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123 Cards in this Set
- Front
- Back
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Vertical and Horizontal Component
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Vertical Component: (Length of hypotenuse)*Sin(Angle)
Horizontal Component: (Length of hypotenuse)*Cos(Angle) |
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Sin30
Cos30 |
Sin30=.5
Cos30=.87 |
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Average Speed
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Total distance traveled/time
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VELOCITY
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VECTOR QUANTITY. If ended up at the starting position, change in distance (displacement) is 0 therefore average velocity is 0.
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A 2kg rock is thrown vertically upward at a speed of 3.2 m/s from the surface of the moon. if it returns to its starting point in 4 seconds, what is the acceleration due to gravity on the moon?
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ROUND TRIP=4sec ---->
ONE WAY = 2sec acceleration=dv/dt =(0-3.2)/2=-1.6 m/s^2 |
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A projectice is launched horizontally from a raised platform. If air resistance is ignored, then as the projectile falls to the earth, the magnitude of the vertical component of the velocity of the projectile:
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INCREASES, WHILE THE HORIZONTAL COMPONENT REMAINS CONSANT.
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Where in its path does a projectile in free fall near the surface of the earth experience the greatest acceleration?
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Acceleration is the same at all points in the path.
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1-kg mass hung at the end of a vertical spring w/ spring constant 20 N/m. When the mass comes to rest, how many meters will the spring have stretched?
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mg=kx
(1)(10)=(20)x x=.5 m |
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The acceleration experienced by a block moving down a frictionless plane inclined at any angle =?
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IS CONSTANT
ma=mgsin(angle) a=gsin(angle) |
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F(normal)
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F(normal)=mgCos(angle)
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An object is moving in a circle at constant speed. Its acceleration vector must be directed?
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RADIALLY and toward the center of the circle. Uniform circular motion.
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Centripital force
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F=mv^2/r
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torque
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torque=radius*Force
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A 200kg airplane flying at 50 m/s is slowed by turbulence to 40m/s over a distance of 150m how much work was done on the plane bu the turbulent air?
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WORK-ENERGY THEOREM.
The work performed on the plane by the turbulent air is = to the change in the planes kinetic energy. dKE=(1/2)m(v^2-v0^2) =1000*(40^2-50^2)=-900 kJ |
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If a hocky puck has mass m and speed v, radius r, and the tension in the string is T. What is the the Tension in the rope when whirled in a circular path?
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Tension in the string prvides the centripetal force.
T=mv^2/r |
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Which of the following statements applies to a system in which 2 objects undergo an ideal ELASTIC COLLISION?
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1) Kinetic energy is conserved (only in elastic)
2) Momentum is conserved: Momentum is ALWAYS conserved (elastic or not) FALSE: The velocity of each object remains unchanged. |
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Potential Energy --> Kinetic Energy
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mgh=(1/2)mv^2
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A particle with an initial v=4 m/s moves along the x axis under constant a. 3 s later its v=14 m/s. How far did it travel during those 3 seconds
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d=(average velocity) * time
d=(1/2)(v0+vf)*t d=(1/2)(4+14)*3 d=27m |
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An object is thrown horizontally w/ an initial speed of 10 m/s. How far will it drop in 4s ?
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ONLY Vertical component
d=(1/2)at^2 d=(1/2)(10)(4)^2 d=80 m |
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From a height of 100 m a ball is thrown horizontally with an initial speed of 15 m/s. How far does it travel horizontally in the first 2 sec?
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ONLY Horizontal component.
d=vt d=(15)*2=30 m |
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Mass Moon=M w/ Radius R
small object w/ mass=m and radius=r What acceleration will it fall? |
ma=GMm/R^2
a=GM/R^2 |
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An object m=36 kg and weighs 360N at the surface of the Earth. If this object is transported to an altitude = to 2Rearth then at this new elevated position the object will have?
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Mass=36 kg because Mass NEVER CHANGES
Rtotal=Rearth + 2Rearth=3Rearth Therefore since R increases by *3, the gravitational acceleration will decrease by a factor of 3^3=9. Therefore the new weight will be 360/9=40N. |
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An object slides down an incline plane with constant speed. If the ramp's incline angle is theta, what must be the coefficient of kinetic friction, u, between the object and the ramp.
