EPI Homework 10
Topics covered: Work and kinetic energy
All problems from the course textbook Physics For Scientists and Engineers, 6th Edition, Tipler
and Mosca unless stated otherwise.
Problems:
1) 6-27
2) 6-28
3) 6—39
4) 6-40
5) 6-68
6) A 6.0-kg block
EP1 Homework 8
Topics covered: Frictional force and spring force
All problems from the course textbook Physics For Scientists and Engineers, 6th Edition, Tipler
and Mosca unless stated otherwise.
Problems:
1) 5-33
2) 5-47
3) 5-52
4) 5-56
5) A block of mas
EP1 Homework 7
Topics covered: Newton's 3rd Law
All problems from the course textbook Physics For Scientists and Engineers, 6th Edition, Tipler
and Mosca unless stated otherwise.
Problems:
1) 4-47
2) 4-50
3) 4-67
4) 4-79
5)Two boxes of mass and connected
EPl Homework 6
Topics covered: Projectile motion
All problems from the course textbook Physics For Scientists and Engineers, 6th Edition, Tipler
and Mosca unless stated otherwise.
Problems:
1) 3-72
2) 3-73
3) 3-74
4) 3-75
5) 3-80
6) A projectile is ﬁred i
EP1 Homework 5
Topics covered: Gravity and Free Fall
All problems from the course textbook Physics For Scientists and Engineers, 6th Edition, Tipler
and Mosca unless stated otherwise.
Problems:
1) 2-69
2) 2-72
3) 2-79
4) 2-73
5) A ball launch straight up
EPl Homework 4
Topics covered: Newton's lst and 2nd Law
All problems from the course textbook Physics For Scientists and Engineers, 6th Edition, Tipler
and Mosca unless stated otherwise.
Problems:
1) 4-31
2) 4-35
3) 4-37
4) 4-57
5) 4-60
6) 4-65
Note: if y
EP1 Homework 3
Topics covered: velocity and acceleration
All problems from the course textbook Physics For Scientists and Engineers, 6th Edition, Tipler
and Mosca unless stated otherwise.
Problems:
1) 2-51 (Chapter 2, Problem 51)
2) 2-62
3) 2-65
4) 2-104
EP1 Homework 2
Topics covered: vectors and scalars
All problems from the course textbook Physics For Scientists and Engineers, 6th Edition, Tipler
and Mosca unless stated otherwise.
Problems:
1) 1-53 (Chapter 1, Problem 53)
2) 1-54
3) 1-55
4) 1-57
5) Let
EP1 Homework 1
Topics covered: the metric system and estimation
All problems from the course textbook Physics For Scientists and Engineers, 6th Edition, Tipler
and Mosca unless stated otherwise.
Problems:
1) 1-23 (Chapter 1, Problem 23)
2) 1-37
3) 1-38
4)
32. From Fgrav = GMm r 2 = mv 2 r we find M v 2 . Thus, the mass of the Sun would be
M s = vMercury vPluto
2
47.9 km s Ms = 4.74 km s
2
M s = 102 M s .
