Recent content by Combinatorics

Thank you! I completely missed this small step. Thank you! It was interesting to read.

Hi rsk, I am not sure if I understand you correctly. Do you mean that I should use that ## \frac{4\pi}{3}g \rho_1 R_1^3= T+\frac{4\pi}{3}g \rho_3 R_1 ^3 ## ## \frac{4\pi}{3}g \rho_2 R_2^3 = \frac{4\pi}{3}g \rho_3 R_2^3 - T ## ? If so, then how can I say something about the relationship between...

Performing force balance on the two balls, I obtain ## T+\frac{4\pi}{3}g \rho_3 R_1 ^3 = \frac{4\pi}{3}g \rho_1 R_1^3 ## ## T+\frac{4\pi}{3}g \rho_2 R_2^3 = \frac{4\pi}{3}g \rho_3 R_2^3 ## from which I obtain ## \rho_3 = \frac{\rho_1 R_1^3 + \rho_2R_2^3 }{R_1^3 + R_2 ^3 }## and ## 2T...

Thank you BVU for helping me understand my misunderstanding regarding the flow rate! Great. This yields answer A as the closest to what I obtain in my calculations. Thank you! Thank you Steve! My original thought was indeed that the taps are connected in series. It is much clearer now.

The possible answers are: (I do not know what is the right one) A. ##H=120m## , ##S=0.0131\frac{m^3}{s}## B. ##H=60m## , ##S=0.0231\frac{m^3}{s}## C. ##H=120m## , ##S=0.0231\frac{m^3}{s}## D. ##H=240m## , ##S=0.0231\frac{m^3}{s}## E. ##H=60m## , ##S=0.0131\frac{m^3}{s}## F. ##H=240m## ...

Amazing! So, $$ ma = N_1 \cos \theta - mg \Rightarrow a_{\text{max}} = \frac{ 150 \cos \theta - mg}{m} = 1.025 $$ Which is indeed one of the answers! Thank you very much! This was very helpful!

Hi, Thank you! Here is my attempt: I defined ##x## and ##y## axis as shown in the image. From this I have that $$ \sum F_x = N_1 \sin \theta - N_2 $$ $$ \sum F_y = N_1 \cos \theta - mg $$ Is this correct? Thank you very much for your help so far.

Hi guys, I went over all the comments here, and understood that my force balance is wrong. Unfortunately, I have no idea where is my mistake exactly. The way I obtained the equation ##\cos\left( \frac{\pi}{6} \right) m(a+g) = 150 ## is: By calculating the forces acting on the sloped part of the...

You are right. The problem statements says it's a mass, and not a point mass. Apologize for this. If ##\theta = 0 ## there is a force exerted on the bottom edge, which would break if ##a## is large enough. Isn't this correct? In any case, isn't the force exerted on the bottom (inclined edge)...

Hi, why should the force tend to zero when ##\theta\downarrow 0\ ## ? The force indeed tends to zero when ##\theta\uparrow \frac{\pi}{2}\ ## which seems reasonable to me (as this expression represents the force exerted on the lower, inclined part of the cube. What am I getting wrong here? Thank...

Using Newton's second law, $$ cos\left( \frac{\pi}{6} \right) m(a+g) = 150 \Leftrightarrow a = \frac{\frac{2\times 150 }{\sqrt{3}}-mg}{m} =4.633. $$ Unfortunately, the possible answers are A. 1.025, B. 0.625, C. 3.75, D. 2.75, E. 1.75, F. 0. What am I getting wrong? Isn't the force exerted on...

Thank you PeroK, you helped a lot!

Thanks for your feedback! here is my attempt. Using conservation of linear momentum, I obtain $$ 2m\cdot \frac{\omega}{L} = m\cdot \frac{\omega_2}{0.5L} + 2m\cdot \frac{0.8\omega}{L} $$ where I used $$v=\frac{\omega}{L}$$. Solving this equation implies $$ \omega_2 = 0.2\omega. $$ When I...

Using conservation of linear momentum, I obtain that $$ \omega_2 = 0.2\omega $$. Substituting in the energy difference gives $$ -0.355 mL^2 \omega^2 $$, which is still not the right answer... What am I doing wrong here? Thank you

Is it true that close enough to the collision I can just use conservation of linear momentum to find the velocity of the second mass after the collision?