Recent content by hquang001

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Yes i have tried this before, but it didn't match the solution that my professor gave, that's why i ask in here to check

Oh, my professor never mention this, i will check this and see Thank you

Given the CTFT X(ω): and here are my solution to find CT signal x(t) associated with the given CTFT , but i got stuck at the limit part

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Should i use P = V.I or P = Vrms.Irms.cosφ ?

The second one

Ok i know the answer, in this case R2 is not in series with R4 so the second answer is correct

Should it be I3 = \frac{(R2+R4)}{(R2+R4) +R3} I or I3 = \frac{R2}{R2 +R3} I

Ohh i got it Thank you

Assume fifth object stay at ro : => w2 = 5.88 rad/s it exceeds the maximum and will slide off but the position of the object is 1.5m, so is it wrong ?

m = 60kg, ω0 = 2.094 rad/s, I of disk = 130 kgm^2 , outer position ro = 1.5m, inner position ri = 0.3m ∴Fifth object : Ffriction = m.ac μ.m.g = m. v^2 / R => vmax = √ 3. (1.5m) . (9.81 m/s^2 ) = 6.64 m/s => ωmax = 4.43 rad/s so when the fifth object move with greater speed than vmax...

I think it should reach the same height because there is no outside force. we can proof it by using conservation of energy PE initial = PE final => mgh1 = mgh2 => h1 = h2 Does it reach greater height than original height ? I'm not sure about this - for part b). yes you're right. As i rounded...

mball = 2 kg, mputty = 0.05 kg, L = 0.5 m, v = 3m/s a) Moment of inertia : I = (2mball + mputty ). ¼ L^2 = 0.253125 kg.m^2 Linitial = Lfinal => mputty. v. r = I.ω => ω = (4.mputty.v.r) / I = 0.148 rad/s b) K initial = 1/2 m v^2 = 0.225 J K final = 1/2 Iω^2 = 2.85.10^(-3) J => Kfinal /...

Oh i missed it, the height should be 0.5 m. and yes the speed does not change that's why the Δp = m( v -(-v) ) but it's not really clear written as you said