At one end of a production line, Isaac puts peaches into cans on a conveyor belt. He fills 18 cans per minute. As soon as Isaac
has filled 670 cans on the conveyor belt, Albert starts putting lids on the cans at the other end of the production line. As Albert finishes closing each can, he moves up along side the conveyor belt to meet the next can. Albert puts lids on 50 cans per minute. (Assume that the conveyor belt initially had no cans on it, and that Isaac stays in place as he continues to fill cans.) How long does it take for Albert to reach Isaac (measured from when Albert starts putting on lids)?
1 answer:
Answer: Long answer...
<em>Explanation:</em> As soon as Isaac has filled 670 cans on the conveyor belt, Albert starts putting lids on the cans at the other end of the production line. As Albert finishes closing each can, he moves up along side the conveyor belt to meet the next can. Albert puts lids on 50 cans per minute.
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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]
