Answer:
C) True. S increases with time, v₁ = gt and v₂ = g (t-t₀) we see that for the same t v₁> v₂
Explanation:
You have several statements and we must select which ones are correct. The best way to do this is to raise the problem.
Let's use the vertical launch equation. The positive sign because they indicate that the felt downward is taken as an opponent.
Stone 1
y₁ = v₀₁ t + ½ g t²
y₁ = 0 + ½ g t²
Rock2
It comes out a little later, let's say a second later, we can use the same stopwatch
t ’= (t-t₀)
y₂ = v₀₂ t ’+ ½ g t’²
y₂ = 0 + ½ g (t-t₀)²
y₂ = + ½ g (t-t₀)²
Let's calculate the distance between the two rocks, it should be clear that this equation is valid only for t> = to
S = y₁ -y₂
S = ½ g t²– ½ g (t-t₀)²
S = ½ g [t² - (t²- 2 t to + to²)]
S = ½ g (2 t t₀ - t₀²)
S = ½ g t₀ (2 t -t₀)
This is the separation of the two bodies as time passes, the amount outside the Parentheses is constant.
For t <to. The rock y has not left and the distance increases
For t> = to. the ratio (2t/to-1)> 1 therefore the distance increases as time
passes
Now we can analyze the different statements
A) false. The difference in height increases over time
B) False S increases
C) Certain s increases with time, v₁ = gt and V₂ = g (t-t₀) we see that for the same t v₁> v₂
Answer: 1.6Hz
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Answer: 8.6 µm
Explanation:
At a long distance from the source, the components (the electric and magnetic fields) of the electromagnetic waves, behave like plane waves, so the equation for the y component of the electric field obeys an equation like this one:
Ey =Emax cos (kx-ωt)
So, we can write the following equality:
ω= 2.2 1014 rad/sec
The angular frequency and the linear frequency are related as follows:
f = ω/ 2π= 2.2 1014 / 2π (rad/sec) / rad = 0.35 1014 1/sec
In an electromagnetic wave propagating through vacuum, the speed of the wave is just the speed of light, c.
The wavelength, speed and frequency, are related by this equation:
λ = c/f
λ = 3.108 m/s / 0.35. 1014 1/s = 8.6 µm.
Answer:
Verified that he oscillations are exactly isosynchronous with frequency ω0 = p g/l, independent of the amplitude.
Explanation:
Starting from the first principle for the derivation and to prove that the oscillations are exactly isosynchronous with frequency ω0 = p g/l, independent of the amplitude. The mathematical manipulations was applied, trigonometric identities was also applied.The steps and explanation are shown in the attachment.
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:
