Recent content by greenrichy

If I can determine the weight of that heavy object placed on the plank, I will be able to determine the stretch of that wire. But, when using the second condition for static equilibrium (torques of the system equal to 0), I always end up with two unknowns, no matter what point of rotation I...

What if, in practice, the momentum after the collision were to be less than the momentum before the collision, would the conversion of kinetic energy into heat/sound energy be the main reason for the loss of momentum (given that there is no friction)? I'm just not sure why the conversion of...

Sorry, I just wrote the numbers incorrectly, please see my last edit. Can the momentum after the collision be greater than the momentum before the collision?

I've come up with the following causes: - air resistance - parallax - during the collision, some of the kinetic energy gets converted into thermal energy. - invisible deformations But I'm not sure which would be the biggest effect on the total momentum change. Are there any other reasons that...

Got it, that makes sense. Thank you for your help.

So I found that the centripetal acceleration is equal to 4.2 m/s^2. Given all the data that I have, the radius of the circular path of the dice turns out to be 171.3 meters. I calculated it using this formula --> a = v^2/r. Does it look right?

So the tangential velocity with which the dice swing is the same as the car's velocity?

Thanks for your reply. From a free body diagram of FD (Fuzzy dice), I know that we can find the centripetal acceleration, right? But how do I find the tangential velocity?

Correct me if I'm wrong, but I think it is not possible to solve (1) with all the data that's given. As for (2), I have come up with the following solutions: (a) - The tension in the string acts as the centripetal force on the fuzzy dice (b) - The frictional force between the road and the car...

Thank you for your reply! Sorry about the confusion in part (b). I interpreted it as follows: After pulling the block 30.5 meters up the incline at a constant acceleration, the person stops (the block's velocity becomes zero) and releases the rope causing the block slide down the hill.

$$\sum F_x = T - w_x - f_k = ma_x $$ $$ T = mg\sin(\theta) + mg\cos(\theta)\mu_k + ma_x$$ $$ T = (9.8 \frac{m}{s^2}) \cdot (\sin(41^{\circ}) + \cos(41^{\circ})) + (75kg)\cdot(0.25\frac{m}{s^2}) $$ $$T = 672.91 N $$ Having found the tension force, I can find the work done by the person who's...