Recent content by bieon

Thank you! I really appreciate the help!

So, ##d=\frac {0.25(0.5)+0.25x} {0.5}## ##d=\frac{1}{4}+\frac {x}{2}## Ahhh I complicated myself with the integrations and others! Thank you!

Sorry I didn't quite catch that, can you elaborate?

Is this correct? Centre of Rod: ##l_1= \frac {\int_0^x l dm}{M}## Where, ##\lambda = \frac {dm} {dl} = \frac {M}{x} ; dm=\lambda dl## ##l_1= \frac {\int_0^x l \lambda dl}{M}## ##l_1= \frac {\int_0^x l \frac {M}{x} dl}{M}## ##l_1= \frac{1}{x} \int_0^x l dl## ##l_1=\frac {1}{x} [\frac...

This is the figure given. My Attempt ##T=2\pi \sqrt\frac {I}{0.5gd}## ##\frac {m_r} {l} ## ##dm= \frac {m} {l}dx## ##dI = dm_r x^2## ##dI=(\frac {m_r} {l}dx)x^2## ##I= \int_l^0 (\frac {m_r} {l}dx)x^2 \, dx ## ##I_c.m=\frac {m_r l^2}{3}## ##I_,, = \frac {m_r l^2}{3}+m_p x^2## Given...

Thank you!

My Attempt So Far: 1. Write down all information: Q=10m³ /min h=75m 2. Change 10m³ /min to m³ /s 10m³ /min × min/60s =1/6 m³ /s 3. Change Volumetric Flow Rate to Mass Flow Rate ṁ=ρQ ṁ=1000kg/m³ × 1/6 m³ /s ṁ=166.67 kg/s 4. Apply Potential Gravitational energy but replacing the Mass with...

Thank you so much!

but why is it L - L cos θ ? Sorry, I just want to understand more

Uhm I think I had trouble, I know the general concept of l-lcos but I am not sure where it came from...

I believe I can! Thank you so much! I am sorry for the trouble!

Like this? ##√2####gh_2## =[M/(M+m)] ##√2####gh_1##

(##M##+##m##)##√2####gh_2##=##M√2####gh_1## This?

Ok, so I put them in terms of velocity now right? M##g##(##h_1##)=½ M##v_1##2 2##g####h_1##= ##v_1##2 With Jane, (M+m)##g##(##h_2##)=½ (M+m)##v_2##2 2##g####h_2##= ##v_2##2

Like this? Without Jane, ##Mg##(##h_1##)=½ ##Mv_1##2 With Jane, ##(M+m)g##(##h_2##)=½ ##(M+m)v_2##2