If one were 230 kilometers above sea level, one would be in the part of the atmosphere known as the ________.
1 answer:
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Answer:
Approximately (assuming that the partner is holding the other end of the coil stationary.)
Explanation:
In a standing wave, an antinode is a point that moves with maximal amplitude, while a node is a point that does not move at all. There is an antinode between every two adjacent nodes. Likewise, there is a node between every two adjacent antinodes.
The side of the spring that is being shaken moving with maximal amplitude. Hence, that point on this spring would also be an antinode. In contrast, the side of the spring that is held still (does not move at all) would be a node.
There would be a node between:
- the antinode at the end of the spring that is being shaken, and
- the antinode between the two ends of this spring.
Overall, the nodes and antinodes on this spring would be:
- node at the end that is being held still,
- antinode (as mentioned in the question),
- node (inferred, not mentioned in the question), and
- antinode at the end that is being shaken.
The distance between two adjacent nodes is equal to one-half (that is, ) the wavelength of the wave. The distance between a node and an adjacent antinode is one-quarter (that is,
) of the wavelength of the wave.
Thus, if the wavelength of the wave in this question is , the length of this spring would be:
.
The question states that the length of this coiled spring is . In other words,
. The wavelength of this wave would be
.
The frequency of this wave is the number of cycles in unit time:
.
Hence, the speed of this wave would be:
.
Answer:
The minimum angle at which the box starts to slip (rounded to the next whole number) is α=19°
Explanation:
In order to solve this problem we must start by drawing a sketch of the problem and its corresponding fre body diagram (See picture attached).
So, when we are talking about friction, there are two types of friction coefficients. Static and kinetic. Static friction happens when the box is not moving no matter what force you apply to it. You get to a certain force that is greater than the static friction and the box starts moving, it is then when the kinetic friction comes into play (kinetic friction is generally smaller than static friction). So in order to solve this problem, we must find an angle such that the static friction is the same as the force applie by gravity on the box. For it to be easier to analyze, we must incline the axis of coordinates, just as shown on the picture attached.
After doing an analysis of the free-body diagram, we can build our set of equations by using Newton's thrid law:
we can see there are only two forces in x, which are the weight on x and the static friction, so:
when solving for the static friction we get:
We know the weight is found by multiplying the mass by the acceleration of gravity, so:
W=mg
and:
we can substitute this on our sum of forces equation:
the static friction will depend on the normal force applied by the plane on the box, static friction is found by using the following equation:
so we can substitute this on our equation:
but we don't know what the normal force is, so we need to find it by doing a sum of forces in y.
In the y direction we got two forces as well, the normal force and the force due to gravity, so we get:
when solving for N we get:
When seeing the free-body diagram we can determine that:
so we can substitute that in the sum of y-forces equation, so we get:
we can go ahead and substitute this equation in the sum of forces in x equation so we get:
we can divide both sides of the equation into mg so we get:
as you may see, the angle doesn't depend on the mass of the box, only on the static coefficient of friction. When solving for we get:
when simplifying this we get that:
now we can solve for the angle so we get:
and we can substitute the given value so we get:
which yields:
α=19.29°
which rounds to:
α=19°
