1. (a) The charge that passes through any cross section is the product of the current and
time. Since 4.0 min = (4.0 min)(60 s/min) = 240 s, q = it = (5.0 A)(240 s) = 1.2 103 C.
(b) The number of electrons N is given by q = Ne, where e is the magnitude of
19
2.
7.501024 . 1 mol 6.021023
(19-2) n
n=
N
7.50 1024
= 12.46
=
NA
6.02 1023
(19.1)
. (74.9 g/mol) (19-3)
Msam = nM = M
N
7.50 1024
= 933.14 g = 0.933 kg
= 74.9
NA
6.02 1023
(19.2)
.
17.
A pA VA = nA RTA B pB VB = nB RTB .
VB = 4VA , nA = pA VA
14.
3.
0.04
7.1m2
0.04atm 7.1m2 = 0.04 105 P a 7.1m2 = 2.8 104 N
14.
P = gh
20m :
Pw = (1024kg/m3 )(9.8m/s2 )(20m) = 2.00 105 P a
7.6km :
Pa = (0.87kg/m3 )(9.8m/s2 )(7600m) = 0.65 105 P a
:
Pw P a = (2.00 + 0.65) 10
13.
3.
:
F =G
m(M m)
r2
m F , .
,
F
=0
m
G
= 2 (M 2m)
r
m
1
=
M
2
5.
A, B :
A, C :
FAB
3m2A
=G 2
d
FAC
75m2A
=G 2
r
.
FAB = FAC
3m2A
75m2A
G 2 =G 2
d
r
r2 = 25d2
r = 5d
C
5d
m1
8.
m2
F1
F2
m5
F4
F
20 2
3.
a). (19-14)
Q = cm(Tf Ti ) = (386 J/kg C) (2.00 kg) (100 C 25.0 C) = 57900 J
(20.1)
.
b). (20-4)
S = nR ln
Tf
Vf
+ nCv ln
Vi
Ti
(20.2)
. ,
. (
(19-39) ),
S = nCv ln
6.
Tf
Tf
373.15 K
= mc ln
= (2 kg) (386 J/kg K) ln
= 173.22 J/K
Ti
Ti
298.15
3.9
(a) Differentiating the CDF gives the PDF,
0
for s 0
s2
s
+
for 0 < s 12
f S ( s ) = 288 24
0
for s > 12
s
The mode ~ is where fS has a maximum, hence setting its derivative to
s
fS( ~ ) = 0
~ /144 + 1/24 = 0
- s
the mode ~ = 6.
s
The mean of S,
E(S
3.7
(a) The mean and median of X are 13.3 lb/ft2 and 11.9 lb/ft2, respectively (as done in
Problem 3-3-3).
(b) The event roof failure in a given year means that the annual maximum snow load
exceeds the design value, i.e. X > 30, whose probability is
P(X >
3.8
(a) Note that we are given the CDF, FX(x) = P(X x) and it is a continuous function. Hence
P(2 X 8) = FX(8) FX(2) = 0.8 0.4 = 0.4
(b) The median is defined by the particular x value where F(x) = 0.5, which occurs somewhere
along x = 2 and x = 6; precis
3.4
(a) The median of X, xm, is obtained by solving the equation that defines xm,
P(X xm) = 0.5
FX(xm) = 0.5
1- (10 xm)4 = 0.5
xm 11.9 (inches)
(b) First, obtain the PDF of X:
fX(x) = dFX/dx
= 4104x-5 for x 10, and zero elsewhere.
Hence the expected am
3.3
(a) Applying the normalization condition
f
X
( x)dx = 1
6
x2 x3
x2
c( x )dx = 1 c = 1
6
2 18 0
0
18
= 1/6
c=
9 36 6 3
6
(b) To avoid repeating integration, lets work with the CDF of X, which is
1 x2 x3
FX(x) =
6 2 18
2
3
= (9x x )/108
(for x b
3.5
Let F be the final cost (a random variable), and C be the estimated cost (a constant), hence
X=F/C
is a random variable.
(a) To satisfy the normalization condition,
a
1
a
3
3
3
dx = = 3 = 1 ,
2
a
x
x 1
hence
a = 3/2 =1.5
(b) The given event is F exc
3.2
Let X be the profit (in $1000) from the construction job.
(a) P(lose money) = P(X < 0)
= Area under the PDF where x is negative
= 0.0210 = 0.2
(b) Given event is X > 0 (i.e. money was made), hence the conditional probability,
P (X > 4 0 | X > 0)
= P(
3.1
Total time T = A + B, which ranges from (3 + 4 = 7) to (5 + 6 = 11).
Divide the sample space into A = 3, A = 4, and A = 5 (m.e. & c.e. events)
P(T = 7) =
P(T = 7 | A = n)P(A = n)
P(B = 7 - n)P(A = n)
n = 3, 4 ,5
=
n = 3, 4 ,5
= P(B =4)P(A = 3) = 0.20