More details on decibels. The relative loudness of sound is called intensity (I) and is measured in decibels (dB). 0 dB = 10 log I/Io Io= 10 -12 N/m2. Remember: 0 to 10 dB is 10 times the intensity; 0 to 20 dB is 100 times the intensity; 0 to 30 dB is 1000 times the intensity; 0 to 40 dB is 10,000 times the intensity; 0 to 50 dB is 100,000 times the intensity; and so on.
Problem 4.51 from Serway, R.A. and Faughn, J.S. College Physics. 3rd Edition
n = m(cos10.5) & f =un so, f = 0.3m(cos10.5)
ma = f -mg(sin10.5) = 0.3m(cos10.5) -mg(sin10.5)
therefore, a = 3.25 m/s/s and since we now know a -- we can use it in v2 = v2o + 2as and solve for s. Where s = 19 m.
Problem 4.93 from Serway, R.A. and Faughn, J.S. College Physics. 3rd Edition
-m1a = T + f1 - m1g(sinx) where f1 = umm1g(sinx)
m2a = T - f2 - m2g(sinx) where f2 = um2g(sinx)
set the above two equation equal to each other by solving for T.
a = g(sinx) {(m1-m2)/(m1+m2)} - ug(cosx) then plug that in for T.
Sample Problem #1 Shear in a steel Plate
shearing strain = delta x/h
Shearing stress = (F/A)/shearing strain
Force = A * Shearing stress
Sample Problem #2 Floating Metal Block
L = edge length; h1 = the depth in mercury; h2 = the depth in gasoline; h3 = the height above the surface
Lecture Problems for Chapters 11 & 12
A length of plastic tubing is sealed at one end and then lowered with a weight (open end down) from the surface of a fresh water lake where the temperature is 25 degrees C to a depth, h, where the temperature is 5 degree C. At this depth, the volume of the air trapped within the tube decreases to one fourth its volume on the surface. Determine the depth, h.
A 500 gram mass of mercury at a temperature of 80 degrees C is poured into a container filled with 300 grams of water at 20 degrees C. The water is contained in a copper cup of mass 75 grams which is also initially at 20 degrees C. What is the final temperature of the system?
A container holds 100 grams of water. Assume that evaporation removes 0.5 grams and estimate the decrease in temperature due to the evaporation process.
A particular house loses about 7.5x107 calories on a typical January day. Estimate the mass and volume of a rock needed as a heat storage reservoir if the solar collectors operate at an average temperature of 65 degree C, assuming that the house is to be maintained at an average temperature of 22 degrees C. The rock has a specific heat of approximately 0.21
cal/goC and a density of about 2.7x103 kg/m3.
An aluminum rod and a copper rod are joined end to end in good thermal contact. The two rods have equal lengths and equal radii. The free end of the aluminum rod is maintained at a temperature of 120 degrees C and the free end of the copper rod is maintained at 10 degrees C. (a) Determine the temperature of the interface where the two rods are joined. (b) If each rod is 0.2 meters long and 6 sq. cm in cross sectional area, what quantity of heat energy is conducted through the combination during a 30-minute interval?
Lecture Problems for Chapters 13
An ideal gas is confined at a high pressure in container A and is allowed to expand into an evacuated container B. Assume the two containers are insulted so that no heat flows into or out of the system and show that the internal energy of the expanding gas does not change. The answer is: equation 13.2 and no change in the internal energy during a free expansion.
A heat engine operates between two reservoirs of 300K and 500 K. During each cycle, it absorbs 200 calories of heat from the hot reservoir. (a) What is the maximum efficiency of the engine? (b) Determine the maximum work, which the engine can perform during each cycle of operation. The answer is: (a) 40% (b) 335 J.
An ideal gas initially in a volume of 1.5 m3 and at a temperature of 250K expands at a constant pressure of 30 N/m2. During the expansion 110 Joules are added to the gas and the internal energy increases by 35 J. Determine the final volume of the sample. The answer is 4m3.
