6.35x10^-4 OR 6.3x10-4 (if only one decimal number is allowed)
Answer:
Your zenith is 43 N of 90 deg (equator)
Thus, your zenith is 90 - 43 = 47 deg
(At the N pole your zenith would be 0 deg from the N pole)
Answer:
a)
b)
c) 
d)
e)
Explanation:
1) Important concepts
Simple harmonic motion is defined as "the motion of a mass on a spring when it is subject to the linear elastic restoring force given by Hooke's Law (F=-Kx). The motion experimented by the particle is sinusoidal in time and demonstrates a single resonant frequency".
2) Part a
The equation that describes the simple armonic motion is given by
(1)
And taking the first and second derivate of the equation (1) we obtain the velocity and acceleration function respectively.
For the velocity:
(2)
For the acceleration
(3)
As we can see in equation (3) the acceleration would be maximum when the cosine term would be -1 and on this case:

Since we know the amplitude A=0.002m we can solve for
like this:

And we with this value we can find the period with the following formula

3) Part b
From equation (2) we see that the maximum velocity occurs when the sine function is euqal to -1 and on this case we have that:

4) Part c
In order to find the total mechanical energy of the oscillator we can use this formula:

5) Part d
When we want to find the force from the 2nd Law of Newton we know that F=ma.
At the maximum displacement we know that X=A, and in order to that happens
, and we also know that the maximum acceleration is given by::

So then we have that:

And since we have everything we can find the force

6) Part e
When the mass it's at the half of it's maximum displacement the term
and on this case the acceleration would be given by;

And the force would be given by:

And replacing we have:

Answer:
0.283m
Explanation:
Speed (v) = wavelength × Frequency (f)
Wavelength = speed/ frequency
Wavelength = 343/ 1210 = 0.283m
Answer:
a) w = 4.24 rad / s
, b) α = 8.99 rad / s²
Explanation:
a) For this exercise we use the conservation of kinetic energy,
Initial. Vertical bar
Emo = U = m g h
Final. Just before touching the floor
Emf = K = ½ I w2
As there is no friction the mechanical energy is conserved
Emo = emf
mgh = ½ m w²
The moment of inertial of a point mass is
I = m L²
m g h = ½ (m L²) w²
w = √ 2gh / L²
The initial height h when the bar is vertical is equal to the length of the bar
h = L
w = √ 2g / L
Let's calculate
w = RA (2 9.8 / 1.09)
w = 4.24 rad / s
b) Let's use Newton's equation for rotational motion
τ = I α
F L = (m L²) α
The force applied is the weight of the object, which is at a distance L from the point of gro
mg L = m L² α
α = g / L
α = 9.8 / 1.09
α = 8.99 rad / s²