A tortoise and a hare engage in a race. A tortoise can run with a speed of 0.15 m/s. A hare can run 25 times as fast as the tort
2 answers:
Answer:
Explanation:
The speed of hare = .15 x 25 = 3.75 m /s . Let tortoise took t second to complete the race .
Distance traveled by it = .15 t
Distance traveled by hare = .15 t - .35 m
Time taken by hare to complete this distance
= t - 5 x 60 s
Speed of hare
= Distance / time
( .15t-.35 ) / t - 300 , so
t = 312.40
= 5 minutes 12.4 seconds
Distance of race
312.4 x speed of tortoise
= 312.4 x .15
= 46.85 m
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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]
