Midterm 2 review. Day 15: Review

Transcription

1 Midterm 2 review Physics 1010: Dr. Eleanor Hodby Day 15: Review Reminders: No new HW this week MIDTERM 2 ON THURSDAY Todays review Qs on website in full after class Reading quiz a week today: see website G116 Computing lab (2 nd floor) for PHET simulations

2 Review session schedule (Note locations carefully) am Tuesday (13 th ) EH - helproom pm Tuesday (13 th ) KH - Phys lounge 11-12pm Wednesday (14 th ) CM Helproom 3-4pm Wednesday (14 th ) LD - Phys lounge 3-4pm Friday (16 th ) Exam debrief CM. - Phys lounge NO PSS or office hours on Monday 19th

3 - 40 multiple choice questions, worth 80 points. - Bring or memorize your 9 digit student ID, and bring your student ID card. - There will be no early or late exams given and no make-up exams. - Exam will be closed book. Mid term on Thursday - TWO 3 by 5 inch formula cards. You can WRITE anything on it BY HAND. - Basic scientific calculator (NOT graphical) and pencil/eraser - Calculator cannot connect to outside world. No calculators on cell phones or laptops allowed. - No sharing of calculators. - No spare calculators available - CLEAN foreign language to English dictionaries allowed. I will be checking them for illegal notes. - Exam grades and solutions will be posted after the exam on D2L - Seating: Seats will be assigned alphabetically. Look for your name on a desk as you enter the room.

4 Midterm preparation Prepare by applying the principles we have learned practice. You CANNOT memorize answers to specific questions. Make a formula card now with the important equations. Go over homeworks, class clicker questions, questions in the book (worked examples). - Not sure how to get the answer take it to the helproom.

5 Conservation of energy Midterm 2 summary W ext - W friction = DPE + DKE - Work done by a force = F d // = energy changed from one form to another - GPE = mgh, KE = ½ mv 2, PPE = PV, SPE= ½ kx 2, Thermal energy = constant T - Ramps, roller coasters, balls, skiers. - Power = energy/time: 1W = 1J/s Bernoulli s equation Sound E tpv = P + ½ rv 2 + rgh - Conservation of energy for an incompressible fluid - Problem solving: E tpv is the same throughout the system (if no ext.work, friction) Calculate E tpv at 2 points and equate - Wave basics: f, A, T, l, v (and relationships) - How to get different notes from a violin - Harmonics on a violin string

6 Conservation of energy I m moving boxes into the back of a truck that sits 1.4m above the ground. I use a 15m long ramp and a small cart to help me. When loaded, the cart has a mass of 20 kg. Friction is negligible. What is the direction of the net force on the cart, when it is on the ramp and noone is pushing it? A

7 Cart mass = 20kg If the cart has a mass of 20kg, how much PE does it gain when you push it up the ramp? a J b. 196 J c. 28 J d J e J

8 Cart mass = 20kg If the cart has a mass of 20kg, how much PE does it gain when you push it up the ramp? a J b. 196 J c. 28 J d J e J GPE = mgh

9 Cart mass = 20kg What is the force required to lift the cart vertically at a steady speed? a. 20 N b. 28 N c. 300 N d. 196 N e. Can t tell without more information

10 Cart mass = 20kg What is the force required to lift the cart vertically at a steady speed? a. 20 N b. 28 N c. 300 N d. 196 N e. Can t tell without more information F lift mg Steady speed a = 0 a = 0 F net = 0 F net = F lift mg F lift = mg = = 196 N

11 Cart mass = 20kg What is the force required to push the cart up the ramp at a steady speed? a. 196 N b N c. 0 N d N e. Something else

12 Cart mass = 20kg What is the force required to push the cart up the ramp at a steady speed? a. 196 N b N c. 0 N d N e. Something else W ext - W fric = DGPE + DKE F ext d = mgh F ext = mgh/d = / 15 = 18.3 N

