fisica1-youn-e-freedman-exercicios-resolvidos

11.30: a) The volume would increase slightly. b) The volume change would be twice
as great. c) The volume is inversely proportional to the bulk modulus for a given pressure
change, so the volume change of the lead ingot would be four times that of the gold.
250 N
6
11.31: a) = 3.33 10 Pa.
- 4 2
×
0.75×
10 m
6
−4
2
b) (3.33×
10 Pa)(2)(200×
10 m ) = 133 kN.
11.32: a) Solving Eq. (11.14) for the volume change,
∆V
= −kV∆P
= −(45.8×
10
−11
3
= −0.0531m .
Pa
−1
3
8
)(1.00 m )(1.16 × 10
Pa −1.0×
10
5
Pa)
b) The mass of this amount of water not changed, but its volume has decreased to
1.000 m
3
3
3
1.03×
10 kg
3 3
− 0.053 m = 0.947 m , and the density is now = 1.09×
10 kg m .
3
3
0.947 m
3
6
( 600 cm )(3.6 × 10 Pa)
9 1
−10
−1
11.33: B =
= 4.8×
10 Pa, k = = 2.1×
10 Pa .
3
(0.45 cm )
B
11.34: a) Using Equation ( 11.17),
5
F|| ( 9×
10 N)
−2
Shear strain = =
= 2.4×
10 .
AS
10
[(.10 m)(.005 m)][7.5 × 10 Pa]
−3
b) Using Equation ( 11.16), x = Shear stain ⋅ h = (.024)(.1m) = 2.4×
10 m.
11.35: The area A in Eq. ( 11.17)
has increased by a factor of 9, so the shear strain for
the larger object would be 1 9 that of the smaller.
11.36: Each rivet bears one-quarter of the force, so
4
F
1
|| ( 1.20×
10 N)
4
8
Shear stress = =
= 6.11×
10 Pa.
−2
2
A π(.125×
10 m)