It is just as belliott4488 described. There is not much more significance to B and C than that.
If you rewrite the equation given in the form y = At^2 + Bt + C, into the form
y = A'(t-tb)^2 + B'(t-tb) + C', where tb is the time of the bounce being considered... then A' is the...
No. don't substitute in an x.
2pi/.80 = 7.854
You need the function:
y = acos(7.854x) + acos(7.854x - 0.7854)
don't substitute in an x... you need to find the amplitude of this function... you can write this sum of two cosines as a single cosine...
try to use this identity. cosA...
I'm sorry. A is not 80... amplitude is unknown variable a. I should have written wavelength is 80cm.
I'm getting a phase difference of -0.7854 radians
Using these and adding
y = a cos (kx) to
y = a cos (kx + phi)
where phi = -0.7854 and k = 2pi/0.80 = 7.854
you can get the...
You can choose any point for the torque pivot... It's best to choose a point that makes your equations simple...
I recommend the upper right point, because then you avoid having the trig stuff in your equation.
Your answer to the first question looks good to me.
But your answer to the second question doesn't. In the first question... you were together with the mirror... whether you were still or moving... you and the mirror were in the same inertial frame of reference. Like you said, the laws of...
Yes, you're right. N must equal L to get the equation you have:
y(N) = 2y sin((Npix)/L)cos((vNtpi)/L)
y = 0.023sin(xpi)cos(0.714 pi t)
So N = 7.
Another way to look at it... Find the angle at the end of the standing wave... The angle is xpi... that's coming from...
V(R) - V(infinity) = -\int_{\infty}^{R}\vec{E}(r)\cdot\vec{dr}
This is just the definition of electric potential.
V(infinity) = 0, so:
V(R) = -\int_{\infty}^{R}\vec{E}(r)\cdot\vec{dr}
Use your formula for E(r) from part a). Solve the integral.
Draw a sketch... first draw the reflections of the point source in the two mirrors... what are the coordinates of these two images. I'll call the I1 (reflection in mirror 1) and I2 (reflection in mirror 2). Now I1 and I2 also have reflections... I1 has a reflection in mirror 2... call it I12...
Yes, use energy... the work done by friction is the same each time the object crosses the rough surface. Use the initial energy... and the work by friction per trip across the rough surface... to get the number of trips before energy of the box becomes zero.
Generally uniform circular motion means a constant speed throughout the circle...
In the case of a ball attaches to a string going through a vertical circle... you don't have uniform circular motion... however at the top and bottom we do have dv/dt = 0 where v is the speed of the ball...
You have the formula for the time constant T = L/R. You need the time constant to become half... so what do you need the R to become?
The steady state value of the current is Vo/R. considering the change in R above, what do you need to do to Vo to keep the same steady state current as before...
Yes, that's the right equation.
I still don't understand... why do you need the second derivative with respect to x, or with respect to time? you need the amplitude at a certain x value right?
you've got the equation of the standing wave.
y(x,t)=(Asinkx)sinwt
The amplitude at any x...
You can use energy conservation... During oscillation, the top and bottoms are when v = 0... ie at those points, kinetic energy is 0.
For simple harmonic motion we need the mass to return to the same maximum height (that's what makes it simple harmonic motion)...
at that maximum height...
Yes, k = 980 N/m as you calculated.
And the minimum x' seems to me to be -0.05 - 0.15 = -0.20m
The equilibrium position is -0.05m. So the max is -0.05+0.15m = 0.10m.
And the minimum position is -0.05-0.15 = -0.20m. Is this the wrong answer?
ok.
do they give any more information? did you state the problem exactly?
it seems to me like the spring constant shouldn't make a difference... ie any spring constant would give the same result...
I'm probably misunderstanding something... but any 10 springs with a mass on top of...
There are 2 mistakes I'm seeing here:
first, you're not taking into account the gravitational potential energy of the block... the block is going to drop a certain height...
second the energy stored in the spring is not (1/2)kA^2 when all the energy is converted to elastic potential...
Everything looks right to me except b). d) is correct... but a couple of little things are there...
For b), they want the height above the plane... not the height above the ground... so you wouldn't add 12...
also, you want: d = v1*t + (1/2)at^2 = 40sin30t + 0.5(-g)t^2... plug in t =...
This is the last question in Elements of Abstract Algebra by Allan Clark.
When is (q) a prime ideal in Z(\rho) (the Kummer ring) where \rho = e^{2\pi i /p}, where p and q are rational primes.
