Simple Nature - Light and Matter

(Martin Jones) #1

the wood. The relationship ism=ρb^3. The graph ofmversusbwould be a curve, and you
would not have any easy way to get the density from such a graph. But by graphingmversus
b^3 , you can produce a graph that is linear, and whose slope equals the density.


Hints for chapter 6
Page 392, problem 4:
How could you change the values ofxandtso that the value ofywould remain the same?
What would this represent physically?


Page 393, problem 8:
(a) The most straightforward approach is to apply the equation ∂^2 y/∂t^2 = (T/μ)∂^2 y/∂x^2.
Although this equation was developed in the main text in the context of a straight string with a
curvy wave on it, it works just as well for a circular loop; the left-hand side is simply the inward
acceleration of any point on the rope. Note, however, that we’ve been assuming the string was
(at least approximately) parallel to thexaxis, which will only be true if you choose a specific
value ofx. You need to get an equation foryin terms ofxin order to evaluate the right-hand
side.
Page 394, problem 12:
The answers to the two parts are not the same.


Hints for chapter 7
Page 463, problem 28:
Apply the equivalence principle.


Hints for chapter 8
Page 526, problem 15:
The force on the lithium ion is the vector sum of all the forces of all the quadrillions of sodium
and chlorine atoms, which would obviously be too laborious to calculate. Nearly all of these
forces, however, are canceled by a force from an ion on the opposite side of the lithium.


Hints for chapter 9
Page 567, problem 20:
The approach is similar to the one used for the other problem, but you want to work with
voltage and electrical energy rather than force.


Hints for chapter 10
Page 658, problem 15:
Use the approximation (1 +)p≈1 +p, which is valid for small.
Page 661, problem 25:
First find the energy stored in a spherical shell extending fromrtor+ dr, then integrate to
find the total energy.
Page 661, problem 26:
Since we havetr, the volume of the membrane is essentially the same as if it was unrolled
and flattened out, and the field’s magnitude is nearly constant.
Page 662, problem 31:
The math is messy if you put the origin of your polar coordinates at the center of the disk. It
comes out much simpler if you put the origin at the edge, right on top of the point at which
we’re trying to compute the voltage.

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