How does gravity affect the motions of Earth and the Moon?
2 answers:
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Answer:
Explanation:
As per the formula of thermal expansion we know that
so here we will have
so here change in the length of the rod is given as
The propagation errors we can find the uncertainty of a given magnitude is the sum of the uncertainties of each magnitude.
Δm = ∑
Physical quantities are precise values of a variable, but all measurements have an uncertainty, in the case of direct measurements the uncertainty is equal to the precision of the given instrument.
When you have derived variables, that is, when measurements are made with different instruments, each with a different uncertainty, the way to find the uncertainty or error is used the propagation errors to use the variation of each parameter, keeping the others constant and taking the worst of the cases, all the errors add up.
If m is the calculated quantity, x_i the measured values and Δx_i the uncertainty of each value, the total uncertainty is
Δm = ∑ | dm / dx_i | Dx_i
for instance:
If the magnitude is a average of two magnitudes measured each with a different error
m =
Δm = | | Δx₁ + |
| Δx₂
= ½
= ½
Δm = Δx₁ + ½ Δx₂
Δm = Δx₁ + Δx₂
In conclusion, using the propagation errors we can find the uncertainty of a given quantity is the sum of the uncertainties of each measured quantity.
Learn more about propagation errors here:
a)
The equation which gives the position of a simple harmonic oscillator is:
where
A is the amplitude
is the angular frequency, with k being the spring constant and m the mass
t is the time
Let's start by calculating the angular frequency:
The amplitude, A, can be found from the maximum velocity of the spring:
So, the equation of motion is
b) t=0.10 s, t=0.52 s
The potential energy is given by:
While the kinetic energy is given by:
The velocity as a function of time t is:
The problem asks as the time t at which U=3K, so we have:
However, , so we have
with two solutions:
c) 3 seconds.
When x=0, the equation of motion is:
so, t=0.
When x=1.00 m, the equation of motion is:
So, the time needed is 3 seconds.
d) 0.097 m
The period of the oscillator in this problem is:
The period of a pendulum is:
where L is the length of the pendulum. By using T=0.628 s, we find