Recent content by Ryan231

The short cct test is just shorting out the Load Impedence (ZL) this allows us to isolate Ze and calculate it by doing: Ze=V/I = 66/8.33 = 7.92 To find the angle we break Ze down into Re (resistance Equivilant) and Xe (inductive). cosX = re/xe = 64 degrees. We found these values using the...

Question: A short circuit test is performed on the low voltage side of a 40-kVA 4800/240V 60HZ transformer and the following data are obtained: W = 240W, V= 66V, and I = rated current. Determine the percent voltage regulation with a 0.88 lagging PowerFactor load connected. On the Short cct...

18L*.96 = 17.28kg of water... So. Q= (17.28)(4186)(70) = 5063385.6J I think that's right?

I've got a question to solve.. Just need 1 thing.. An automobile cooling system holds 18L of water. How much heat does it absorb if its temperature rises from 20ºC to 90ºC? I need to find how many kg of water I have... and then plug into Q =mc(T1-T2) Q = (m)(4186)(-70) Look...

so... D2 = V(s) * T2 = 488.6m D1 = 4000 * T1 = ? I'm not following?

OK, the problem is correct.. 4000m/s for concrete is what was given so that's what I'm using... Forget what I put.. I was just brainstorming.. Can you maybe get me started in the right direction? Maybe switch D2 with T2.. As in when the person hears the first sound the other one is at the...

help anyone? :frown:

Ive been trying this one for a little while now and keep getting myself stumped here... So the question is... A person sees a heavy stone strike the concrete pavement. A moment later two sounds are heard from the impact: one travels in air and the other in the concrete, and they are 1.4 s...

I'm no expert but I did it like this.. cos70 = 7500/x x= hypot cos70 = .3420 .3420 = 7500 /x .3420x = 7500 x = 7500 / .3420 x = 21929.82456

i guess the amplitude of vibration is the full length of compression to the spring, in which case there will be potential energy of .5kx^2 stored in the spring. that's equal to the initial kinetic energy of the bullet since there's no initial kinetic energy in the block. so .5kx^2=mv^2...

Ok I'm not great at doing it the other way but I'll give it a shot so... m1v1 = (m1+m2)v2 1/2 ( m1 + m2)v2^2 = 1/2kA^2 Sub m1v1 = (m1+m2)v2 into 1/2 (m1+m2)v2^2 = 1/2kA^2 = 1/2 m1v1^2 = 1/2kA^2 Cancel out ^2's and 1/2's m1v1 = kA .025kg v1 = 6700(.215m) v1 = 57,620? This is an...

I'm pretty bad at physics overall so I'm not sure what you mean at this point: "You have momentum of the bullet as m1v1." I would assume you mean the momentum is 25g*velocity1 thus making it... m1v1 = (.025kg+.6kg)v2 .5*(.025+.6)v2^2 = .5kA^2 .5*6700*(.215m^2) = 154.85...

One the first question I tried ma = -kx where k is the amplitude.. but it gave me an answer of a = -5,762,000 :frown: Really stumped here..

I'm really stumped on 3 of my physics questions.. I've been able to find a few things but there's so many equations and variables I've gotten myself lost.. if anyone could help it would be greatly appreciated! A 25.0 g bullet strikes a 0.600 kg block attached to a fixed horizontal spring...