Answer:
A sound wave can be affected by a lot of different variables. As an audio engineer some of the more common things we deal with involve air temperature, humidity and even wind. The first two affect the speed at which the wave travels, while wind can actually cause a phase like effect if it is blowing hard enough. Another big one though not directly related to the air is walls and other solid objects that cause the sound wave to bounce off of them and reflect. This causes a secondary wave that isn’t as strong as the first wave but is the cause of “muddy” sounding venues when you are indoors.
Explanation:
The total mass of the mixture would not change.
Think about it as a mix of marbles. Suppose you have a group of 10 marbles. 5 marbles are red, and 5 marbles are blue. If you separate the marbles into 2 different groups by color, you still have 10 marbles in total, but they are separated and no longer a mixture of colors.
Similarities: They are waves, they both displace particles, both transfer energy
Differences: Transverse- particles move perpendicularly, waves are perpendicular to direction, and they are 2 dimension . Longitudinal- particles are displaced same direction as a wave, move left and right, and are one dimension
How to measure:
Transverse- measure 2 adjacent crests or troughs
Longitudinal- measure from one rare faction to another or one compression to another
Example: Transverse- vibrations in string . Longitudinal- sound waves
Answer:
3.88 × 10^-4 m
Explanation:
Given that a person who weighs 614 N is riding a 85-N mountain bike. Suppose the entire weight of the rider plus bike is supported equally by the two tires. If the gauge pressure in each tire is 9.00 x 10^5 Pa. What is the area of contact between each tire and the ground?
The total weight = 614 + 85
The total weight = 699N
Let the total area of contact = A
Pressure = Force / A
Substitute all the parameters into the formula
900000 = 699 /A
A = 699 / 900000
A = 7.77 × 10^-4 m
The area of contact between each tire and the ground will be = A/2
That is, 7.77 / 2 = 3.88 × 10^-4 m
Therefore, the area of contact between each tire and the ground is 3.88 × 10^-4 m approximately.
