Answer:
Amplitude, A = 0.049 meters
Explanation:
Given that,
A harmonic wave travels in the positive x direction at 6 m/s along a taught string. A fixed point on the string oscillates as a function of time according to the equation :
.......(1)
The general equation of a wave is given by :
.......(2)
A is amplitude of wave
On comparing equation (1) and (2) we get :
A = 0.049 meters
So, the amplitude of the wave is 0.049 meters.
Complete Question
The complete question is shown on the first uploaded image
Answer:
The temperature change is 
Explanation:
From the question we are told that
The velocity field with which the bird is flying is 
The temperature of the room is 
The time considered is t = 10 \ seconds
The distance that the bird flew is x = 1 m
Given that the bird is inside the room then the temperature of the room is equal to the temperature of the bird
Generally the change in the bird temperature with time is mathematically represented as
![\frac{dT}{dt} = -0.4 \frac{dy}{dt} -0.6\frac{dz}{dt} -0.2[2 * (5-x)] [-\frac{dx}{dt} ]](https://tex.z-dn.net/?f=%5Cfrac%7BdT%7D%7Bdt%7D%20%3D%20-0.4%20%5Cfrac%7Bdy%7D%7Bdt%7D%20-0.6%5Cfrac%7Bdz%7D%7Bdt%7D%20-0.2%5B2%20%2A%20%20%285-x%29%5D%20%5B-%5Cfrac%7Bdx%7D%7Bdt%7D%20%5D)
Here the negative sign in 
is because of the negative sign that is attached to x in the equation
So
![\frac{dT}{dt} = -0.4v_y -0.6v_z -0.2[2 * (5-x)][ -v_x]](https://tex.z-dn.net/?f=%5Cfrac%7BdT%7D%7Bdt%7D%20%3D%20-0.4v_y%20%20-0.6v_z%20-0.2%5B2%20%2A%20%20%285-x%29%5D%5B%20-v_x%5D)
From the given equation of velocity field
So
![\frac{dT}{dt} = -0.4[0.2t] -0.6[-1.4] -0.2[2 * (5-x)][ -[0.6x]]](https://tex.z-dn.net/?f=%5Cfrac%7BdT%7D%7Bdt%7D%20%3D%20-0.4%5B0.2t%5D%20%20-0.6%5B-1.4%5D%20-0.2%5B2%20%2A%20%20%285-x%29%5D%5B%20-%5B0.6x%5D%5D)
substituting the given values of x and t
![\frac{dT}{dt} = -0.4[0.2(10)] -0.6[-1.4] -0.2[2 * (5-1)][ -[0.61]]](https://tex.z-dn.net/?f=%5Cfrac%7BdT%7D%7Bdt%7D%20%3D%20-0.4%5B0.2%2810%29%5D%20%20-0.6%5B-1.4%5D%20-0.2%5B2%20%2A%20%20%285-1%29%5D%5B%20-%5B0.61%5D%5D)
Magnetic flux can be calculated by the product of the magnetic field and the area that is perpendicular to the field that it penetrates. It has units of Weber or Tesla-m^2. For the first question, when there is no current in the coil, the flux would be:
ΦB = BA
A = πr^2
A = π(.1 m)^2
A = π/100 m^2
ΦB = 2.60x10^-3 T (π/100 m^2 ) ΦB = 8.17x10^-5 T-m^2 or Wb (This is only for one loop of the coil)
The inductance on the coil given the current flows in a certain direction can be calculated by the product of the total number of turns in the coil and the flux of one loop over the current passing through. We do as follows:
L = N (ΦB ) / I
L = 30 (8.17x10^-5 T-m^2) / 3.80 = 6.44x10^-4 mH
<span>Depends on the precision you're working to.
proton mass ~ 1.00728 amu
neutron mass ~ 1.00866 amu
electron mass ~ electron mass = 0.000549 amu
Binding mass is:
mass of constituents - mass of atom
Eg for nitrogen:
(7*1.00728)-(7*1.00866)-(7*0.000549)
-14.003074 = 0.11235amu
Binding energy is:
E=mc^2 where c is the speed of light. Nuclear physics is usually done in MeV[1] where 1 amu is about 931.5MeV/c^2. So:
0.11235 * 931.5 = 104.6MeV
Binding energy per nucleon is total energy divided by number of nucleons. 104.6/14 = 7.47MeV
This is probably about right; it sounds like the right size!
Do the same thing for D/E/F and recheck using your numbers & you shouldn't go far wrong :)
1 - have you done this? MeV is Mega electron Volts, where one electronVolt (or eV) is the change in potential energy by moving one electron up a 1 volt potential. ie energy = charge * potential, so 1eV is about 1.6x10^-19J (the same number as the charge of an electron but in Joules).
It's a measure of energy, but by E=mc^2 you can swap between energy and mass using the c^2 factor. Most nuclear physicists report mass in units of MeV/c^2 - so you know that its rest mass energy is that number in MeV.</span>
Based on the given values above, in order for us to get the answer, we need to convert the units first. So in 1 kilogram, there is 1,000,000 micrograms. In this case, 1.6 kilograms is 1,600,000 micrograms. For the week to seconds, 1 week is equivalent to 604,800 seconds. Therefore, 1,600,000 micrograms/604,800 seconds. So we are going to simplify this. So it would be 2.65<span>µg/s. Hope this answers your question.</span>
