Recent content by peter010

Thanks indeed for you all :)))

ok, I am sorry if i am not clear,, any body at rest, like me now setting on my chair, my weight is then -> F=mg. Now, let's say that i thew myself from the window at rest, and i fell over a car, i presume then, that my force just over the surface of this car is bigger than F=mg at rest, due to...

but how this would answer me :))) ?

in the mud puddle, clearly, the surface will get changed .. we will get a depth as an impact. for steel surface, the effect will be less, and force will be discharged then as vibration.

oh! do you want to say, that its the same reacting force at final stage.. the steel plate will react on its surface, while the mud puddle will counteract the rock weight by making it accelerate inside it (as same as in my case)?

if you meant by the supporting force is the reaction force, then it should be the same in both cases.

Absolutely not! this is wt I am talking about! here i have same ground, but comparing between the body's force at rest VS a force of same body just hitting the ground after being fell...

F_new: the foce of the body when hitting the ground after being fell F=mg ...... (1) F=m[v2-v1]/t //v1=0 F=[m * v2]/t Hi, i did applied Newton law -> v2^2 = 2gS and i got finally, the rule: F=[ m * (2*g*S)^0.5] /t ...... (2) lets assume: m=100 kg, g=9.81 m/s^2, s=0.3 meter, t=1sec,, the...

1. problem: a body of mass started to fall from rest, and from S elevation. what is the equation that calculate the force of what this body will hit the ground? Homework Equations :[/B] F=mg // which is used to calculate the force of the body on the ground when it is already on the ground...

Thanks for responding Actually, I can best describe my case by saying: After reducing the source voltage, I have then two unknowns: the current and the EMF. [Before reducing the voltage, I was able to identify the current value for my motor's circuit from this motor's characteristic graph...

Actually, is it possible to calculate the new current value and the new back emf value after decreasing the source voltage? I tried to figure this out, but i could not predict or calculate the new value of the current since i can't depend then on the original characteristic chart of this motor.

Acually, now i am discussing it please from another perspective, what if I used a pulse width modulation device to reduce the main votage to a certain limit, let's say Vnew=70% Vold. What will happen to the back emf? Can I then predict or calculate precisly the value of back emf at same load ?

oh thanks alot. I thought you was referring to double the mechanical load by expressing it metaphorically by a resistor. :)

oh ok.. May I ask you a favour please,, can u add this resistance R of load to the attached DC serries cercuit and then attach it. This will help me so much for imaging its place in the curcuit :)))

back emf will drop down and thus geting more current in the curcuit. AWWW Thank you very much :)))))))