Lecture Problems for Chapters 29
Compton Scattering
A photon is scattered from a stationary electron imparting kinetic energy of 5 keV to the electron. The scattered photon moves along a line, which makes and angle of 60 degrees relative to the direction of the incident photon. Calculate the wavelength of the incident photon.
Approach:
Express KE of the electron in joules
Use equ. 29.12.
Write the equation, which expresses conservation of energy for a scattering event.
Use the value calculated for delta lambda.
Rearrange the equation into a standard quadratic equation.
Solve for the incident photon's lambda.
Photoelectric Effect
The cutoff wavelength for lithium is 5400 A. What is the stopping potential for electrons emitted by lithium when the incident light has a wavelength of 3200 A?
Approach:
Use equ. 29.8.
Use equ. 29.7 for KEmax.
Express the kinetic energy in units of eV.
de Broglie Wavelength
(a) Calculate the de Broglie wavelength of electrons which have been accelerated through a potential difference of 160 V. (b) At what absolute temperature will electrons have the thermal energies which result in the de Broglie wavelength equal to 10 times the value found in part (a)?
Approach:
(a) Find the velocity of the electron.
Use equ. 29.16.
(b) Solve for the absolute temperature.
Solve for velocity.
Use equ. 29.16.
Lecture Problems for Chapters 30
An electron in a hydrogen atom undergoes a transition from the n=5 to the n=2 state. Find the frequency and the wavelength of the photon emitted in this process.
Hint use equations 30.17 and 30.19
Consider an electron transition between adjacent energy level in the hydrogen atom. Determine the smallest value of the quantum number, n, such that the energy difference between the nth level and the highest level will be (a) 10% of the total energy, (b) 1% of the total energy.
Hint use equations 30.16 and the quadratic equation.
Free Body Diagrams
A Free Body Diagram is a simplified picture showing a single object or collection of objects and all external forces acting on the system.
- All forces should be labeled with cleearly identifying symbols or names.
- It is a good idea to show the length in approximate relative proportion to the other forces in the diagram when this is possible.
-You may find the following conventionss useful in drawing the forces but they are not always used
or followed.
- Place the tail of the arrow representting the force on the object to which the force is applied. For example a block being pushed to the right will have the same diagram as one being pulled to the right as shown here.
- The tail of the force of gravitation,, mg, may be placed at the center of gravity of the object.
- Other forces such as the normal whichh acts over a surface can be placed at the "average" point of application.
o Forces that act parallel to a surface, such as the force of friction, should extend along the surface as shown here. The following diagram would do until we consider torque, or in cases where no rotation of the object is involved.*
o If torque is involved, the tail of the force vector should pass through the point of application of the force.
* See note on next page.
*This diagram, to be correct for rotational equilibrium would have to have the normal shifted to the right because the applied force F and the friction would tend to cause a clockwise rotation increasing the contact force toward the right edge of the block.
Two further examples:
#1.
#2
CROSS PRODUCTS AND THE RIGHT HAND RULE
The cross (or vector) product of two vectors, A and B, is written as A X B and the resulting vector, let us call it C is given according to the following rules:
if A X B = C then:
1) The magnitude of C is given by A X B = ABsin?
where the angle is the smallest angle between the two vectors.
2) The direction is given by the right hand rule.
The right hand rule works as illustrated below for a sample pair of vector. First extend the fingers of the right hand in the direction of the first vector. Turn your hand so that the fingers may be bent toward the second vector. The extended thumb then points in the direction of the product.
RHR practice: Note a vector out of the paper is represented by a dot in a circle, and a vector into the paper is represented by a cross in a circle.
Surface Vectors and Gauss's Law By Joe Stieve
The first point to consider is the surface vector. The area of a surface can be represented by a vector drawn perpendicular to the surface and of magnitude equal to that area. Study the following examples.
