Member: Rom

Quote:
If dust falls at approximately .5meters per second off a 200ft high wall. the wind is blowing at a 45 degree angle up with a force of 1 meter an hour. when does the dust hit the ground?

The point of this problem is to find out the resulting downwards velocity of the dust given that the wind is blowing at an upward angle.
Note that since we are not told the mass of the dust it is assumed we can neglect the force of gravity for this problem and that the particles velocity will be constant.

The velocity upwards that the wind gives the dust is 1 m/h * sin(45 degrees) = sqrt(2)/2 m/h = sqrt(2)/7200 m/s; (3600 seconds to an hour)

The upwards velocities just add so the resulting velocity is

0.5 + sqrt(2)/7200 m/s = (3600 + sqrt(2))/7200 m/s ~ 0.5 m/s; (note the wind has almost no effect, are you sure it wasn't 1 m/s windspeed?)

The time it will take to fall 200 ft is 200 feet divided by the velocity (in feet/second, we'll have to convert from m/s to ft/s)

There are about 3.05 feet/meter so the velocity in feet/s is 0.5m/s * 3.05 ft/m = 1.525 ft/s

So the time is now given by 200ft/1.525 ft/s = 131.15 seconds

RomFeb 10, 2014

+6251

all rational numbers that don't have a finite decimal representation are repeating.

RomFeb 10, 2014

+6251

Clipboard01.jpg

from the picture it should be clear that

tan(theta) = 74/70

theta = tan ^-1(74/70) = 49.6 degrees

RomFeb 9, 2014

+6251

Quote:
9x^2 + x^2 = 576

9x ² + x ² = 576; add the x ² terms together

10x ² = 576; divide both sides by 10

x ² = 576/10; take the square root of both sides

x = +/- sqrt(576/10); compute the square roots

x = +/- 24/sqrt(10); multiply top and bottom by sqrt(10) to get it proper form

x = +/- 24sqrt(10)/10; depending on what your teacher likes to see you can either leave it like this or do the division by 10 to obtain

x = +/- 2.4sqrt(10)

RomFeb 9, 2014

+6251

I assume you mean on the calculator here.

Selecting sin should be pretty obvious, just click the sin button.

To get sin ^-1 you can either

1) click 2nd then click sin

2) click and hold sin until it changes to sin ^-1 then release

RomFeb 9, 2014

+6251

Quote:
r=sqrt(20-r)

r = sqrt(20-r)

r ² = 20 - r

r ² + r - 20 = 0

(r + 5)(r - 4) = 0

r = -5, r = 4

but we're not done yet, since we squared both sides we have to check to see that all the solutions are valid for the original equation

4 = sqrt(16) = sqrt(20-4) so r=4 is indeed valid

on the other hand since a square root is never negative -5 is not a valid solution.

Thus the only solution to the original equation is r=4

RomFeb 8, 2014

+6251

Quote:
X-1 * X = X^2-X .. why not X^2-1 ???

as you've written it

X - 1 * X = X - X = 0

you probably meant

(X-1) * X

so you much multiply each term inside the parens by the X, thus

(X-1) * X = X*X - 1*X = X ² - X

RomFeb 8, 2014

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