<h2>
Answer: Toward the center of the circle.</h2>
This situation is characteristic of the uniform circular motion , in which the movement of a body describes a circumference of a given radius with constant speed.
However, in this movement the velocity has a constant magnitude, but its direction varies continuously.
Let's say
is the velocity vector, whose direction is perpendicular to the radius
of the trajectory, therefore
the acceleration
is directed toward the center of the circumference.
<span>The answer is low gear. To dry the brakes of a vehicle, one must driv slowly in low gear and apply light pressure on the vehicle's brake pedal. Dry brakes are less expensive and are cheap when it needs replacement, yet under pressure they may cause more damage to the system than wet brakes, which are better in terms of emergency.</span>
Answer:
The time where the avergae speed equals the instaneous speed is T/2
Explanation:
The velocity of the car is:
v(t) = v0 + at
Where v0 is the initial speed and a is the constant acceleration.
Let's find the average speed. This is given integrating the velocity from 0 to T and dividing by T:
![v_{ave} = \frac{1}{T}\int\limits^T_0 {v(t)} \, dt](https://tex.z-dn.net/?f=v_%7Bave%7D%20%3D%20%5Cfrac%7B1%7D%7BT%7D%5Cint%5Climits%5ET_0%20%7Bv%28t%29%7D%20%5C%2C%20dt)
v_ave = v0+a(T/2)
We can esaily note that when <u><em>t=T/2</em></u><u><em> </em></u>
v(T/2)=v_ave
Now we want to know where the car should be, the osition of the car is:
![x(t) = x_A + v_0 t + \frac{1}{2}at^2](https://tex.z-dn.net/?f=x%28t%29%20%3D%20x_A%20%2B%20v_0%20t%20%2B%20%5Cfrac%7B1%7D%7B2%7Dat%5E2)
Where x_A is the position of point A. Therefore, the car will be at:
<u><em>x(T/2) = x_A + v_0 (T/2) + (1/8)aT^2</em></u>
Due to the fact that no one can consume .04 of a tablet, we can round down this answer to 1. This means that Mr. Jones should take C- 1 tablet per day.
I hope I've helped! :)
Answer:
![\frac{d^2x}{dt^2}+\frac{\beta}{m}\frac{dx}{dt}+\frac{k}{m}x=0](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5E2x%7D%7Bdt%5E2%7D%2B%5Cfrac%7B%5Cbeta%7D%7Bm%7D%5Cfrac%7Bdx%7D%7Bdt%7D%2B%5Cfrac%7Bk%7D%7Bm%7Dx%3D0)
Explanation:
let
be the mass attached, let
be the spring constant and let
be the positive damping constant.
-By Newton's second law:
![m\frac{d^2x}{dt^2}=-kx-\beta \frac{dx}{dt}](https://tex.z-dn.net/?f=m%5Cfrac%7Bd%5E2x%7D%7Bdt%5E2%7D%3D-kx-%5Cbeta%20%5Cfrac%7Bdx%7D%7Bdt%7D)
where
is the displacement from equilibrium position. The equation can be transformed into:
shich is the equation of motion.