Force is directly proportional to mass according to the second law of Newton, meaning that the greater the mass is, bigger the force should be in order to move the object. In this case, Mutt's wagon has a mass two times greater than Jeff's and they have to be equal. So either Jeff must slow down twice as much or Mutt has to speed up twice as much. The only option we can choose according to our reasoning is that Mutt must use twice as much force to push his cart, because his mass is two times bigger. According to me the answer is C).
Answer:
The answer is based on the conservation of energy law; something you should really understand by now.
For convenience we can hold one of the two charges still; it becomes the frame of reference. And everything we say is in reference to the designated static charge, call it Q.
So the moving charge, call it q, has total energy TE = PE. It's all potential energy as we start with q not moving.
It has potential energy because in order to separate q from Q, we had to do work, add energy, on q. And from the COE law, that work added is converted into PE.
It's a bit like lifting something off the ground. That's work and it becomes GPE. So there's some work, in separating the two charges in the first place.
But there's more.
Now we let q go. As opposites attract, q is pulled to Q. And that force from Q is working on q, force over distance. Which means the potential energy q started with is being converted into kinetic energy. q is accelerating and picking up speed.
And there's more work, done by the EMF on charge q. That converts the PE into KE and the q charge smashes into Q with some kinetic energy.
Answer:
C. 85%
Explanation:
A cylinder fitted with a piston exists in a high-pressure chamber (3 atm) with an initial volume of 1 L. If a sufficient quantity of a hydrocarbon material is combusted inside the cylinder to produce 1 kJ of energy, and if the volume of the chamber then increases to 1.5 L, what percent of the fuel's energy was lost to friction and heat?
A. 15%
B. 30%
C. 85%
D. 100%
work done by the system will be
W=PdV
p=pressure
dV=change in volume
3tam will be changed to N/m^2
3*1.01*10^5
W=3.03*10^5*(1.5-1)
convert 0.5L to m^3
5*10^-4
W=3.03*10^5*5*10^-4
W=152J
therefore
to find the percentage used
152/1000*100
15%
100%-15%
85% uf the fuel's energy was lost to friction and heat