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Fnet=Rgravdownramp-Ffriction
Fnet=mgsin(theta)-umgcos(theta)=0 mgsin(theta)=umgcos(theta) u=sin(theta)/cos(theta) |
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Work/KE/PE
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Work=PE=KE
Fdcos(theta)=mgh=(1/2)mv^2=(1/2)kx^2 |
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A car weights 8500N and traveling 20 m/s engages its brakes. The car skids for 200m b4 coming to rest. What is the coefficient of friction b/w the road and the car's tires?
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Ff=uFn=umg
Ff(d)=umgd umgd=(1/2)mv^2 u=(1/2)v^2/gd u=(1/2)(20)^2/(10)*200 u=.1 |
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POWER
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P=Fv=W/t
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2 objects are submerged in a fluid at the same depth. Compared to the smaller of the 2 objects, the larger one will experience?
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EQUAL FLUID PRESSURE!
The pressure due to a fluid of density p surrounding an object submerged at depth d below the surface is given by the expression P=pgd which is clearly independent of the size of the object |
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An object is floating on a fluid. The weight of the fluid displaced by the floating object is?
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EQUAL to the weight of the object.
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The atmospheric pressure at a height of 2 km above the surface of the earth is?
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LESS than the atmospheric pressure at the surface
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Bouyant force
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B=(density of liquid)*Volume of object * g
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A circular plate with an area of 1 m^2 covers a drain hole at the bottom of a tank of water which is 1m deep. Approx how much Force is required to lift the cover if it weighs 2000N.
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F=PA
P=pgd F=pgdA F=(1000)(10)(1)(1)=10000 Ftotal=10000+2000=12,000 N |
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Flow rate
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flow rate= Area * velocity
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An object that weighs 200N floats in a tank of water. how much of the object's volume is actually submerged?
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Since the object floats, the buoyant force it feels must balance its weight. Therefore, B=200N.
By Archimedes Principle: B=(pfluid)*(Vsub)*g B=200N Vsub=200/(1000*10) V=.02 m^3 |
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Total Forces when an object sinks
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Fbouyant+Fnormal=Fgravity
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Youngs Modulus
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E=pressure/fractional change in length
E=Stress/strain |
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If a metal pole can be subjected to great stress while undergoing relatively little fractional change in lenght, then young's modulus has a?
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LARGE POSITIVE VALUE
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Work and Normal Force
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The work done by the normal force is automatically zero since N is perpendicular to the displacement of the object.
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How much work is done by the gravitational force as a 10kg object is lifted from a height of 1 m above the ground to a height of 3 m above the ground?
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Since the object's displacement is upward, but the force of gravity is downward, the work done by gravity on this object must be negative.
W=-F(d2-d1)=-mg(d2-d1) W=-(10)(10)(2)=-200J |
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Work and Kinetic energy
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Work=Change in KE
W=(1/2)m(V2^2-V1^2) |
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Units of momentum
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1) p=E/c (for photons): [p]=J/(m/sec)=J*sec/m
2)p=Ft: [p]=N*sec 3)p=mv: [p]=kg*m/sec |
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When particles of neon gas at room temperature collide, the collisions are usually perfectly elastic. This is because?
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Most of the collisions do not cause electron transitions b/w energy levels. The only way to have an inelastic collision is to have a way to absorb some o the energy. Thats pertty easy to arrange with macroscopic objects: the objects can deform, fall apart, stick together, etc. The only way to change the energy of a neon atom is to excite one of its electrons into a higher energy level. If this does not happen, then the collision has to be perfectly elastic.
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In kan elastic collision b/w 2 balls, the avg magnitude of the internal forces exerted by each ball on the other is:
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INVERSELY proportional to the contact time.
p=Ft---> F=p/t |
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Direction of motion when rope is cut when roating in a circle?
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If the centripetal force holding the object in its circular path is suddenly taken away, the object will fly off in whatever direction it was deaded at the instant the centripetal force was removed. Velocity is always tangent to the objects path.
*Acceleration is always directed radially inward |
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T (period) of rotation of a rotating platform?
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T=d/v
T=2(pi)r/v |
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An object of mass 2kg floats MOTIONLESS in a fluid of specific gravity 0.8. What is the magnitude of the buoyant force?
