To find the answer, take 55 and divide it by 1.85 to get the thickness of one card. In this case the answer would be 29.72973 cm. each.
Answer:
Rate of change of area will be 
Explanation:
We have given rate of change of radius 
Radius of the circular plate r = 52 cm
Area is given by 
So 
Puting the value of r and 

So rate of change of area will be 
Both organisms attempt to use the same limited sources
I'll go ahead and answer the ones here without an answer. For reference, the half-life formula is <em>final amount = original amount(1/2)^(time/half-life)</em>
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4) 12.5g
x = 100(1/2)^(63/21)
5) 50g
3.125 = x(1/2)^(0.1/0.025)
6) 500g
x = 4000(1/2)^(525/175)
7) 0.24g
0.06 = x(1/2)^(11430/5730)
8) 125g
x = 1000(1/2)^(17100/5700)
Hope this helps! :)
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]