13 Cart mass = 20kg What length of ramp would I need if I could only push at a maximum force of 12 N? a m b m c m d m e. Can t tell without more information

14 Cart mass = 20kg What length of ramp would I need if I could only push at a maximum force of 12 N? a m b m c m d m e. Can t tell without more information F ext d = mgh d = mgh / F ext = / 12 = 22.9 m

15 Cart mass = 20kg A cart is at the top of the ramp and I accidently let it go. It rolls down the ramp and then into thick grass on a horizontal lawn. The lawn rubs on the cart and produces a force of friction of 80N. How far across the lawn does the cart travel before coming to a stop? a. 2.3 m b. 6.9 m c. 4.4 m d. 3.4 m e. Can t tell without more information

16 Cart mass = 20kg A cart is at the top of the ramp and I accidently let it go. It rolls down the ramp and then into thick grass on a horizontal lawn. The lawn rubs on the cart and produces a force of friction of 80N. How far across the lawn does the cart travel before coming to a stop? a. 2.3 m b. 6.9 m c. 4.4 m d. 3.4 m e. Can t tell without more information GPE KE Thermal (= work done by friction) GPE = W friction = F friction d d = mgh/f friction = 274.4/80 = 3.4 m

17 Cart mass = 20kg Now assume that the cart breaks and I have to push each 20 kg box up the ramp. Friction is now significant and is 50N. What is the total work that I do pushing a box to the top of the ramp at a steady speed a. 750 J b J c J d J e. Something else

18 Cart mass = 20kg Now assume that the cart breaks and I have to push each 20 kg box up the ramp. Friction is now significant and is 50N. What is the total work that I do pushing a box to the top of the ramp at a steady speed a. 750 J b J c J d J e. Something else W ext - W fric = DGPE + DKE W ext = DGPE + W fric = mgh + F fric d = J = J

19 Cart mass = 20kg Now assume that the cart breaks and I have to push each 20 kg box up the ramp. Friction is now significant and is 50N. Work to push one box up ramp = J. If I have an average mechanical power output of 60W, how many boxes do I slide up the ramp in one hour? a. 210 b. 4 c. 17 d. 43 e. 170

20 Cart mass = 20kg Now assume that the cart breaks and I have to push each 20 kg box up the ramp. Friction is now significant and is 50N. Work to push one box up ramp = J. If I have an average mechanical power output of 60W, how many boxes do I slide up the ramp in one hour? a. 210 b. 4 c. 17 d. 43 e. 170 Work done in one hour = 60 * 60 * 60 = J Number of boxes = J/ J per box = 210

21 8m h = 0 I m designing a fountain for a city park. I want the water in the fountain to squirt 8m vertically in the air. What pressure (above atmospheric) will I need in the large pipe just before the nozzle? (Assume that friction can be ignored) a Pa b Pa c Pa d Pa e. More information required

22 8m h = 0 I m designing a fountain for a city park. I want the water in the fountain to squirt 8m vertically in the air. What pressure (above atmospheric) will I need in the large pipe just before the nozzle? (Assume that friction can be ignored) a Pa b Pa c Pa d Pa e. More information required E tpv = P + ½ rv 2 + rgh Top of water jet: v = 0, P = AP, h = 8m In pipe: v = 0, h = 0, P = AP + P pipe E tpv is same in both locations AP + P pipe = AP + rgh jet P pipe = rgh jet = 1000*9.8*8 = Pa

23 8m h = 0 What is the speed of the water immediately after the nozzle? a m/s b m/s c. 8.7 m/s d. 4.3 m.s e. Water does not come out

24 8m h = 0 What is the speed of the water immediately after the nozzle? a m/s b m/s c. 8.7 m/s d. 4.3 m.s e. Water does not come out E tpv = P + ½ rv 2 + rgh Inside pipe: v = 0, P = AP Pa, h = 0 After nozzle: v =?, h = 0, P = AP E tpv is same in both locations P pipe = ½ r v 2 v = sqrt (2P pipe /r) = sqrt ( 2* / 1000 ) = 12.5 m/s

