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The due dates/day for the homework, along with the problem numbers, are listed below. Problems in bold red are graded in detail out of 5 points each. The rest are graded out of 2 points. An almost correct solution or a valiant attempt gets 2 points while a decent attempt with the right method and equations and some progress, gets 1 point.

Week	Hw #	Date	Day	Problems
Week 1		June 4	Monday
		June 5	Tuesday
	1	June 6	Wednesday	1.3, 1.20, 1.31, 1.54, 2.1, *1.5, 1.55*
	2	June 7	Thursday	2.5, 2.11, 2.12, 2.21, 2.27, 2.42, *2.43, 2.63*
	3	June 8	Friday	3.1, 3.9, 3.25, 3.31, 3.36, 3.47, *3.34, 3.59*

Week 2	4	June 11	Monday	1.18, 1.50, 2.6, 2.10, 2.40, 4.1, 4.2, 4.5, 4.10, 4.19, 4.58, 4.67, 4.21, 4.54
	5	June 12	Tuesday	5.1,5.2,5.3,5.7,5.9,5.11,5.14 (Refer to hints under the section "Answers to EVEN numbered problems"
		June 13	Wednesday
	6	June 14	Thursday	5.18,5.19,5.22, 5.23, 5.41, 5.45
	7	June 15	Friday	5.24, 5.25, 5.37, 5.44, 5.46

Week 3	8	June 18	Monday	4.33, 4.34, 4.35, 6.1, 6.5, 6.9, 6.11, 6.13, 6.15, 6.19, 6.20, 6.61, 6.63, 6.65, 6.71
	9	June 19	Tuesday	7.5, 7.10, 7.11, 7.19, 7.25, 6.58
	10	June 20	Wednesday	5.54(ignore d),5.57,7.31,7.33,7.34,8.5,8.26
	11	June 21	Thursday	7.53, 8.10, 8.13, 8.24, 8.31, 8.36, 8.37
	12	June 22	Friday	6.18, 6.17, 8.30, 8.33, 7.56, 8.11, 8.17, 8.30, 8.33

Week 4	13	June 25	Monday	8.48, 8.57, 8.59, 8.61, 8.63, 8.65, 8.73, 8.66
	14	June 26	Tuesday	9.1, 9.3, 9.7, 9.9, 9.10, 9.17, 9.18, 9.20
	15	June 27	Wednseday	9.24, 9.28, 9.38, 9.41, 9.67,
	16	June 28	Thursday	15.2, 15.3, 15.9, 15.11, 15.13, 15.16, 15.18, 15.21, 15.23
	17	June 29	Friday	15.5, 15.53, 15.28, 15.32, 10.1, 10.3, 10,4, 10.7, 10.20, 10.21

Week 5	18	July 2	Monday	10.31, 10.34, 10.37, 10.38, 10.46, 10.70, 10.71, 11.3, 11.11, 11.30, 11.28, 11.36, 11.50
	19	July 3	Tuesday	12.43, 12.45, 12.49, 12.55, 13.2, 13.3, 13.15, 13.19, 13.39, 13.43 (a,b,c only)

	20	July5	Thursday	14.6, 14.8, 14.12, 14.18, 14.20, 14.24, 14.25, 14.27, 14.29, 14.48, 14.51
	21	July6	Friday	16.2, 16.9, 16.17, 16.27, 16.31