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B=w=mg=(2)*(10)=20
B=pVg Since the object is motionless, its acceleration is zero, so the net force on it must also be zero. Therefore, the magnitude of the buoyant force upward must = the magnitude of the weight downward. |
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A block of some unknown material is floating in a fluid of specific gravity 1.5. If 1/2 of the block is submerged, what is its density?
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w=B
pblockVg=pfluidVsubg pblockV=pfluid(1/2V) pblock=(1/2)pfluid =(1/2)(1.5)(1000) =750 kg/m^3 |
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An object whose specific gravity is 2.0 weighs 200N less when it is weighed while totally submerged in water than when it is weighed in air. What is the weight of this object in air?
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B=200N
w/B=(pVg)/(pwaterVg) =2*pwater/pwater=2 w=2B=2(200)=400N |
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Specific gravity
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Specific gravity=
=w/B=pobject/pwater |
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Let P1 be the total pressure at depth d below the surface of a large body of water, and let P2 be the total pressure at depth 2d. Assuming that the variation in density is negligible, how do P and p compare?
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P=Patm + pgd
P1=Patm + pgd P2=Patm + 2pgd Therefore P2<2P1 |
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A fluid of density p flows through a horizontal pipe with negligible viscosity. The flow is streamline with constant flow rate. IF the speed at Point 1 is v and the speed at Point 2 is 3v, then the pressure at Point 2 is?
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LESS than the pressure at point 1 by 4pv^2:
P1+(1/2)pv1^2=P2+(1/2)pv2^2 P2=P1+(1/2)p(v1^2-v2^2) P2=P1+(1/2)p(v^2-9v^2) P2=P1-4pv^2 |
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Poiseuille's Law
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Poiseuille's Law: smooth/Laminar flow
Volume Flow rate=F=(p1-p2)/R =[pi*(change in P)r^4]/[8*viscosity*length] Resistance to flow=[8*viscosity*L]/[pi*r^4] |
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Archimede's principle
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Archimede's principle:
Biody immersed in a fluid is buoyed up by a force = to the weight of the displaced fluid |
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Pascal's Law
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Pascal's Law:
Change in pressure applied to an enclosed incompressible fluid is transmitted undeminished to every portion of the fluid to the walls of its container -incompressible fluid transmits pressure -Pistons P1=P2 ----> F1/A1=F2/A2 A1d1=A2d2 |
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Bernouili's Principle
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Bernouili's Principle:
Pressure+(KE/volume)=constant As speed of a moving fluid )liquid/gas) increases, Pressure w/in the fluid decreases P1+pgh2+(1/2)pv1^2=P2+pgh2+(1/2)pv2^2 |
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fluids
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pgh=(1/2)pv^2
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Capacitors
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ENERGY Stored=PE=(1/2)CV^2
Q=CV Work=qV C=Ke0A/d (e0=8.85*10^-12 C2/N=m^2) In order for charged parallel plates to act as capacitors the material b/w the plates must not conduct electricity. |
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The direction of the force exerted on the wire by the field is?
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PERPENDICULAR TO THE WIRE AND PERPENDICULAR TO THE FIELD LINES:
Since F=IL*B, the F is, by definition, perpendicular to both L and B, that is, perpendicular to both the wire and to the magnetic field lines. |
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Magnetic field
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B=u0I/(2*pi*r)
u0=permeability of free space = 1.26 * 10^-6 Tm/A) |
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Magnetic force
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F=ILBsin(Theta)
I=Current Theta=angle between L and B B=Magnetic field L=length of wire |
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Electrostatic field due to a single point charge
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E=kQ/r^2
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Force between 2 particles
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F=kq1q2/r^2
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Electric Field Vectors:
As the electrostatic field strength is measured at points that progressively approach a negatively-charged particle, the field vectors will point? |
TOWARDS THE PARTICLE AND EXHIBIT PROGRESSIVELY INCREASING MAGNITUDE.
Electric Field vectors always point toward negative source charges. SINCE ELECTRIC FIELD STRENGTH IS INVERSELY PROPORTIONAL TO THE SQUARE OF THE DISTANCE FROM THE SOURCE CHARGE, THE MAGNITUDE OF THE ELECTRIC FIELD WILL INCREASE AS WE MOVE CLOSER TO THE SOURCE CHARGE. |
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An object with a mass of 1.25kg and a charge of -1C is at rest 2m above a large, positively-charged plate. the object experiences a constant electric field of 5 N/C. If the object is free to move, waht will be its approx velocity at the instant it strikes the plate?