This seems to be a difficult question to answer in general... since considerable effort goes into...
yes, v=sqrt(3Lg) looks right to me. Yes, direction of Fnet is towards the center... in other words downwards since the mass is at the top of the circular path.
length = 0.24m
width = 0.16m
height = 0.10m
can you find the area of the 6 sides of the cube?
side 1 and 2... length*width
side 3 and 4... width*height etc...
It's in the DTFT pairs here:
http://www.neng.usu.edu/classes/ece/5630/notes_transforms.pdf [Broken] on page 20.
but maybe the prof wants you to derive it from the properties you already have in your text?
as n->infinity n^n>n! so \frac{4^n}{n!} > \frac{4^{n}}{n^{n}}.
I think probably the idea being expressed was to use a constant in the denominator instead of n. \frac{4^n}{8^n}. (or any constant^n in the denominator where the constant>4)
\frac{4^n}{n!}< \frac{4^n}{8^n} for large enough n.
Yeah, but be careful of shortcuts... be aware of what you're doing...
we can't use p^2/(2m) before the collision... p^2/(2m) is only for 1 mass (in other words we can't use p as the sum of the two momenta and m as the sum of the two masses... that won't give the right answer. but for each...
Think about the 3 equations involved...
sum of torques about any point = 0 (1)
sum of forces in x-direction = 0 (2)
sum of forces in y-direction = 0 (3)
equation (2) should let you answer a).
equation (1) gives you part b). try to go on from there to get the other parts...
the sin(theta) shouldn't be there. torque is the force times the perpendicular distance from the pivot to the line of the force... you had the T part right. so it should be:
that's one equation.
you get 2 more equations by setting sum of forces in x-direction = 0 and sum of forces in...
depends on the R you're using...
if you learned
R = 8.314472 m3·Pa·K-1·mol-1
then you need to use pressure in Pa. and volume needs to be in m^3. which R did you learn?
if you learned
R = 0.08205784 L·atm·K-1·mol-1
then you need to use pressure in atm. and volume needs to be in L...
first get the force due to B on A... then from that get the force due to C on A...
once you have the force due to C on A... just use coordinates of C.. (x,y). what is the force of C on A in terms of x and y (set this force equal to the force you got above)... you'll get 2 equations in 2 unknowns.
what is the pressure outside the can? what is the surface area of the can. I assume the can is a cube given the dimensions...
what is the relationship between force, pressure and area?
I think they mean percentage error maybe... one way of doing it is taking +/- 1 of the last digit.
so 125g has a percentage error of [1g/(125g)]*100% = 0.8%
but don't know if this is how you're expected to do it.
I actually just used google.com. I typed in 3000 ft^3 and it automatically converted it.
But we can do it this way:
we know that 1 ft = 0.3048m
(1ft)^3 = (0.3048m)^3
so 1ft^3 = 0.0283168466 m^3
so 3000 ft^3 = 84.95 m^3
you need the torque about that point to be zero... so you need:
(Force of sign)(Radius of sign)+(Force of rod)(Radius of rod)-(vertical component of Tension)*distance=0
(since horizontal component of tension exerts 0 torque about the pivot)
what is the vertical component of Tension in...
First find the mass of the rod, and the location of the center of mass of the rod...
The measurement the scale reads, is the magnitude of the normal force that the scale exerts...
You can find one of these forces using net torque about the appropriate point = 0...
Then the other force...
They want the magnitude and direction of the net force in terms in terms of m, g and L... so v shouldn't be in your answer...
When the ball is at the top of the loop T and mg are in the same direction:
Fnet = T + mg = mv^2/L
so what is Fnet in terms of m and g... what is the direction...
yes, that method also makes sense... I think that if Tf comes out greater than or equal to 0... then everything is good and the ice has melted... but I might have missed something...
Qice is the energy the ice gains in going from -20 to 0... and then melting (heat of fusion)
So that comes...
If Tf comes out >=0 ... that means all the ice melted...
If Tf comes out <0... then the ice didn't melt... (reason is we have an internal contradiction... assuming the ice melts leads to Tf<0 which implies the ice doesn't melt... that means our initial assumption is wrong... and Tf may be...
Well... I'm confused by what you were doing in class... the ice won't change temperature unless it goes below 0 right? from 0 onwards the ice is in its melting phase... ?
here's how I'd do the problem... don't know if it's completely right...
msoda*csoda*(Tf-20) + mglass*cglass*(Tf-20) +...