At rest, initial speed zero
x=v(initial) t+ 1/2 at^2
-1000m=0(10) + 1/2 a 10^2
-1000m=50a
a = -20 m/s^2
Answer:
Choice A: approximately 
, assuming that the two pistons are connected via some confined liquid to form a simple machine.
Explanation:
Assume that the two pistons are connected via some liquid that is confined. Pressure from the first piston:
.
By Pascal's Principle, because the first piston exerted a pressure of 
on the liquid, the liquid will now exert the same amount of pressure on the walls of the container.
Assume that the second piston is part of that wall. The pressure on the second piston will also be . In other words:
.
To achieve a force of 
, the surface area of the second piston should be:
.
Answer: a) io=233.28 A ( initial current); b) τ=R*C= 22.31 ms; c) 81.7 ms
Explanation: In order to explain this problem we have to use, the formule for the variation of the current in a RC circuit:
I(t)=io*Exp(-t/τ)
and also we consider that io=V/R=(1.5/6.43*10^3)
=233.28 A
then the time constant for the RC circuit is τ=R*C=6.43*10^3*3.47*10^-6
=22.31 ms
Finally the time to reduce the current to 2.57% of its initial value is obtained from:
I(t)=io*Exp(-t/τ) for I(t)/io=0.0257=Exp(-t/τ) then
ln(0.0257)*τ =-t
t=-ln(0.0257)*τ=81.68 ms
Answer:
It is calculated by dividing Resistance, R, by Inductive reactance, XL.
Explanation:
Q is called the Q factor of a resonance circuit. In a parallel resonance circuit, it is calculated by finding the ratio of the power stored in the circuit to the power distributed in the circuit. It is a way of measuring the quality of a circuit or how effective the circuit is.
Q factor is the inverse in the resonance series circuit.
Q factor of a resonance parallel circuit,
<h3>
Q = R/XL</h3>
R = Resistance
XL = Inductive reactance