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Work=Felec*d=qEd
Work-Energy Theorem qEd=(1/2)mv^2 plug in : q=1C E=5 N/C d=2m m=1.25kg Therefore: v=4 m/s |
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An object with mass m and a net charge of -q moves in a circular orbit of radius r around a fixed object of mass m with a net charge of +q. What is the spreed of the orbiting object?
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The attractive Coulomb force provides the centripetal force;
kqq/r^2=mv^2/r Therefore: v=sqroot(kq^2/mr) k=Coulomb's constant =9*10^9 N-m^2/C^2 |
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Force on a charge q by the electric field E
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F=qE
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Electric current will flow?
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ONLY FROM A POINT OF HIGH POTENTIAL TO A POINT OF LOWER POTENTIAL AND ONLY THROUGH A POTENTIAL DIFFERENCE.
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Vrms?
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Vrms=Vmax/sqroot(2)
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A positive test charge moves perpendicular to a uniform magnetic field directed into the page. Which of the following best depicts the subsequent path followed by the charge if no other factors affect its motion?
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COUNTER CLOCKWISE ROTATION:
The force F on a charge q moving with velocity v through a magnetic field B is given by the equation F=qV*B. If q is positive, then the direction of F is the same as the direction of v*B. If B points into the page and v is perpendicular to it, then F will always be directed 90degrees counterclockwise from v, implying that the charge will execute counterclockwise circular motion. |
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A motionless positive-charged particle is situated in an unknown medium. The magnitude of the magnetic field generated by the particle will be?
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EQUAL to ZERO:
A magnetic field is created only by electric charges in motion. Since the charged particle described in this question is motionless, it generates no magnetic field. |
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2 objects A and B have = charge and = mass. Neither body is in motion because the gravitational and electrostatic forces b/w them are =. If the mass of object A is 1/2, equilibrium would b maintained if?
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THE CHARGE on OBJECT B were HALFED!
Gm1m2/r^2=kq1q2/r^2 [Gm1(1/2m2)]/r^=[kq1(1/2)q2]/r^2 |
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Power
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Power=Energy dissipated per until time=E/t
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Resistance
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R=p(L/A)
p=resistivity L=Length A=Area |
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Change in Potential Energy of a charge when moved from one position to another
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detaPE=qV=q[(kQ/r1)-(kQ/r2)]
=qkQ[(1/r1)-(1/r2)] |
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Power and (Heat) Energy
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P=I^2R
E=Pt |
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How much work is required to insert the dielectric in?
dielectric constant =4 =4.3*10^/4 J 2 uF parallel plate capacitor 12V battery |
Wagainst_field=change in PE
=(1/2)C'V^2-(1/2)CV^2 =(1/2)V^2(4C-C) =(1/2)(12)^2[3*(2*10^-6)] =4.3*10^/4 J |
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Springs
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Elastic Potential Energy=(1/2)kx^2
(1/2)kx^2=(1/2)mv^2 Hook's law: F=-kx |
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Light Waves
SNELL's Law |
n1sin(theta1)=n2sin(theta2)
n=the refractive index Theta1=angle of refraction Theta2=angle of incidence n=c/v=(3*10^8)/v n increases with increasing f Critical Angle: n1>n2 where refractive index of incidence is creater than refractive index of refraction, sin(thetac)=n2/n1 |
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On passing from a medium of lower refractive index, a light wave will experience?
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decreased wavelength and decreased velocity.
n=c/v when n increases v decreases. v=wavelength*frequency v decreases then wavelength also decreases since f is constant |
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A transverse wave and a longitudinal wave are compared. They are found to have amplictudes of = magnitude. WHat can be said about their speeds.
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THE SPEEDS OF THE @ WAVES ARE UNRELATED TO THEIR AMPLITUDES. The speed of a wave is determined by the characteristics of the medium and the type of the wave. It is independent of amplitude.
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Souhnd would be expected to travel most slowly in a medium that exhibited?
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Low resistance to compression and high density:
v=sqroot(B/p) where B=bulk modulus (resistance to compression) p=desnity |
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An increase in which of the following freatures of a sound wave would necessarily result in an increase in its decibel level?
