Answer:
Components: 0.0057, -0.0068. Magnitude: 0.0089 m/s
Explanation:
The displacement in the x-direction is:

While the displacement in the y-direction is:

The time taken is t = 304 s.
So the components of the average velocity are:


And the magnitude of the average velocity is

Answer:
A) 4037.2[km]; B) 21472[m/s]
Explanation:
A)
This part can be solved using the principle of energy conservation, where kinetic energy will be equal to potential energy. We will define the kinetic energy and the potential energy.
![Ek= 0.5*m*v^2\\where:\\m = mass [kg]\\v = 8.90*10^3[m/s]\\Ep=m*g*h\\where:\\g = gravity = 9.81[m/s^2]\\h = elevation or height [m]\\](https://tex.z-dn.net/?f=Ek%3D%200.5%2Am%2Av%5E2%5C%5Cwhere%3A%5C%5Cm%20%3D%20mass%20%5Bkg%5D%5C%5Cv%20%3D%208.90%2A10%5E3%5Bm%2Fs%5D%5C%5CEp%3Dm%2Ag%2Ah%5C%5Cwhere%3A%5C%5Cg%20%3D%20gravity%20%3D%209.81%5Bm%2Fs%5E2%5D%5C%5Ch%20%3D%20elevation%20or%20height%20%5Bm%5D%5C%5C)
Now we have to match both equations in this way the mass value is canceled and we can clear h.
![0.5*m*(8.9*10^3)^{2}=m*9.81*h\\ h=4037206.93[m] = 4037.2[km]](https://tex.z-dn.net/?f=0.5%2Am%2A%288.9%2A10%5E3%29%5E%7B2%7D%3Dm%2A9.81%2Ah%5C%5C%20h%3D4037206.93%5Bm%5D%20%3D%204037.2%5Bkm%5D)
B)
To solve this problem we can use the kinematic equations, but first we must identify the initial data:
yo = 2.35*10^7[m]
y = 0 [m] when hit the ground [m]
vo = 0 [m/s], It is essentially at rest...
![v_{f} ^{2} =v_{o} ^{2} +2*a*(y)\\v_{f} =\sqrt{2*9.81*2.35*10^{7} } \\v_{f} =21472.54[m/s] o 21.47[km/s]](https://tex.z-dn.net/?f=v_%7Bf%7D%20%5E%7B2%7D%20%3Dv_%7Bo%7D%20%5E%7B2%7D%20%2B2%2Aa%2A%28y%29%5C%5Cv_%7Bf%7D%20%3D%5Csqrt%7B2%2A9.81%2A2.35%2A10%5E%7B7%7D%20%7D%20%5C%5Cv_%7Bf%7D%20%3D21472.54%5Bm%2Fs%5D%20o%2021.47%5Bkm%2Fs%5D)
She needs to lose an average of 10 pounds per week in order to fulfill her expectation. Go Becky !
Since the discrete Fourier series, the Sampling rate, would be the equivalent of the inverse of the passage of time, that is, to the frequency, mathematically this can be written as,

In turn, the time can be described depending on the period and the amount of data samples taken. This would be,

Here,
m = Data Samples
T = Period
Replacing,

Replacing the value of the time from the first equation,


At the same time, the range then will be given between the basic frequency to the half of the sample, that is,



Therefore the lowest frequency is 5000Hz and highest 9.165Hz