Guillaume Amontons first took a stab at measuring absolute zero in 1702. What would be the most reasonable way it would have bee
1 answer:
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50N is your force and the acceleration is -9.8m/s^2 due to gravity.
So, you just plug that in.
BUT you know that mass cannot be negative, so you just disregard the negative sign and the mass of the rock is 5.102 grams.
50N is your force and the acceleration is -9.8m/s^2 due to gravity.
So, you just plug that in.
BUT you know that mass cannot be negative, so you just disregard the negative sign and the mass of the rock is 5.102 grams.
Answer:
k = 6,547 N / m
Explanation:
This laboratory experiment is a simple harmonic motion experiment, where the angular velocity of the oscillation is
w = √ (k / m)
angular velocity and rel period are related
w = 2π / T
substitution
T = 2π √(m / K)
in Experimental measurements give us the following data
m (g) A (cm) t (s) T (s)
100 6.5 7.8 0.78
150 5.5 9.8 0.98
200 6.0 10.9 1.09
250 3.5 12.4 1.24
we look for the period that is the time it takes to give a series of oscillations, the results are in the last column
T = t / 10
To find the spring constant we linearize the equation
T² = (4π²/K) m
therefore we see that if we make a graph of T² against the mass, we obtain a line, whose slope is
m ’= 4π² / k
where m’ is the slope
k = 4π² / m'
the equation of the line of the attached graph is
T² = 0.00603 m + 0.0183
therefore the slope
m ’= 0.00603 s²/g
we calculate
k = 4 π² / 0.00603
k = 6547 g / s²
we reduce the mass to the SI system
k = 6547 g / s² (1kg / 1000 g)
k = 6,547 kg / s² =
k = 6,547 N / m
let's reduce the uniqueness
[N / m] = [(kg m / s²) m] = [kg / s²]
Answer:
Explanation:
a ) angular frequency ω =
k is spring constant and m is mass attached
ω =
= 3.6515 rad / s
frequency of oscillation n = 3.6515 / (2 x 3.14)
= .5814 s⁻¹
x = .1 mcos(ωt)
= .1 mcos(3.6515t)
b ) maximum speed = ωA , A is amplitude
= 3.6515 x .1
= .36515 m /s
36.515 cm /s
maximum acceleration = ω²A
= 3.6515² x .1
= 1.333 m / s²
c ) Kinetic energy at displacement x
= 1/2 m ω²( A²-x²)
potential energy =1/2 m ω²x²
so 1/2 m ω²( A²-x²) = 1/2 m ω²x²
A²-x² = x²
2x² = A²
x = A / √2