Answer: Transverse waves have motion perpendicular to velocity, while longitudinal waves have motion parallel to velocity.
Explanation:
Transverse waves are characterized by the fact that the particles of the medium in which they propagate move transversely to the direction of propagation of the wave.
In other words,<u> its displacement is perpendicular to the direction of propagation of the wave</u>, being a good example the circular waves in the water.
On the other hand, Longitudinal waves are characterized by the fact that <u>the oscillation of the particles in the medium is parallel to the direction of propagation of the wave.</u> A good example of this is the sound wave.
I think you forgot to give the choices along with the question. I am answering the question based on my research and knowledge. <span>If a layer was deposited but does not appear in the rock record, the thing that happened is erosion. I hope that this is the ans wer that has actually come to your desired help.</span>
Answer:
If there is no damping, the amount of transmitted vibration that the microscope experienced is = 
Explanation:
The motion of the ceiling is y = Y sinωt
y = 0.05 sin (2 π × 2) t
y = 0.05 sin 4 π t
K = 25 lb/ft × 4 sorings
K = 100 lb/ft
Amplitude of the microscope ![\frac{X}{Y}= [\frac{1+2 \epsilon (\omega/ W_n)^2}{(1-(\frac{\omega}{W_n})^2)^2+(2 \epsilon \frac{\omega}{W_n})^2}]](https://tex.z-dn.net/?f=%5Cfrac%7BX%7D%7BY%7D%3D%20%5B%5Cfrac%7B1%2B2%20%5Cepsilon%20%28%5Comega%2F%20W_n%29%5E2%7D%7B%281-%28%5Cfrac%7B%5Comega%7D%7BW_n%7D%29%5E2%29%5E2%2B%282%20%5Cepsilon%20%20%5Cfrac%7B%5Comega%7D%7BW_n%7D%29%5E2%7D%5D)
where;


= 
= 4.0124
replacing them into the above equation and making X the subject of the formula:



Therefore; If there is no damping, the amount of transmitted vibration that the microscope experienced is = 
Answer:
The work flow required by the compressor = 100.67Kj/kg
Explanation:
The solution to this question is obtained from the energy balance where the initial and final specific internal energies and enthalpies are taken from A-17 table from the given temperatures using interpolation .
The work flow can be determined using the equation:
M1h1 + W = Mh2
U1 + P1alph1 + ◇U + Workflow = U2 + P2alpha2
Workflow = P2alpha2 - P1alpha1
Workflow = (h2 -U2) - (h1 - U1)
Workflow = ( 684.344 - 491.153) - ( 322.483 - 229.964)
Workflow = ( 193.191 - 92.519)Kj/kg
Workflow = 100.672Kj/kg