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A drinking fountain projects water at an initial angle of 50 degrees above the horizontal, and the water reaches a maximum height of .150 m above the point of exit a. Calculate the speed at which the water leaves the fountain. b. The radius of the fountain's exit hole is 4.00 x 10^-3 m. Calculate the volume rate of flow of the water. c. The fountain is fed by a pipe that at once point has a radius of 7.00 x 10^-3 m and is 3.00 below the fountain's opening. The density of water is 1.0 x 10^3 kg/m^3. Calculate the gauge pressure of the feeder pipe at this point. The fuck is the volume flow rate? How do we find the velocity it exits the pipe at? How is it possible to find the pressure only knowing the radius the depth? This doesn't make any sense. |
Volume flow rate is the rate of flow of a volume of a liquid as a function of time (I believe). For finding the initial velocity, you know the angle and the maximum height. Use kinematics to find the y component of velocity needed to get it to that height against the acceleration due to gravity. From there, you can use trig to determine the initial velocity. I didn't really look to hard at the rest. |
This may help for the Bimodal part. I'll try and find the others. Surprisingly, I've never covered these in the topics of Histograms... |
Polynomials Question: 7. Draw sketch graphs of y=2sinx and y=x and show that the equation x - 2sinx = 0 has a root near x=2. Find this root correct to two decimal places. Working out: ![]() f(x) = x - 2sinx f'(x) = 1 - 2cosx Therefore x2 = 2 - [(2-2sin(2))÷ (1-2cos(2))] Which = 3.93... Which doesn't sound right. The answer should be around 1.9 But when I try the same thing in Radians instead of Degrees mode, I get 2 - 0.099 which = 1.90. I just want to know why I must use Radians to get the right answer, and not Degrees? |
sin, cos and tan are functions that repeat every 2pi. When you put it in "degree mode", you basically stretch them out to repeat every 360 degrees. It's telling the calculator, "Okay, whatever I put in there, divide by 180/pi before actually calculating it." It's kind of like if you had a function "d = mpg(x)" that would tell you how many miles d you could go on x gallons of gas - if you give the gas as milliliters, the answer isn't going to be nearly right. |
Hi, I have a bunch of calculus problems that I don't fully understand. Would someone mind helping me get these answers? I have to draw them up in paint as I am uncertain as to how to make a differentiation sign in html. calculus problem 2.png [Mod-Edit: Please use the link option in the image uploader for large, screen-stretching images] |
No. It still works. The the integrand you get is something along these lines: New function of x - some constant that happens when you plug 2 into the indefinite integral. Then, you derive that and the constant disappears. From there, basically, it just undoes the integral (more or less) but the chain rule has to be used when you derive it. |
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