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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

Oh if so "v" in n.m.v is the change in velocity i just showed the full formula with the part n.m.v but didn't know how to deal with this part so i used instead Δp / Δt

I'm not sure about it, but as i read, n is number of beads. I think because F = n.m.v F has SI units of kg.m/s^2 on the right side, m.v is [kg.m/s] so i think n should be 1/s so that it has equal unit to force ? or is it number of beads per second ? F = Δp / Δt As i said above, 100/s = 1/ Δt...

n is the number of particles or in this case glass beads and its unit is 1/s i think ? and so i think 100/s = 1/ Δt => Δt = 1s/100 => F = 0.313 N => M = 0.313/ 9.81 = 31.9 g

To find the mass in other pan, i need to find the force caused by beads on the pan ∴ KEinitial + PEinitial = KEfinal + PEfinal 0 + mgh = ½ mv^2 => v = 3.13 m/s ∴ The change in momentum : p2 - p1 = m ( v2-v1) = m( v - (-v)) = 2mv ∴ F = Δp / Δt = n. m. v How can i apply the rate of 100...

Oh ok Thank you

So work done equal negative change of potential energy and in case of kinetic energy, work done is equal to the change of kinetic energy am i correct ?

Sorry but this is quite complicated for me, I've never seen this

Here it writes Wby spring= Uinitial - Ufinal. So is it equal to opposite of the change in potential energy ?

Thank you

1 Block m2: m2g - 2T = m2a2 (1) Block m1: ∑F= ma T= m1a1 = m1.2a2 (2) Cancel T in (1) and (2) I have : m2g = a2 (4m1 + m2) => a2 = m2g / (4m1 + m2) => a1 = 2m2g / (4m1 + m2) Is my answer correct ? Thank you

Yes i have calculated all the quantities you mentioned, and find each distance x1,x2,x3 which add up 60km. So i think this should be correct. Thank you very much for your help

Yes, Vmax = 137.91 is possible but i don't know if it's correct or not, i don't know how to check it

Only one solution is possible When i substitute V = 522.09 km/h to find t3, t3 = 0.87h which will make t2 negative

I have done it again and also include a simple graph

This is the hint which i forgot to put in the question statement: It could be that you get more than one answer . After considering your help and calculating again, i really got two possible maximum speed value like the hint stated I want to ask how i can check if my results are correct in...

total time: t = 36 mins = 0.6h = t1 + t2 => t2 = t - t1 = 0.6h - 0.1h = 0.5h Vmax = a1 x t1 Vat C = Vmax + a2t2 substitute Vmax in Vat C we have : 0 = a1 x 0.1h + (-600 km/h²) x 0.5h => a1 = 3000km/h² Vmax = a1 x t1 = 3000 x 0.1 = 300km/h I check the result by: x1 = ½ a1 t1² = ½ . (3000)...