EXAMPLE 1: A flat disk of radius 0.50 m. The area, 0.79 m2, can be represented by a vector such as that shown in the diagram below. In this case the vector could be drawn either upward or downward.
EXAMPLE 2: The area of each surface of the rectangular solid shown above can be represented by a surface vector. By convention these vectors are always taken as outward from the enclosed volume.
EXAMPLE 3: For a curved surface, such as the sphere shown at the above right, the incremental surface area dA can be represented by a incremental vector dA drawn normal to the surface at that spot.
Once you understand the idea of a surface vector you can move on to the idea that electric field may be viewed as penetrating a given surface. It is useful to think of the field as a "flow" across the surface and the strength of the field as the amount of flow per unit area at that location. This "flow" is properly called flux. If we do this, we can calculate the total flux through the surface by multiplying the field strength E by the area. By using the scalar product of E and the surface vector A we automatically take only that component of the field that is perpendicular to the surface. Study the examples on the following page.
EXAMPLE 4: The electric field through the surface shown at the left in the diagram below is perpendicular to the surface and therefore in the same direction as the surface vector. Here the flux is simply EoA which equals 6 Nom2/C. Note that flux is a scalar quantity.
EXAMPLE 5: The field in the case shown at the right is not perpendicular to the surface. If we think of the field as broken into components along the surface and perpendicular to the surface, we can see that only the perpendicular component "takes field through the surface". The scalar product EoA takes this into account for us because EoA = EAcos?.
Since electric fields always begin and end on electric charges, if an enclosed volume contains no net charge the flux taken over the entire closed surface must be equal to zero. This follows because whatever flux enters must also leave. Entering flux will be negative because the perpendicular component of the entering flux will be 180° from the outward directed surface vector, that is the scalar product EodA = EdAcos180° = -EdA. The outward flux will be positive.
It would seem reasonable that the net flux through a closed surface containing a charge will not be zero and will in fact be proportional to the size of the charge contained within. This fact is stated in what is known as Gauss's Law.
Here the circle on the integral sign is an indication that the surface is closed. The symbol ?o is the constant of proportionality. Remember that q is the charge contained within the surface.
The Derivative
By Joe Stieve
The following discussion is meant to provide an introduction and elementary working knowledge of the derivative. It will give a sufficient background for the initial work done at the start of this course. The derivations and background provided in you calculus course will enable you to gain a better comprehension of how the derivative works and how it is extended to functions other than polynomials.
In physics a concept that is of constant interest is how a physical quantity such as position, velocity or momentum changes with time. As an example, let us examine the position vs. time graph shown here.
You will notice that the average rate of change in position for the first 9 seconds can be found by taking the over all change for some interval and dividing it by the elapsed time. This quantity is called the average velocity and is given by vav = ?x/?t. You should notice that this is the same as the slope of a straight line drawn from the beginning of the interval to the end of the interval. In a like manner the slope of a straight line connecting any two points on the graph represents the average velocity over that time interval. If the quantity had been something other than position the slope of such lines would clearly represent the average rate of change of that quantity with respect to time.
Now, the role of the differential calculus is to find the instantaneous rate of change of a function. We will leave the derivations for your calculus class and simple state that the method of achieving this involves inspection of the limit of the average rate of change over smaller and smaller intervals. Shown on the next page is the same position vs. time graph however the lines representing the average velocity for intervals to either side of the time
t = 5.0s have been shown.
Notice that as the intervals become shorter to either side of this point the slope gets closer to 10. The slope of the graph at t = 5.0s, and therefore the velocity at the instant t = 5.0s, is said to be 10.m/s. A line drawn through this point and having this slope is defined as a tangent drawn to the curve at this point. The instantaneous rate of change is represented by the symbol d/d associated with the appropriate variables. Here we have v = dx/dt. Because, in general, dx/dt will be a function which is derived from the original function, we refer to it as the derivative.
