fisica1-youn-e-freedman-exercicios-resolvidos

F (90.8 N)
7
7
11.37: = 3.41×
10 Pa, or 3.4 10 Pa to two figures.
=
3 2
×
A
−
π(0.92×
10 m)
−3
10
−6
2
3
11.38: a) (1.6 × 10 )(20×
10 Pa)(5×
10 m ) = 1.60×
10 N.
the wire would stretch 6.4 mm.
b) If this were the case,
−3
10
−6
2
3
c) (6.5×
10 )(20×
10 Pa)(5×
10 m ) = 6.5×
10 N.
8
−4
2
Ftot ( 2.40 × 10 Pa)(3.00 × 10 m ) 3
2
2
11.39: a = =
− 9.80 m s = 10.2 m s .
m
(1200 kg)
350 N
−7
2
11.40: A = = 7.45×
10 m , so d = 4A
π 0.97 mm.
8
=
4.7×
10 Pa
11.41: a) Take torques about the rear wheel, so that fω d = ωx , or cm
xcm
= fd .
b) ( 0.53)(2.46 m) = 1.30 m to three figures.
11.42: If Lancelot were at the end of the bridge, the tension in the cable would be
(from taking torques about the hinge of the bridge) obtained from
2
2
T(12.0
N) = (600 kg)(9.80 m s )(12.0 m) + (200 kg)(9.80 m s )(6.0 m),
so T = 6860 N. This exceeds the maximum tension that the cable can have, so Lancelot is
going into the drink. To find the distance x Lancelot can ride, replace the 12.0 m
3
multiplying Lancelot’s weight by x and the tension T by Tmax = 5.80×
10 N and solve
for x;
3
2
(5.80×
10 N)(12.0 m) − (200 kg)(9.80 m s )(6.0 m)
x =
= 9.84 m.
2
(600 kg)(9.80 m s )