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AMPLITUDE:
The sound level (in dB) of a sound wave of intensity I is given by the equation B=10log(I/I0),where I0=10^-12 W/m^2, the threshold of hearing. Clearly, B is independent of v, wavelength, or frequency. The answer must be amplitude because an increase in amplitude is a result of an increase in energy, and thus in the intensity of the wave. |
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The discovery of "red shift" was a major scientific breakfhrough. Upon detecting the light waves emitted by stars, it was discovered that the measured frequency of the light was lower than was known to be the case, One explanation for this phenomenon is that?
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The stars are moving away from the earth: The Doppler Effect for light is similar to that for sound; that is, motion of the source away from the observer results in a decrease in the perceived frequency.
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A screeching bat, emitting sounds with a frequency of 60 kHz, is moving at a speed of 10 m/s toward a stationary observer. what is the apparent frequency of the sound waves detected by the observer? (speed of sound in air=340m/s)
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Since the bat is moving twd the observer, the percieved frequence, f', must be higher than the actual source frequency, f.
Doppler-effect equation: f'=[v/(v-10)]f f'=[340/330](60) f'=61.8khz |
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Which of the following is true of the properties of a light wave that is traveling in vacuum?
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INCREASED FREQUENCY RESULTS IN DECREASED WAVELENGTH
c=(wavelength)*frequency |
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Which of the following is true of the properties of a light wave as it moves from a medium of lower refractive index to a medium of higher refractive index? And the angle of refraction will be?
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SPEED WILL DECREASE
v=c/n THE ANGLE OF REFRACTION WILL BE LESS THAN THE ANGLE OF INCIDENCE AND REFLECTION: Snells low implies that a light beam entering a medium with a greater refractive index than the incident medium will refract toward the normal. Thus, the angle of refraction will be less than the angles of incidence and reflection. |
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MIRRORS
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REAL---> INVERTED (-m)
VIRTUAL---> UPRIGHT (+m) |
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The image of an object placed outside the focal point of a concave mirror will be?
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REAL AND INVERTED
1/0+1/i=1/f 1/i=1/f-1/o 1/i is positive therefore--> m is negative--> Inverted and if invered then it is real |
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The image of an object placed in front of a convez mirror will be?
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VIRTUAL and UPRIGHT
1/0+1/i=1/f 1/i=-1/f-1/o 1/i is negative --> m is posisitve--> upright --> Vertual |
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Lens Power
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P=1/f
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Lens Equation
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1/o+1/i=1/f
m=-i/o |
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If the intensity of a sound wave is decreased by a factor of 10, by how much will the sound level of the wave decrease?
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B=10log(I/I0)
every factor of 10 decrease ina sound wave's intensity results in a decrease by 10dB in its sound level. |
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Speed of sound through an ideal diatomic gas
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v=sqroot(1.4RT/M)
R=gas constant T=absolute temp M=molecular mass |
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Frequency of Beats
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fbeat=f1-f2
The difference between the frequencies f1 and f2 of the 2 parent waveform |
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Frequency of the composite waveform
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fcomposite=(1/2)(f1+f2)
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Amplitude of the resulting waveform
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A(t)=2A0cos[pi*(f1-f2)t]
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Two tuning forks, whith f=260 and 265 Hz are struck, and heats are heard. What is the frequency of the resultant sound wave and frequenciy of the beats respectively?
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fcomplosite=(1/2)(f1+f2)
=(1/2)(265+260)=262.5 Hz. Beat frequency is fbeat=f1-f2=265-260=5Hz |
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A sound wave (f1=400 Hz, A1=5units) interferes with another sound wave (f2=410 Hz, A2=5 units). What is the max amplitude of the resultant sound wave?
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When 2 waves interfere, the max possible amplitude of the resultant wave is = to the sum of the 2 indidivual amplitudes (this is completely constructive interference). In this case, Amax=A1+A2=5+5=10units
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The following are transverse waves
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1) A wave traveling down a stretched string
2) Waves from a pebble dropped in a pool of water 3) light waves 4) NOT SOUND WAVES: These are longitudinal. |
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Radius of curvature for mirror
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r=2f
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When the eye is focused on a distant object, the image will appear?
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AT THE FOCAL POINT:
1/i+1/o=1/f o=infinity 1/i=1/f |
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The lens used to correct hyperopia (farsightedness) will be?