25 8m If the fountain squirts 2m 3 of water in the air every second, what is the power produced by the pump that is supplying the pressurized water to the nozzle? a W b. 2 W c W d. 237,806 W e. 156,800 W

26 8m If the fountain squirts 2m 3 of water in the air every second, what is the power produced by the pump that is supplying the pressurized water to the nozzle? a W b. 2 W c W d. 237,806 W e. 156,800 W Power = energy supplied to water / sec = GPE gained by water /sec = mgh/s = (m/s) g h Mass of water squirted per second = 2m kg/m 3 Power = = 156,800 W = 2000kg

27 Air pressure Below is a plot of air pressure outside your ear when you hear a sound wave 0 4 Time (ms) 8 What is the frequency of the note? a Hz b. 0.5 Hz c. 500 Hz d. 125 Hz e. Something else

28 Air pressure Below is a plot of air pressure outside your ear when you hear a sound wave 0 4 Time (ms) 8 What is the frequency of the note? a Hz b. 0.5 Hz c. 500 Hz d. 125 Hz e. Something else Period = 4ms = 0.004s f = 1/T = 1/0.004 = 250 Hz

29 Air pressure Below is a plot of air pressure outside your ear when you hear a sound wave 0 4 Time (ms) 8 If the distance between adjacent pressure maxima is 1.34m, what is the speed of the sound wave? a. 330 m/s b. 332 m/s c. 335 m/s d. 338 m/s e. 340 m/s

30 Air pressure Below is a plot of air pressure outside your ear when you hear a sound wave 0 4 Time (ms) 8 If the distance between adjacent pressure maxima is 134cm, what is the speed of the sound wave? a m/s b m/s c. 335 m/s d. 338 m/s e. 340 m/s v = f l = = 335m/s

31 Air pressure Below is a plot of air pressure outside your ear when you hear a sound wave 0 4 Time (ms) 8 On the same axis, draw the air pressure when you hear a note of the same volume but 3 times the frequency

32 Air pressure Below is a plot of air pressure outside your ear when you hear a sound wave 0 4 Time (ms) 8 On the same axis, draw the air pressure when you hear a note of the same volume but 3 times the frequency

33 Air pressure Below is a plot of air pressure outside your ear when you hear a sound wave 0 4 Time (ms) 8 On the same axis, draw the air pressure when you hear the sound of someone talking

34 Air pressure Below is a plot of air pressure outside your ear when you hear a sound wave 0 4 Time (ms) 8 On the same axis, draw the air pressure when you hear the sound of someone talking Any random squiggle around the background pressure

35 Air pressure Below is a plot of air pressure outside your ear when you hear a sound wave 0 4 Time (ms) 8 Draw a sound with a frequency of 125 Hz that is louder than the note shown

36 Air pressure Below is a plot of air pressure outside your ear when you hear a sound wave 0 4 Time (ms) 8 Draw a sound with a frequency of 125 Hz that is louder than the note shown T = 1/f = 1/125 = 0.008s = 8ms Louder bigger amplitude

37 Air pressure Below is a plot of air pressure outside your ear when you hear a sound wave 0 4 Time (ms) 8 The sun goes down and the air cools down a bit, causing the air pressure to drop. What does the plot look like now for the same 250 Hz note?

38 Air pressure Below is a plot of air pressure outside your ear when you hear a sound wave 0 4 Time (ms) 8 The sun goes down and the air cools down a bit, causing the air pressure to drop. What does the plot look like now for the same 250 Hz note? Although not relevant to this plot, the speed and hence wavelength of the wave will also change slightly if the background pressure drops. However the frequency and hence period of the wave (plotted above) is determined by oscillation rate of speaker/instrument and so will NOT change at all.