AP Physics preparation

Test Preparation AP Test Preparation
1

<p>I have a few questions that I can;t figure out and I’m doing my AP PHysics studying. If anyone can help me out with any of these and show me how they are doen I would be tremendously appreciative.</p>

<li><p>A stream of water(initially horizontal) flows from a small hole in the side of a tank. The stream is 2.5 m from the side of the tank after falling .7m. How far below the surface of the water in the tank is the hole?</p></li>
<li><p>A straight horizontal pipe with a diameter of 1 cm and a length of 50 m carries oil with a density of 930 kg/m^3 (kilogram per meter cubed) and a viscosity of .12 Pa.s. The discharge rate is .8 kg/s at atmospheric pressure. Find the gauge pressure at the pipe input.</p></li>
<li><p>An 80 kg person weighs 40 N in fresh water and finds it difficult to stay afloat. In the Great Salt Lake the density is much higher and the person can stay afloat just submerged to the top of his head. What is the density of the Great Salt Lake?</p></li>
</ol>

2

<p>Bump for assistance</p>

3

<p>Anyone know how to do any of these?</p>

4

<p>I am not responsible if these are wrong!
1.
-.7=.5(-9.8)tsquared
t=.142857
if it went 2.5 m in that t its horiz. speed is 17.5 m/s
bernoulli's equation says
1000(9.8)h+atm=atm+.5<em>1000</em>17.5squared
atm cancels out
h=15.625m</p>

5

<p>4.
Sum of forces in y in fresh water=40 N
so:
80<em>9.8-Volumeofperson</em>1000=40(this is ASSUMING by fresh water it means PURE water)
so V=.744
Sum of forces in y in salt water=0
so:
80<em>9.8-.744</em>D=0
D=1053.763441 kg/m^3</p>

6

<p>Well, you set me in the right direction, but when you solved for t i think you forgot to take the square root. I got that t is .377 seconds, which gives me a final height of 2.2m. But thanks so much for the help.</p>

<p>Any of the other problems?</p>

7

<p>Thanks again for number four as well.</p>

8

<p>for the middle one, M/V=D
so .8/V=930
so the volume flow rate is 8.60215E-4 mcubed per second
using Poiseuille's Law
8.60215E-4=(PI(.005)^4(Pressure))/(8(.12)(50))
solve for pressure and I think you get 21028988.61 Pa, I'm not sure how confident I am on this one though, it seems pretty big.</p>

9

<p>I don't understand that though. Mass/volume = density, but .8 isn't the mass. Isn't .8 the volume flow rate already?</p>

10

<p>.8 is the MASS flow rate, I'm converting .8 to mcubed to get the volume flow rate</p>

11

<p>.8 is the mass its the mass leaving the tube per second, and .8 kg/V=932</p>

12

<p>Oh OK. Thanks so much. You really helped me out.</p>

13

<p>i dont think these are even SAT II questions.. i knew almost no equations, just concepts and got an 800</p>

14

<p>look at the title man, AP not SAT II</p>