For this case, we observe that in the interval of 5 to 10 seconds the behavior of the function is linear. Therefore, the average speed in that interval will be given by the slope of the line. This can be calculated as follows
m = (y2-y1) / (x2-x1)
Substituting the values of points A and C
m = ((30) - (10)) / ((10) - (5)) = 4
answer
the average velocity between 5 s and 10 s is 4m / s
B. +4.0 m / s
Given:
Stopping distance range is d = (65, 70) ft.
The stopping distance, d, obeys this formula.
d = v²/(2μg)
where
v = speed of the vehicle
μ = 0.8, coefficient of static friction under good road conditions
g = acceleration due to gravity, 32.2 ft/s²
Therefore
v = √(2*0.8*32.2*d) = 7.178√d
Test d = 65 ft.
v = 7.178√(65) = 57.87 ft/s = (57.87/88)*60 = 39.5 mph
Test d = 70 ft.
v = 7.178√(70) = 60.05 ft/s = 40.9 mph
To be safe, the lower speed of 39.5 mph is preferred.
Answer: 40 mph
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I think its 1x^103 (kg/m.3) I hope this help
<span>B. Controls: The known powders attributes are already studied, so the are used as control measures to compare with the findings in the experiment with the unknown powders.</span>
