I don't understand it either. If ##s_1## and ##s_2## are the lengths of both springs then ##a = s_1 + s_2##.
Now, if one spring stretches by ##x## then the other spring must be compressed by the same amount, ## (s_1 + x) + (s_2 - x) = a##
And if ##V_1 = \frac12 k x^2## then ##V_2 = \frac12 k...
our university is transitioning to online delivery. However, physics labs, as of Friday last week, are still going ahead as normal. So I'll go into work unless labs are cancelled.
it means that you evaluate the differentiation and keep ##T## constant. Similarly you could find ##\left(\frac{\partial P}{\partial T}\right)_V## where you would treat ##V## as a constant
The Langrangian is an equation of the generalised coordinates ##q## and velocities ##\dot q##. To get from the Lagrangian to the Hamiltonian you perform a Legendre Transform and the Hamiltonian becomes a function of ##q## and ##p = \frac{\partial L}{\partial \dot q}##. So you need to consider...
what I learned only recently and find very interesting is the fact that even after the accident in unit 4 they continued construction on units 5 & 6 for two years until 1988 and operation of units 2, 1, 3 until 1991, 1996, 2000
you want to know what the mass ##M## must be to balance the beam on the blue pivot point. That means the sum of torques must be zero or, since the the torques act on either side of the pivot point, they must be equal in magnitude.
So you have one mass at ##3m## to the right of the pivot point...
unless the beam is attached to the blue structure it won't supply any support at all. if you look at your png again, you can see that the beam is lifted left of the the blue point.
The beam is lifted left of the blue point and lowered on the right of the blue point. so the beam rotates around...
in whatever way you define your domain it is obvious that $$f(x) = \frac{x^2 - 1}{x - 1}$$ is not defined at ##x = 1##. Don't know what book the question comes from, however, I would take the position without having any further information that the function is defined everywhere where it can be...
MATLAB, Mathematica, Maple run natively on Linux. I guess because they were all developed in an academic environment and predate Windows by at least a decade.
How easy? very easy!
the only real compatibility issues there are is that games usually are not (yet) ported to linux. but steam is doing a great service to promote games on linux as well.
other compatibility issues might occur when using proprietary file types like any of the windows office...
it could certainly be done in more than one step. However, now that you have the time it takes to get up to the highest point of her jump you'd need to calculate how high she's up in the air. you need the information to calculate the time it takes to fall from this height, given that the speed...
In other words, you have calculated the time after which she is at the highest point. Now you have to add the time it takes to fall down into the pool.
she won't be at rest when she hits the water. she'll be at her highest speed just before she hits the water.
I can't really follow all your steps because it's almost 2am over here and it's been a long day. However, your second line can be written as
##\frac12 \log_2(2x - 1) < \frac12 \log_2(x)##
from here it should be solved in 3 steps
what you need to consider is that there is only one external force ##F## acting on ##M## to the right. this this force ##F## is accelerating ##(M + m)## to the right at ##a = \frac{F}{M + m}##
If there is no slip then ##m## is accelerated with the same ##a_m = a##. But what force is it that...
Looks like you added the total forces. However, they point in different directions.
You were asked just to add the force components that point to the north.
##0.99999... = 0.\bar9= 3 \times \frac{1}{3} = \frac{3}{3} = 1##
I don't know anything about finitism. And I also don't know why one needs to be all philosophical about the question.
As I see it the problem just lies in the way numbers are represented in the decimal number system. Nobody would...
if one ##Q## was positive, the other negative, let's just say ##Q_1## positive and ##Q_2## negative, then the force of ##Q_1## on ##P## would push ##P## to the right and the force of ##Q_2## on ##P## would pull it to the right. So both forces would act in the same direction and the net force on...
the normal force that counteracts the birds weight is provided by the wire. The weight only has a component in -y, the two forces along the wire to each side of the bird have a component each in the +y direction, the -x direction and the +x direction respectively. So the force components in the...
the difference, however, is that nobody uses a logarithmic temperature scale whereas only the Americans don't use the metric system. so the comparison is kinda apples and oranges.
it says in the problem statement that the freight train moves at a constant speed. So how is it possible that the freight train ends up with a speed ##v = 0##?
if the train moves at constant speed, is ##v^2 = a(d - d_0) + v_0^2## really a relevant equation?
if you're a beginner programmer I don't see a problem trying to use the tools a programming language provides to solve simple problems. Once you solve the problem using a for loop you might see that a for loop isn't the optimal tool to solve this particular problem since the for loop continues...
the for loop takes in a parameter, checks if a condition is met, and if so it keeps looping until the condition is no longer met
So in your case it would set c to 100 and then check if c is equal to -273. that is not the case, so the loop exits immediately.
you want to check if c is greater...
first of all you can check whether your equations are correct if you do a dimensional analysis.
Look at ##\ddot R = 2b\dot\theta##
##[\ddot R] = ms^{-2}##
##[2b\dot\theta] = ms^{-1}##
So you see something is not quite right.
When you take the 2nd derivative of ##\dot R =...
is it possible that you have to calculate the impulse? Impulse is the change of momentum. Momentum is, as already mentioned, the product of mass and velocity ##p = mv##. Since the velocity is ##0## after the impact the impulse is ##I = mv - 0 = mv##.
with mass and velocity of the object given you can calculate its momentum, ##p = mv##. All you need to know then is that force is the change in momentum per time. The problem is if we assume that the object hitting the moon comes to rest instantaneously the time it would take is ##t = 0s## and...