Week 6!	22	July 9	Monday	16.30, 16.39, 16.47,16.50, 18.5, 18.13, 18.19, 18.21,18.27 (REVIEW PROBLEMS: 2.63(with the exam twist),2.43, 4.21, 4.54,4.67, 5.54,5.44)
	23	July 10	Tuesday	16.32,16.57,16.62,18.14, 18.17, 18.37, 18.41,18.43,17.37 (Review problems: 6.61, 6.63, 6.65, 7.33, 8.61,8.65,15.53,14.20)
	24	July 11	Wednseday	14.48, 14.51, 14.38, 14.48, 17.19, 17.18, 17.39, 17.60, 19.1, 19.2, 19.5, 19.6, 19.15, 19.16, 19.18 (REVIEW PROBLEMS: 12.43,11.36,example 11.8, 10.70, 9.67, 9.24)
	25	July12	Thursday	19.45, 19.46, 19.50, 19.56, 20.7, 20.11, 20.18, 20.16, 20.24, 20.38, 20.23 (REVIEW 8.33, 8.57,15.11, 15.19,6.60)
	26	July13	Friday	13.2, 13.15, 13.19, 13.39, 20.30, REVIEW problems (Derivation eq-10.29, eq-10.30, quick quiz 10.12,10.13, 10.14, 11.30,

Some students are using the 7th addition. The table below has the links to the pdf files of the questions and answers to each homework.

1.18. 35.7 m2 .
4.10. 7.23× 103 m, 1.68× 103 m
4.58. 7.50 m/s (in what direction? You have to specify)
4.54 10.7 m/s

5.2. 1.66× 10^6 N
5.14. a) 181 degrees Counteclockwise from X axis, b) 11.2 Kg, c) 37.5 m/s at 181 degrees
5.11 (Hint) Write the forces in vector notation. Use F=ma and find a in vector notation. FInd magnitude
and direction of a. Or you could find the resultant of F1 and F2 and then find acceleration.

5.24. s=6.3 m/s^2 T=31.5 N
5.44. a=2.31 m/s^2 down for m1, left for m2, up for m3, T between m1 and m2=30.0 N, Tm2<->m3=24.2 N
5.46. Pmax=48.6 N, Pmin=31.7 N

4.33.   1.48 m/s^2, 29.9 degrees
4.34  a) 0.6 m/s^2 b)Radial acce =0.8 m/s^2,c) Resultant acc= 1.00m/s^2, 53.1 degrees
6.20  a) 8.62 m b) Mg downward, c) 8.45 m/s^2
6.61 (Hint)c) The freebody diagram is more like in the case of a banked curve where the forces acting are just N and mg. But N is at an angle so that the horzontal component of N is along the radius an dthe vertical component of N balances Mg. so figure theta out.

6.65 Time period of revolution is distance/velocity T= (2*Pi*r)/v. Frequency is 1/T and is in units revolutions/second. convert that to revolutions/min.

5.54.   a) a=2.00 m/s^2, b) 8.00 N on 4kg block, 6.00 N on 3kg, 4.00 N on 2kg c) 8.00 N and 14.0 N
7.34  a) 329J b)0J,c) 0 d) 185J e)+144J
8.26  a) 22J, 40 J

9.10.   a) I=5.40i N.s, b) W=-27.0J
9.18  a) Vf=2.50 m/sb)Kf-Ki=-3.75*10^4 J
9.20  a) hmax=0.556 m (the velocity of m1 just after collision works out to be -3.30m/s)

9.24. v=(4M)Sqrt(gl)/m
9.28 a) theta=32.3 deg, V=2.88 m/sb)Kf-Ki=783J
9.38 a) Xcm=0, Ycm=1.00m

15.28 gc/gt= Lc/Lt =1.0015
15.32 T=3.65s b) 6.41 s c)4.24s
10.4 a) 1200 rad/s b)25.0 s
10.20 a) 92 kg m^2m KErot=184 J b) v1=6m/s v2=4m/s, v3=8 m/s. KE1=72 J
KE2= 16.0 J KE3= 96.0 J Ktot= 184 J

10.34 a) 1.03 seconds, b) 10.3 rev
10.38 coefficient of kinetic fric= 0.312 (USE torque equation since friction provides the torque that slows it down)
10.46 Use conservation of energy along with rotation to find v=2.36 m/s
10.70 Omega= Sqrt[(2*m*g*d*sin(theta)+k*d^2)/(I+m*R^2)]
11.28 ωf =(I1/( I1 +I2))* ωi . b) Kf/Ki = I1/( I1 +I2)
11.30 a) 1.91 rad/s b) Ki=2.53 J Kf= 6.44 J
11.36 apparent acceleration is “Radial or centripetal acceleration” = 12.3 m/s^2
11.50 WILL DO IT IN CLASS TODAY

