It took a crew 9 h 36 min ( = 9 3/5 = 48/5) to row 8 km upstream and back again. Therefore:

8/(x - 2) + 8/(x + 2) = 48/5 (notice that: time = distance/speed)

Multiplying by x² - 2², which is equivalent to (x-2)*(x+2)

8*(x+2) + 8*(x-2) = (48/5)*(x² - 4)

Dividing by 8

(x+2) + (x-2) = (6/5)*(x² - 4)

2*x = (6/5)*x² - 24/5

0 = (6/5)*x² - 2*x - 24/5

Using quadratic formula

$x = \frac{2 \pm \sqrt{(-2)^2 - 4(6/5)(-24/5)}}{2(6/5)}$

$x = \frac{2 \pm 5.2}{2.4}$

$x_1 = \frac{2 + 5.2}{2.4}$

$x_1 = 3$

$x_2 = \frac{2 - 5.2}{2.4}$

$x_2 = -1\; 1/3$

A negative result has no sense, therefore the rowing speed in still water was 3 km/h

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1) Important concepts

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2) Part a

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$v_{max}=A\omega =0.002mx2000\frac{rad}{s}=4\frac{m rad}{s}=4\frac{m}{s}$

4) Part c

In order to find the total mechanical energy of the oscillator we can use this formula:

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5) Part d

When we want to find the force from the 2nd Law of Newton we know that F=ma.

At the maximum displacement we know that X=A, and in order to that happens $cos(\omega t +\phi)=1$ , and we also know that the maximum acceleration is given by::

$|\frac{d^2X}{dt^2}|=A\omega^2$

So then we have that:

$F=ma=mA\omega^2$

And since we have everything we can find the force

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6) Part e

When the mass it's at the half of it's maximum displacement the term $cos(\omega t +\phi)=1/2$ and on this case the acceleration would be given by;

$|\frac{d^2X}{dt^2}|=A\omega^2 cos(\omega t +\phi)=A\omega^2 \frac{1}{2}$

And the force would be given by:

$F=ma=\frac{1}{2}mA\omega^2$

And replacing we have:

$F=\frac{1}{2}(0.01Kg)(0.002m)(2000\frac{rad}{s})^2 =40N$

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