The red, yellow, and green wavelengths of sunlight are absorbed by water molecules in the ocean. ... In coastal areas, runoff from rivers, resuspension of sand and silt from the bottom by tides, waves and storms and a number of other substances can change the color of the near-shore waters.
Answer:
a) x = 4.33 m
, b) w = 2 rad / s
, f = 0.318 Hz
, c) a = - 17.31 cm / s²,
d) T = 3.15 s, e) A = 5.0 cm
Explanation:
In this exercise on simple harmonic motion we are given the expression for motion
x = 5 cos (2t + π / 6)
they ask us for t = 0
a) the position of the particle
x = 5 cos (π / 6)
x = 4.33 m
remember angles are in radians
b) The general form of the equation is
x = A cos (w t + Ф)
when comparing the two equations
w = 2 rad / s
angular velocity and frequency are related
w = 2π f
f = w / 2π
f = 2 / 2pi
f = 0.318 Hz
c) the acceleration is defined by
a == d²x / dt²
a = - A w² cos (wt + Ф)
for t = 0
, we substitute
a = - 5,0 2² cos (π / 6)
a = - 17.31 cm / s²
d) El period is
T = 1/f
T= 1/0.318
T = 3.15 s
e) the amplitude
A = 5.0 cm
Answer:
m³/(kg⋅s²)
Explanation:
Hello.
In this case, since the involved formula is:

By writing a dimensional analysis with the proper algebra handling, we obtain:
![N[=]G*\frac{kg*kg}{m^2}\\ \\kg*\frac{m}{s^2}[=]G *\frac{kg*kg}{m^2}\\\\G[=]\frac{kg*m*m^2}{kg^2*s^2}\\ \\G[=]\frac{m^3}{kg*s^2}](https://tex.z-dn.net/?f=N%5B%3D%5DG%2A%5Cfrac%7Bkg%2Akg%7D%7Bm%5E2%7D%5C%5C%20%5C%5Ckg%2A%5Cfrac%7Bm%7D%7Bs%5E2%7D%5B%3D%5DG%20%2A%5Cfrac%7Bkg%2Akg%7D%7Bm%5E2%7D%5C%5C%5C%5CG%5B%3D%5D%5Cfrac%7Bkg%2Am%2Am%5E2%7D%7Bkg%5E2%2As%5E2%7D%5C%5C%20%5C%5CG%5B%3D%5D%5Cfrac%7Bm%5E3%7D%7Bkg%2As%5E2%7D)
Thus, answer is:
m³/(kg⋅s²)
Note that the [=] is used to indicate the units of G.
Best regards
Answer: A
Explanation:Earthquakes occur on faults - strike-slip earthquakes occur on strike-slip faults, normal earthquakes occur on normal faults, and thrust earthquakes occur on thrust or reverse faults. When an earthquake occurs on one of these faults, the rock on one side of the fault slips with respect to the other.
Answer:
x = 0.9 m
Explanation:
For this exercise we must use the rotational equilibrium relation, we will assume that the counterclockwise rotations are positive
∑ τ = 0
60 1.5 - 78 1.5 + 30 x = 0
where x is measured from the left side of the fulcrum
90 - 117 + 30 x = 0
x = 27/30
x = 0.9 m
In summary the center of mass is on the side of the lightest weight x = 0.9 m