Answer:
X = 2146.05 m
Explanation:
We need to understand first what is the value we need to calculate here. In this case, we want to know how far from the starting point the package should be released. This is the distance.
We also know that the plane is flying a certain height with an specific speed. And the distance we need to calculate is the distance in X with the following expression:
X = Vt (1)
However we do not know the time that this distance is covered. This time can be determined because we know the height of the plain. This time is referred to the time of flight. And the time of flight can be calculated with the following expression:
t = √2h/g (2)
Where g is gravity acceleration which is 9.8 m/s². Replacing the data into the expression we have:
t = √(2*2500)/9.8
t = 22.59 s
Now replacing into (1) we have:
X = 95 * 22.59
<h2>
X = 2146.05 m</h2>
This is the distance where the package should be released.
Hope this helps
Answer: B (on the rough side)
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Answer:
Explanation:
1, i, π, e, and φ
dozen, gross, googol, and Avogadro's number etc, are pure numbers
hope this helps
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Answer:
Sand
Explanation:
While the term buoyancy may confuse the reader, the question here is referring to buoyant force.
Buoyant force is the amount of pressure exerted on an object by the liquid it is in. Given by the formula

Where
= Buoyant Force
= Volume of Object submerged in the liquid
= Density of Liquid
= Force of gravity
Since in this question the swimmer with the sand is completely submerged in water, more of the volume of his preserver is under the water hence the buoyant force is greater on it than the swimmer with the Styrofoam (as part of him is not submerged in water)
Answer:
(A) Fo = 72 Hz
(B) The pipe is open at both ends
(C) The length of the pipe is 2.38m
This problem involves the application of the knowledge of standing waves in pipes.
Explanation:
The full solution can be found in the attachment below.
For pipes open at both ends the frequency of the pipe is given by
F = nFo = nv/2L where n = 1, 2, 3, 4.....
For pipes closed at one end the frequency of the pipe is given by
F = nFo = nv/4L where n = 1, 3, 5, 7...
The full solution can be found in the attachment below.