The question remains how do we find the exact value for this instantaneous rate of change of the function? Fortunately, even though the proofs are rather involved, the answer is relatively easy for most functions. For our immediate purposes we will consider only monomials and polynomials. Three rules will serve:
i. For a constant x = C, dxdt = 0.
ii. For a monomial x = Cxn dxdt = nCxn-1
iii. For a polynomial simply take the sum of the derivatives of the individual monomial parts. That is: the derivative of a sum is
the sum of the derivatives.
For the graph shown above the function is x = t2, and the derivative is dx/dt = 2t. At the point we considered t = 5 therefore the velocity was 2 X 5 or 10.
The units were left out of the above for simplicity. Actually the original equation should be written as x = (1m/s2)t2 and the derivative will follow automatically as v = dx/dt = 2X(1m/s2)t. Putting t = 5.0s in the last expression gives us that v = 10.m/s.
What if the original equation had expressed the velocity as a function of time? The slope of a straight line drawn between any two points on the graph would express the average rate of change of the velocity with respect to time. This quantity is referred to as the average acceleration. It follows then that the slope of a tangent drawn to a point on such a graph has a slope, which describes the instantaneous rate of change of velocity with respect to time, that is the instantaneous acceleration. In terms of the function then, a = dv/dt.
In summary then, we start with x =f(t), v = dx/dt and a = dv/dt and we can think of a as the derivative of the derivative or the second derivative of position with respect to time. This is written
as a = d2x/dt2.
One final word about notation. The following shows some of the various notations that are used for the first and second derivative of a function. The final ones, referred to as "x dot" and "x double dot", are Newton's notation and are used in physics for the derivative with respect to time and are not used if the independent variable is some other expression.
Area & The Definite Integral
As a first example, suppose we wish to find the area bounded by the line f(x) = 3x, the x axis, x = 0 and x = 2. This area is shown in figure 1. We recognize this as a triangle and can compute its area by the simple formula: A = 12 bh. The answer is, of course, 6.
Now let us find the area using calculus. The same graph is shown in figure 2, but this time an extremely narrow rectangle is also shown. The height of the rectangle is simply the value of the function at that point, f(x). The width we will represent by the symbol dx which stands for the minute difference in the x values on the left and right sides of the rectangle. The area of this small rectangle we will symbolize by dA, therefore dA = f(x)dx. Now to get the area of the entire triangle we must take the sum of all such rectangles which can be drawn to cover the given area. To indicate this process we use the distorted S, for Sum, known as the integral sign, ? . We take this sum for values of x from O to 2. An expression for this is:
To evaluate this expression we resort to the following theorem (The Fundamental Theorem of Integral Calculus) : If the function f is continuous on the closed interval [a,b] and if F is any antiderivative of f on [a,b], then
For the above example we have:
P. 2 of 2
The answer is of course the same as that achieved using the area of a triangle, however this method works for all functions that meet the criterion of continuity expressed above. Clearly this is not a rigorous mathematical treatment of the definite integral. That is left to your mathematics course. The following examples should help you apply this powerful tool to physics problems.
EXAMPLE 1: Find the area bounded by the graph of f(x) = x2, the x axis, x = 1 and x = 2.
Note that if we take the limits of integration in the reverse order we get the negative of the area.
This may be interpreted as the result of taking the area of the minute rectangle as f(x)dx, because dx, the difference in x, if you insist on moving from right to left, i.e. from 2 to 1, is negative making the entire product f(x)dx negative. Then in turn the sum of these will come out negative.
EXAMPLE 2: Find the area bounded by the curve f(x) = x3, the x axis, x = -2 and x = 0.
Note that here the area comes out negative in spite of the fact that we move from left to right, i.e. from x = -2 to x = 0. The dx is positive here, but the height of the minute rectangle, f(x), is negative producing a negative product for f(x)dx. Signed areas have meaning in physics. For example, the area under a velocity vs. time graph represents displacement. A negative area represents a negative displacement.