fisica1-youn-e-freedman-exercicios-resolvidos

11.23:
Fl0 ( 400 N)(2.00 m)
−6
2
A = =
= 1.60×
10 m ,
10
− 2
Y∆l
(20×
10 Pa)(0.25×
10 m)
and so d =
−3
4A
π = 1.43×
10 m, or 1.4 mm to two figures.
∆l F
11.24: a) The strain, from Eq. (11.12), is = . For steel, using Y from Table (11.1)
2
d
−4
2
and A = π = 1.77 × 10 m ,
4
∆l
l
0
=
(2.0 × 10
l0 YA
( −4
11
4000 N)
= 1.1×
10
− 4 2
Pa)(1.77 × 10 m )
11
−4
Similarly, the strain for copper ( Y = 1.10×
10 Pa)
is 2.1× 10 . b) Steel:
−4 −5
−4
−4
(1.1 × 10 ) × (0.750 m) = 8.3×
10 m . Copper: (2.1×
10 )(0.750 m) = 1.6 × 10 m ⋅
.
11.25: From Eq. (11.10),
5000 N)(4.00 m)
Y = = 2.0×
10
−4 2
−2
(0.50×
10 m )(0.20 × 10 m)
( 11
Pa.
11.26: From Eq. (11.10),
2
65.0 kg)(9.80 m s )(45.0 m)
Y = = 6.8×
10
−3
2
( π (3.5×
10 m) )(1.10 m)
( 8
Pa.
2
11.27: a) The top wire is subject to a tension of (16.0 kg)(9.80 m s ) = 157 N and
hence a tensile strain of
(157 N)
10
−7
2
(20×
10 Pa)(2.5×
10 m )
= 3.14×
10
−3
, or 3.1×
10
−3
to two figures. The
−3
bottom wire is subject to a tension of 98.0 N, and a tensile strain of 1.96 × 10 , or
−3
3
2.0× 10 to two figures. b) (3.14× 10
− )(0.500 m) = 1.57 mm,
(1.96
10
3
× −
)(0.500 m) = 0.98 mm.
(8000 kg)(9.80 m
s
2
6
11.28: a) = 1.6 × 10 Pa.
−2
2
π(12.5×
10 m)
−5 −5
c) (0.8×
10 ) × (2.50 m) = 2×
10 m.
)
6
1.6× 10 Pa
−5
b) = 0.8×
10 .
10
2.0×
10 Pa
5
2
6
11.29: (2.8 − 1)(1.013×
10 Pa)(50.0 m ) = 9.1×
10 N.