fisica1-youn-e-freedman-exercicios-resolvidos

11.87: Use subscripts 1 to denote the copper and 2 to denote the steel. a) From Eq.
(11.10), with ∆ l
1
= ∆l2
and F1
= F2
,
2
10
⎛ A2Y2
⎞ ⎛ (1.00 cm )(21×
10 Pa) ⎞
L
2
= L1
⎜
⎟ = ( 1.40 m)
⎜
= 1.63 m.
2
10
1 1
(2.00 cm )(9 10 Pa)
⎟
⎝ A Y ⎠ ⎝
× ⎠
b) For nickel, F 8
= 4.00 × 10 Pa and for brass, F
8
= 2.00×
10 Pa. c) For nickel,
8
4.00× 10 Pa
−3
10 1.9 × 10
21×
10 Pa
A1
8
2.00× 10 Pa
−3
= and for brass, = 2.2×
10 .
10
9×
10 Pa
A2
⎛ ∆l
⎞
10
−4
2
4
11.88: a) F
max
= YA
⎜
⎟ = (1.4 × 10 Pa)(3.0 × 10 m )(0.010) = 4.2 × 10 N.
⎝ l0
⎠
max
b) Neglect the mass of the shins (actually the lower legs and feet) compared to the
rest of the body. This allows the approximation that the compressive stress in the shin
bones is uniform. The maximum height will be that for which the force exerted on each
lower leg by the ground is Fmax
found in part (a), minus the person’s weight. The impulse
that the ground exerts is
4
2
3
J = (4.2 × 10 N − (70 kg)(9.80 m s ))(0.030 s) = 1.2 × 10 kg ⋅ m s. The speed at the
ground is
2 gh,
so 2J
= m 2gh
and solving for h,
but this is not recommended.
h =
2
1 ⎛ 2J
⎞
⎜ ⎟
2g
⎝ m ⎠
= 64 m,
2
11.89: a) Two times as much, 0.36 mm, b) One-fourth (which is ( 1 2) ) as much,
0.045 mm.c) The Young’s modulus for copper is approximately one-half that for steel, so
10
20×
10 Pa
the wire would stretch about twice as much. (0.18 mm) = 0.33 mm.
11.90: Solving Eq. (11.14) for ∆ V ,
mg
∆V
= −kV0∆P
= −kV0
A
−11
−1
(1420 kg)(9.80 m s
= −( 110×
10 Pa )(250 L)
2
π(0.150 m)
= −0.0541L.
2
)
10
11×
10 Pa
The minus sign indicates that this is the volume by which the original hooch has
shrunk, and is the extra volume that can be stored.