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CONVERGING LENS, RESULTING IN A DECREASED FOCAL LENGTH. Adding a corrective converging lens to the converging crystalline lens of the eye results in an increase in lens power, which is necessary to correct hyperopia. An increase in lens power implies a decrease in focal length, since power is the reciprical of focal length.
Hyperopia occurs when light entering the eye focuses behind the retina, instead of directly on it. This is caused by a cornea that is flatter, or an eye that is shorter, than a normal eye. |
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2 Transverse waves travel in the same direction along a stretched string. Individually, they shave the same amplitude and frequency. However at t=0, where Wave #1 has its max positive displacement, Wave #2 has 0 displacement; and where Wave #1 has zero displacement, Wave #2 has its max + displacement. what is the phase difference between them?
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90 degrees
They are off by (1/2) of a wavelength, Therefore ( 1/2)*360=90 decrees |
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A 12m string fixed at both ends is resonating at a frequency that produces exactly 4 nodes. Calculate the wavelength of the standing wave.
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Since the 2 ends count as nodes, tehre will be 3 antinodes. Therefore there are 3 1/2 wavelengths supported on the length L of the string.
3*lamda/2=L Wavelength=2L/3=(2*12)/3=8m |
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The distance b/w the positions of an antinode and the nearest node on a standing wave (produced on a vibrating wire fixed at both ends) is d. Determine the # of nodes on teh string in terms of d and the length L of the wire.
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Since we have a standing wave, we know that the harmonic wavelength must satisfy the following equation, where n is the # of antinodes:
L=n(lambda/2) or Lambda=(2L)/n The distance between an antinode and the nearest node is always 1/4 of a wavelength. So since d is = 1/4 wavelength and the # of nodes is 1 more than the # of antinodes: d=(1/4)lamda=(1/4)*(2L/n)=L/2n n=L/2d Number nodes=(L/2d)+1 |
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A sound wave w/ frequency f travels through air at speed v. With waht speed will a sound wave with frequency 4f travel through air?
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v : Since the wave speed is independent of the frequency, the speed is still v. Of course the wave with frequency f4f will have 1/4 the wavelength, but the wave speed will stay the same.
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When 2 tuning forks are struck simulaneously, beats are heard every 250ms. If one of these tuning forms produces a 298 Hz sound wave, what could be the frequency of the sound wave produced by the other tuning fork?
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If the beats are heard every 250ms then there would be 4 beats per sec. Since fbeat=4hz, the frequencies of the 2 tuning forms must differ by 4 hz. if one fork has a frequency of 498 then the other must have a frequency of either 498-4=292Hz or 298+4=502Hz.
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nth harmonic
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Whether the pipe is open or closed, the nth harmonic frequency is n times the fundamental frequency.
fn=nf1=n(v/4L) |
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A person blowing a whistle is running twd anothe rperson who is standing still. As the runner approaches(at constant velocity), the strationary listener will hear a sound that is?
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HAS A FIXED PITCH BUT INCREASES IN INTENSITY.
The Doppler Effect tells us that if the source moves toward the detector, then the perceived frequency will be higher than that of the source. But if the source's speed is constant, this higher pitch will remain constant as well. Finally , its clear that if the source of sound gets closer, the intensity increases. |
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A sound wave travels through a metal rod with wavelength lambda and frequency f. When this sound wave passes into the air, the wave will then have a wavelength:
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SHORTER THAN LAMBDA AND FREQUENCY EQUAL TO f.
When a wave passes from 1 medium to another (or is reflected at a boundary between 2 media), the frequency of the transmitted or reflected wave is no different from that of the original wave. Since sound travels more slowly through air than it does through metal, the equation lambda=v/f tells us that the wavelength in air will be shorter than in metal. |
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An airplane is traveling at constant speed in a circular path of radius 400m parallel to the groupd whose center is 300 m above an air traffic control tower. Its engine is the source of audible sound waves of a fized frequency. To a stationary listener on the tower, how would the detected frequecy of the engine differ from the actual source frequency while the plane circled above her?
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THE DETECTED FREQUENCY WOULD REMAIN CONSTANT AND = TO THE SOURCE FREQUENCY.
