Simple Nature - Light and Matter

(taking into account all the preparation, etc.), then this would be 10^10 man-hours.

Page 53, problem 44:

Directly guessing the number of jelly beans would be like guessing volume directly. That would

be a mistake. Instead, we start by estimating the linear dimensions, in units of beans. The

contents of the jar look like they’re about 10 beans deep. Although the jar is a cylinder,

its exact geometrical shape doesn’t really matter for the purposes of our order-of-magnitude

estimate. Let’s pretend it’s a rectangular jar. The horizontal dimensions are also something like

10 beans, so it looks like the jar has about 10× 10 ×10 or∼ 103 beans inside.

Solutions for chapter 1
Page 71, problem 12:
To the person riding the moving bike, bug A is simply going in circles. The only difference
between the motions of the two wheels is that one is traveling through space, but motion is
relative, so this doesn’t have any effect on the bugs. It’s equally hard for each of them.

Solutions for chapter 2
Page 120, problem 1:
(a) The energy stored in the gasoline is being changed into heat via frictional heating, and also
probably into sound and into energy of water waves. Note that the kinetic energy of the propeller
and the boat are not changing, so they are not involved in the energy transformation. (b) The
crusing speed would be greater by a factor of the cube root of 2, or about a 26% increase.
Page 120, problem 2:
We don’t have actual masses and velocities to plug in to the equation, but that’s OK. We just
have to reason in terms of ratios and proportionalities. Kinetic energy is proportional to mass
and to the square of velocity, so B’s kinetic energy equals (13.4 J)(3.77)/(2.34)^2 = 9.23 J.
Page 120, problem 3:
Room temperature is about 20◦C. The fraction of the energy that actually goes into heating
the water is
(250 g)/(0.24 g·◦C/J)×(100◦C− 20 ◦C)
(1.25× 103 J/s) (126 s)

= 0.53

So roughly half of the energy is wasted. The wasted energy might be in several forms: heating

of the cup, heating of the oven itself, or leakage of microwaves from the oven.

Page 120, problem 5:

Etotal,i=Etotal,f

PEi+ heati=PEf+KEf+ heatf

1

2

mv^2 =PEi−PEf+ heati−heatf

=−∆PE−∆heat

v=

√

2

(

−∆PE−∆heat

m

)

= 6.4 m/s