14.6 Guage pressure is the difference between the absolute pressure (which could include the atmospheric pressure) and the atmospheric pressure. A) P= 1.01 *10^7 Pa B) 1.0*10^7Pa
14.8 F2= 225 N
14.10 Fb= 5.88 *10^6 N F= 196 kN on each end. F= 588 kN on each side
14.12 a)20.0 cm, Use density = mass/volume ,and volume = area*height
b) Use the fact that the amount of water below the horizontal level is the same as the volume of mercury above the level. So A1*h= A2*h_below
and then use the fact that the pressure at the lowest point of water is the same as the pressure at the same level on the other side. (ALL mercury below that level). Pressure on right is just due to water above and on left is just due to mercury. Using these two equations solve for h. =0.490 cm
14.20 h1=12.6 cm similar logic as above. Call the height in the middle as some h and that will cancel out when you equate pressure on both sides. Choose a level where the liquid below that level is the same on both sides.
14.24 A= m/[(Rho_water-Rho_styrofoam)*h]
14.48 P2= 6.8*10^4 Pa (DUE FRIDAY)

Homework 22.
16.32 v= sqrt(m L tan(theta)/4Mg) b) find t using distance/vel.
16.62 a) 5 m/s +x b) 5 m/s in –x c) 7.5 m/s – x dirn. d) 24.0 m/s +x dirn.
18.14 a) find where the nodes are in terms of pi. B) ymax = 0.0294 m

Homework 24
14.48 6.8*10^4 Pa
14.38 12.8 kg/s use dm/dt= density*Volume flow rate
14.18,19, NOT DUE
17.39 write two equations for observed frequency. One for approach and one for going away. Use the two equations and find the siren frequency and ambulance speed.
17.60 First step calculate the frequency that the insect hears. Who is the observer who is the source? Next step insect’s body acts as the source and reflects the sound. Calculate the frequency of the echo the bat hears. Who is the observer ?
19.6 Tc=1.33Ts+20
19.16 delta A= 0.109 cm^2,
19.18 T=437oC.b)3000oC will it work? What could happen before that?

Homework 25.
19.45. Find out how much the rod contracts by and do the same for the tape. The difference is the change in the reading. So if rod contracts more by x amount, then the reading is .950-x

19.46. Find out what the mass m of 10 gallons is at 0 degrees. At 20 degrees the 10 gallons (Vo) becomes some new V gallons. So V gallons at 20 degrees are m kg. so
How much mass do 10 gallons have? What is the difference between this and m?
0.523 kg.

19.50 use P1V1/T1 = P2V2/T2 .
The new volume is V1 + Ah. New pressure has increased since the compressed spring exerts a force. Find the excess pressure . = P1+ kh/A. Solve for h.= 0.169 m, P2=1.35 *10^5 Pa.

19.56. Length expands. Find new time. Find the time lost per sec. a) delta T= 9.5*10^(-5) seconds. find the time lost per week. 57.5 seconds.

20.7. Is interesting! You do not know if Aluminum will rise in temperature or drop in temperature. Find the heat needed for water to go up to 26 degrees. Call it Qw. Find the heat lost by copper to come to 26 degrees. Call it Qc. If Qw >Qc then the copper cannot take the water beyond 26 degrees. So the temperature is less than 26 and aluminum cools.
If Qw<Qc, then the temperature is more than 26 degrees. So once you know what the situation is you know if aluminum gains or loses heat!

20.18 Temperature is 40.4 degrees. SO all ice melts. You have do show it as done in class. B) 8.04 grams melts.