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23 Nov 2019

so for the solution for ch2. #37 in concepts of modern physics(Beiser) cramster's solution starts the problem with
tan?= (sinf)/(?c/c)(1-cosf)+(1-cosf)
it looks much prettier when they've written it. but I have NOIDEA how they got this equation for angle of recoil. anybodyknow?

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Related textbook solutions

Physics: Principles with Applications
7th Edition, 2013
Giancoli
ISBN: 9780321625922

Related questions

For this problem. I understand how to set it up and how to conserve the mechanical energy so we will have two sides an equation. But when I try to get the potential energy at the final state, where the string forms an angle Θ, the equation you have to use is mgL(1-cosΘ) instead of just mgL(cosΘ). So how do you figure out to use the equation mgL(1-cosΘ)?

Calculus Based Physics

Physics For Scientists And Engineers withModern Physics (7th)by Serway, Jewett

46

55AP

6/10/2010 at 11:55:00 AM

Question Details: Ans. given below isnotadequate.

problem 55 ch 46 page 1390 asks for "the upper limit thattheobservation sets on the mass of a neutrino".

Obs. was a supernova 170 000 ly away.

8 neutrinos in 6s, (energies betw. 8 MeV and 40 MeV)

and 11 n's in 13 s ( 10 MeV n. needs 10 s more than photontotravel this distance)

The expert solution does not give this ans.

It only converts light years to seconds!

The book gives the ans. as " 19 eV/(c squared)"

Please explain this ans.

Brian.

Response Details:

19eV is energy in electron volts. 1 electron volt is1.6e-19Joules, so for very small energies it is more convenient touse theelectron volt rather than Joules.

From Einstein's equation e=mc*c, the solution you gave hasjustrearranged this to m=e/c*c

Was this what you needed clarification on?

My Response Details:

That is true but does this give the ans. 19 eV/c*c ?

m = e/c*c

m= 1.6 e -19/c*c but this is not = 19??

Sorry this is wrong. I have confused e for energy with eforelectronic charge.

Energy was 10 MeV,so m = 10 e^6/c*c.

This is still not = 19/c*c.

Please show how to get 19/c*c.

Brian.

Learning Goal: Understand how to determine the constants in thegeneral equation for simple harmonic motion, in terms of giveninitial conditions.
A common problem in physics is to match the particular initialconditions - generally given as an initial position x_0 andvelocity v_) at t=0 - once you have obtained the general solution.You have dealt with this problem in kinematics, where theformula

1. x(t) = x_0 + v_0t + at^2

has two arbitrary constants (technically constants of integrationthat arise when finding the position given that the acceleration isa constant). The constants in this case are the initial positionand velocity, so "fitting" the general solution to the initialconditions is very simple.

For simple harmonic motion, it is more difficult to fit the initialconditions, which we take to be

x_0, the position of the oscillator at t = 0, and
v_0, the velocity of the oscillator at t = 0.

There are two common forms for the general solution for theposition of a harmonic oscillator as a function of time t:

2. x(t) = A\cos(omega t + phi) and
3. x(t) = C\cos(omega t) + S\sin(omega t),

where A, phi, C, and S are constants, omega is the oscillationfrequency, and t is time.

Although both expressions have two arbitrary constants--parametersthat can be adjusted to fit the solution to the initialconditions--Equation 3 is much easier to use to accommodate x_0 andv_0. (Equation 2 would be appropriate if the initial conditionswere specified as the total energy and the time of the first zerocrossing, for example.)

Find C and S in terms of the initial position and velocity of theoscillator.
Give your answers in terms of x_0, v_0, and omega. Separate youranswers with a comma.

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