<Mentor moved to Physics>
My teacher talks about turbulence (2D and 3D), but I don't quite understand this. How is the turbulence different in the two buckets, and why does my teacher talk about turbulence but not diffusion? Is not diffusion the reason why the dye spreads in the water? I have...
So I have always been thinking that equilibrium means that an object is not moving or having constant acceleration. On a webside they said: " A rigid body is in equilibrium when it is not undergoing a change in rotational or translational motion. " To me it sounds like the object then must not...
Hmm, I see. But if I were to use the two equations, I would have to make sure that ##\sum F_y=mg*cos(15)+T_b sin(75)-mgcos(15)=0## ?
And that would make the m=408.3 kg:oops:
The buoyant force is ##F_b=\frac{1000}{2400}kg*g##
The effective force is ##m=1kg-##mass of water##=(1-\frac{1000}{2400})kg##
But what is the importance of the effective weight in this problem? Are these equations wrong by the way
I have been looking at the units but I still don't see it :confused:
##2400 kg/m^3=\frac{1kg}{V_{block}}##
##V_{block}=\frac{1}{2400}m^3##
##p_{water}=1000 kg/m^3=\frac{mass}{\frac{1}{2400}m^3}##
##mass=p_{water}*V_{water}=1000kg/m^3* \frac{1}{2400}m^3= \frac{1000}{2400}kg##
Yeah, I also think it was weird that they used the word weight, because I also thought kg/m^3 is density. So I just assumed the person who wrote the problem mixed up the words. But I struggle with finding the acceleration, do you know how?
Hmm, I'm sorry, but I don't understand. Why does the water displaced have the mass of 1000 kg? I thought like this
##p_{concrete}=m_{block}/V_{block}##
##V_{block}=1/2400 m^3=V_{water}##
##p_{water}=m_{water}/V_{water}##
##m_{water}=1000/2400 kg##
So I have made force diagram
And I think that I should find the acceleration by using these equations:
##\sum Fx=w\sin(15)-f_k-T_{x-buoyancy} ##
##\sum F_y=N+T_{y-buouancy}-w ##
I know that the volume of water displaced must be ##V=\frac{1}{2400}m^3## and the mass of the water is then...
Thank you so much everybody for your relplies, I really appreciate it!:smile: I will now try to solve some more problems like this with your comments in mind, and see if I understand it better now.
So here are two problems I have been working with lately:
I have solved both, so I don't need the help to find the answers. The thing that confusses me is which object one should choose to apply the equations above. When it comes to the first problem, we apply the equations to the pole, but...
Ok, I think I understand a bit more now. So for a particle, the equation would then result in a double count. But when do I know if I can treat the object as a particle or a rigid body? Because if it was a rigid body, then I would have to use ##T = T_{CM} + T_{rot}## right? I'm sorry for all the...
So the rotational energy is the same as the kinetic energy in this example? But in several problems, for example when a ball dropps from a height. They wright the equation as this ##mgh= \ {1\over 2} mv^2\ + \frac{1}{2}I\omega^2 ## So I thought they where two different things?
When does an object have rotational energy? Is it only if it rotates around an axis within the object? Does for example a ball attached to a string with a uniform circular movement have rotational kinetic energy?
I have done question 1. But I'm struggling with the other one. So since the only thing I know about the rocket is the mass and the velocity, I guess I have to use momentum to solve this problem. From the first question, I found out that the x-velocity of the projectile is ##v_x=5...
Sorry, for not answering before now. My internet has been down the last couple of days. I understand the minus sign now. Thank you so much for your answers, I really appreciate it!
I have tried but I still dont get it. Maybe I'm doing it wrong? So box B is affected by the weight of box A equal to ##m_A*g## and the normal force from the ground ##m_B*g## and then the pull equal to 12 N.
Box A is only affected by the friction force going in opposite direction of the...
I don't understand the problem. Does not block A and B make a system, so they should have the same velocity and acceleration at all time? If not, why do they have different accelerations? I don't understand this part of the problem either: "pull applied to block B equals 12.0 N, then block B has...
So thing that confusses me, is what is the difference regarding the calculation of speed and velocity? In think I should use the formula I wrote above to find the average velocity, but I don't understand what the formula for average speed will be then. By the way, when using the formula above...