31. (a) From Eq. 41-29, we know that N 2 N1 = e E kT . We solve for E: E = kT ln
N1 1 0.25 = ( 8.62 105 eV K ) ( 2.7K ) ln N2 0.25
= 2.56 104 eV 0.26 meV. (b) Using the result of problem 83 in Chapter 38,
= hc 1240eV nm = = 4.84 106 nm 4.8mm. E 2.56 104 e
30. (a) Letting v(r ) = Hr ve = 2G M r , we get M r 3 H 2 2G . Thus,
=
M 3 M 3H 2 = . 4 r 2 3 4 r 3 8G
(b) The density being expressed in H-atoms/m3 is equivalent to expressing it in terms of 0 = mH/m3 = 1.67 1027 kg/m3. Thus, 3 ( 0.0218 m s ly ) 1.00 ly
29. (a) From f = c/ and Eq. 37-31, we get
0 = 1 1 = ( 0 + ) . 1+ 1+
Dividing both sides by 0 leads to
1 = (1 + z ) We solve for :
1 . 1+
=
(b) Now z = 4.43, so
(1 + z ) 2 1 z 2 + 2z =2 . (1 + z ) 2 + 1 z + 2 z + 2
b4.43g + 2b4.43g = 0.934 . = b4.43g + 2
28. First, we find the speed of the receding galaxy from Eq. 37-31:
=
=
1 ( f f0 ) 2 1 ( 0 ) 2 = 1 + ( f f0 ) 2 1 + ( 0 ) 2 1 (590.0 nm 602.0 nm )2 = 0.02013 1 + (590.0 nm 602.0 nm )2
where we use f = c/ and f0 = c/0. Then from Eq. 44-19,
8 v c ( 0.02013
27. We apply Eq. 37-36 for the Doppler shift in wavelength:
v = c
where v is the recessional speed of the galaxy. We use Hubbles law to find the recessional speed: v = Hr, where r is the distance to the galaxy and H is the Hubble constant 21.8 103 sm . T
25. From = 1 + K/mc2 (see Eq. 37-52) and v = c = c 1 2 (see Eq. 37-8), we get
v = c 1 1+
F H
K mc 2
I K
2
.
(a) Therefore, for the *0 particle,
1000MeV v = (2.9979 10 m s) 1 1 + 1385MeV
8 2
= 2.4406 108 m s.
For 0, 1000 MeV v = (2.9979 10 m s) 1 1 + 1192.
24. If we were to use regular rectangular axes, then this would appear as a right triangle. Using the sloping q axis as the problem suggests, it is similar to an upside down equilateral triangle as we show below.
The leftmost slanted line is for the 1 cha
23. (a) Looking at the first three lines of Table 44-5, since the particle is a baryon, we determine that it must consist of three quarks. To obtain a strangeness of 2, two of them must be s quarks. Each of these has a charge of e/3, so the sum of their c
22. (a) Using Table 44-3, we find q = 0 and S = 1 for this particle (also, B = 1, since that is true for all particles in that table). From Table 44-5, we see it must therefore contain a strange quark (which has charge 1/3), so the other two quarks must h
21. (a) We indicate the antiparticle nature of each quark with a bar over it. Thus, u u d represents an antiproton. (b) Similarly, u d d represents an antineutron.
20. (a) The combination ddu has a total charge of 1 1 + 2 = 0 , and a total strangeness 3 3 3 of zero. From Table 44-3, we find it to be a neutron (n).
b
g
(b) For the combination uus, we have Q = + 2 + 2 1 = 1 and S = 0 + 0 1 = 1. This is the 3 3 3 + par
19. (a) From Eq. 37-50, Q = mc 2 = ( m0 m p m )c 2 = 1115.6 MeV 938.3 MeV 139.6 MeV = 37.7 MeV. (b) We use the formula obtained in problem 44-9 (where it should be emphasized that E is used to mean the rest energy, not the total energy):
Kp = = 1 E E p 2
17. (a) As far as the conservation laws are concerned, we may cancel a proton from each side of the reaction equation and write the reaction as p 0 + x. Since the proton and the lambda each have a spin angular momentum of 2, the spin angular momentum of x
16. The formula for Tz as it is usually written to include strange baryons is Tz = q (S + B)/2. Also, we interpret the symbol q in the Tz formula in terms of elementary charge units; this is how q is listed in Table 44-3. In terms of charge q as we have u
15. (a) See the solution to Problem 11 for the quantities to be considered, adding strangeness to the list. The lambda has a rest energy of 1115.6 MeV, the proton has a rest energy of 938.3 MeV, and the kaon has a rest energy of 493.7 MeV. The rest energy
13. For purposes of deducing the properties of the antineutron, one may cancel a proton from each side of the reaction and write the equivalent reaction as + p = n. Particle properties can be found in Tables 44-3 and 44-4. The pion and proton each have ch