The distance b/w the source (the plane) and the listener, 500m, never changes. Since tehre is no relative motion between the source and the tetector, the detected frequency will be no different than that of the source. |
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Spreading Angle
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detatheta=wavelength/diameter
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Sound waves are similar to waves on a string in the following ways:
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1)Both transer energy (True for all waves)
2)Both involved a series of masses connected by spring-like forces. True for both because they both involve the displacement of matter (as opposed to light waves) 3)The frequency is equal to the ratio of wave speed to wavelength (True for all waves) 4)The difference is that sound waves are longitudinal waves and waves on a string are transverse waves. |
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Transverse waves
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Displacement of the medium is perpendicular to the direction of propagation of the wave.
For example: ripple in pond or wave on a string. They can occur on a sting, surface of liquid and through a solid. They cannot propage in a gas of liquid. |
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Longitudinal waves
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Displacement of medium is parallel to propagation of waves.
For example sound waves. |
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Total internal Reflection
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n1 has to be greater than n2 for TIR to occur. Therefore n1 has to be optically denser than the refracting medium. But the angle of incidence must be large enough:
Theta1>sin-1(n2/n1) |
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An incident ray (in air) makes an angle of 30 degrees with the surface of a medium whose index of refraction is squareroot(3). Find the angle that the refracted ray makes with the surface:
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The angle of incidence is 60 degrees and NOT 30 degrees. REMEMBER: Even if we are given info about the angle made with the surface, we must turn that into information about the angles made WITH THE NORMAL. Snell's Law then says:
n1sin(theta1)=n2Sin(theta2) 1sin60=sqroot(3)sin(theta2) sin(theta2)=1/2. Theta2=30 degrees. So the angle the refracted ray makes WITH THE SURFACE is 60 degrees. |
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Let n1 be the index of refraction of the incident medium and let n2 be the index of refraction of the refracting medium. If the angle that the refracted ray makes with the boundary is less than the angle that the incident ray makes with the boundary, then what has to be true about n1 and n2?
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n1>n2:
The angle of refraction is GREATER than the angle of incidence since the angle the refracted ray makes WITH THE BOUNDARY is less than the angle that the incident ray makes WITH THE BOUNDARY. EVERYTHING MUST BE IN TERMS WITH THE NORMAL. Recall that the refracted ray bends away from the normal when the refracting medium is optically less dense than the incident medium. Since this is the case here, we must have n1>n2. Since Theta1<Theta2 |
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Converging lens are placed in contact. If one has a focal length of 10cm and the other has a focal length of 20cm, what is the power of the combination?
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f1=10cm-->.1m
f2=20cm-->.2m P1+P2=Ptotal (1/.1)+(1/.2)=10+5=15D Power=15D |
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If size of the entrance were increased by a factor of 2 and the frequency of the wave were increased by a factor of 2, then how would the spreading of angle change?
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deltatheta=wavelength/d
v=wavelength*frequency wavelength=v/frequency deltatheta=v/(frequency*d) therefore, deltatheta'=v/(2frequency*2d) deltatheta'=(1/4)deltatheta IT WOULD DECREASE BY A FACTOR OF 4 |
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The Energy ( i electron volts, eV) of a photon of electromagnetic radiation
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E=hf
h=Planck's constant h=4.1*10^-15 eV-s f=wave's frequency |
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In fiber optics systems, infrared light is used instead of visible light because?
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VISIBLE LIGHT HAS A SHORTER WAVELENGTH AND IS MORE READILY ABSORBED BY THE FIBER.
Infrared light contains less energy than visible light. Infrared light has a lower index of refraction than visible light. In general the longer the wavelength, the lower the index of refraction. In general, electromagnetic radiation with a short wavelength is more easily absorbed, bent and scattered. |
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If a ray of light were to travel through the air and then enter a glass fiber, its wavelength would?
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DECREASE: As the light ray enters a more optically dense medium, its speed decreases (by definition of the index of refraction). Since the frequency of the wave does not change, the wavelength must decrease (since wavelength*frequency=velocity)
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2 Balls are dropped from a tall tower. THe balls are the same size, but Ball X has greater mass than Ball Y. When both balls have reached terminal velocity, what is true?
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Ball X has greater Velocity because at terminal velocty, acceleration is zero. The force of air resistance counters gravity exactly so the force is = to the weight for both balls. Ball X requires more collisions with air molecules to compensate for the larger gorce of gravity. More collisions means greater air resistance.
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