Answer:
Explanation:
Take at look to the picture I attached you, using Kirchhoff's current law we get:
This is a separable first order differential equation, let's solve it step by step:
Express the equation this way:
integrate both sides, the left side will be integrated from an initial voltage v to a final voltage V, and the right side from an initial time 0 to a final time t:
Evaluating the integrals:
natural logarithm to both sides in order to isolate V:
Where the term RC is called time constant and is given by:
Answer:
The first law, also called the law of inertia, was pioneered by Galileo. This was quite a conceptual leap because it was not possible in Galileo's time to observe a moving object without at least some frictional forces dragging against the motion. In fact, for over a thousand years before Galileo, educated individuals believed Aristotle's formulation that, wherever there is motion, there is an external force producing that motion.
The second law, $ f(t)=m\,a(t)$ , actually implies the first law, since when $ f(t)=0$ (no applied force), the acceleration $ a(t)$ is zero, implying a constant velocity $ v(t)$ . (The velocity is simply the integral with respect to time of $ a(t)={\dot v}(t)$ .)
Newton's third law implies conservation of momentum [138]. It can also be seen as following from the second law: When one object ``pushes'' a second object at some (massless) point of contact using an applied force, there must be an equal and opposite force from the second object that cancels the applied force. Otherwise, there would be a nonzero net force on a massless point which, by the second law, would accelerate the point of contact by an infinite amount.
Explanation:
..... It would possibly she eenejjsjejeej 1.4
Answer:
vp = 0.94 m/s
Explanation
Formula
Vp = position/ time
position: Initial position - Final position
Position = 25 m - (-7 m) = 25 m + 7 m = 32 m
Then
Vp = 32 m / 34 seconds
Vp = 0.94 m/s
Set up the problem with the conversion rates as fractions where when you multiply the units cancel out leaving the desired units behind.
