Answer:
6.75 seconds
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration = 16 m/s²
g = Acceleration due to gravity = 9.81 m/s²
Let y be the distance the rocket is accelerating
960-y is the distance traveled in free fall

In free fall

The distance the rocket will keep accelerating is 364.881828749 m
After which it will travel 960-364.881828749 = 595.118171251 m in free fall

The time the rocket is accelerating is 6.75 seconds
The U.S. Environmental Protection Agency (EPA) standard for nitrate in drinking water is 10 milligrams of nitrate (measured as nitrogen) per liter of drinking water (mg/L). * Drinking water with levels of nitrate at or below 10 mg/L is considered safe for everyone.
The answer to your question is A.
45N and 91W
Answer:
(D) 4
Explanation:
The percentage error in each of the contributors to the calculation is 1%. The maximum error in the calculation is approximately the sum of the errors of each contributor, multiplied by the number of times it is a factor in the calculation.
density = mass/volume
density = mass/(π(radius^2)(length))
So, mass and length are each a factor once, and radius is a factor twice. Then the total percentage error is approximately 1% +1% +2×1% = 4%.
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If you look at the maximum and minimum density, you find they are ...
{0.0611718, 0.0662668} g/(mm²·cm)
The ratio of the maximum value to the mean of these values is about 1.03998. So, the maximum is 3.998% higher than the "nominal" density.
The error is about 4%.
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<em>Additional comment</em>
If you work through the details of the math, you will see that the above-described sum of error percentages is <em>just an approximation</em>. If you need a more exact error estimate, it is best to work with the ranges of the numbers involved, and/or their distributions.
Using numbers with uniformly distributed errors will give different results than with normally distributed errors. When such distributions are involved, you need to carefully define what you mean by a maximum error. (By definition, normal distributions extend to infinity in both directions.) While the central limit theorem tends to apply, the actual shape of the error distribution may not be precisely normal.