This is the answer on Edgenuity 2.25*10^17
Everything in the universe that is not a star reflects light from Stars. Otherwise you can't see it.
Answer:
<em>Since I can see no choices, I answered it in my own understanding.</em>
Brian - amplitude and frequency
Marcia - amplitude and longitudinal wave
Explanation:
"Sound" and "sound waves" are essential part of a person's life. They can be used for<u> communicating</u> and <u>detecting some object</u>s.
Brian loves singing in the shower which means that he is using a greater amplitude. Amplitude refers to the<em> intensity of the sound </em>or the amount of energy that a sound carries. When one sings in the shower, the sound cannot travel very far. It bounces immediately back to the person singing thus, making the sound bigger. Brian is also using a <em>different range of </em><em>frequency</em><em> compared to his normal way of talking.</em> The frequency of a normal male voice is normally 85 to 180 Hz. A person singing may have a frequency as high as 1,500 Hz.
Marcia talks loudly on the phone. This means that she is also using a greater amplitude because the intensity of her voice is big. Since she is using the telephone, this means that her voice travels in a longitudinal wave through the telephone. This allows her voice to reach to the person on the other end of the line.
go with D. cyanobactweia,sea scorpion,birds,saber-toothed cats
To solve this problem it is necessary to apply the kinematic equations of motion and Hook's law.
By Hook's law we know that force is defined as,

Where,
k = spring constant
x = Displacement change
PART A) For the case of the spring constant we can use the above equation and clear k so that




Therefore the spring constant for each one is 11876.92/2 = 5933.46N/m
PART B) In the case of speed we can obtain it through the period, which is given by

Re-arrange to find \omega,



Then through angular kinematic equations where angular velocity is given as a function of mass and spring constant we have to




Therefore the mass of the trailer is 4093.55Kg
PART C) The frequency by definition is inversely to the period therefore



Therefore the frequency of the oscillation is 0.4672 Hz
PART D) The time it takes to make the route 10 times would be 10 times the period, that is



Therefore the total time it takes for the trailer to bounce up and down 10 times is 21.4s