# Search results

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2. ### Engineering Signal and System CTFT

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
3. ### Engineering Signal and System CTFT

Oh, my professor never mention this, i will check this and see Thank you
4. ### Engineering Signal and System CTFT

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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6. ### Real power in AC Circuit

Should i use P = V.I or P = Vrms.Irms.cosφ ?
7. ### Current divider

The second one
8. ### Current divider

Ok i know the answer, in this case R2 is not in series with R4 so the second answer is correct
9. ### Current divider

Should it be I3 = \frac{(R2+R4)}{(R2+R4) +R3} I or I3 = \frac{R2}{R2 +R3} I
10. ### Masses Moving Radially on a Rotating Disk

Ohh i got it Thank you
11. ### Masses Moving Radially on a Rotating Disk

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 ?
12. ### Masses Moving Radially on a Rotating Disk

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...
13. ### Rotational motion and angular momentum

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...
14. ### Rotational motion and angular momentum

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 /...
15. ### Linear momentum problem with n particles

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
16. ### Linear momentum problem with n particles

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
17. ### Linear momentum problem with n particles

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...
18. ### Linear momentum problem with n particles

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
19. ### Linear momentum problem with n particles

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...
20. ### Work by a spring force

Oh ok Thank you
21. ### Work by a spring force

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 ?
22. ### Work by a spring force

Sorry but this is quite complicated for me, I've never seen this
23. ### Work by a spring force

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

Thank you
25. ### Pulley and Blocks

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
26. ### Motion in one dimension

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
27. ### Motion in one dimension

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
28. ### Motion in one dimension

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

I have done it again and also include a simple graph
30. ### Motion in one dimension

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...
31. ### Motion in one dimension

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)...