fisica1-youn-e-freedman-exercicios-resolvidos

6.90: a) The hummingbird produces energy at a rate of 0.7
J/s to 1 .75 J/s. At
10 beats/s, the bird must expend between 0.07 J/beat and 0.175 J/beat.
b) The steady output of the athlete is 500 W/70 kg = 7 W/kg, which is below the
10 W/kg necessary to stay aloft. Though the athlete can expend
1400 W/70 kg = 20 W/kg for short periods of time, no human-powered aircraft could
stay aloft for very long. Movies of early attempts at human-powered flight bear out this
observation.
d d
dm
6.91: From the chain rule, P = W = ( mgh)
= gh,
for ideal efficiency. Expressing
dt dt
dt
the mass rate in terms of the volume rate and solving gives
6
(2000×
10 W)
3
= 1.30×
10
2
3
(0.92)(9.80 m/s )(170 m)(1000 kg/m )
m
s
3
.
6.92: a) The power P is related to the speed by 1 2
2Pt
Pt = K = mv , so v = .
b)
dv d 2Pt
2P
d 2P
1 P
a = = = t = = . .
dt dt m m dt m 2 t 2mt
a)
2P
2P
2 8P
2
3
2
2
2
x − x
1
0
= ∫ v dt = t dt t t .
m
∫ = =
m 3 9m
3
2
m
3 3
3 3
2
5
6.93: a) (7500× 10
− kg )(1.05×
10 kg/m )(9.80 m/s )(1.63 m) = 1.26×
10 J.
5
b) (1.26×
10 J)/(86,400 s) = 1.46 W.