To develop this problem it is necessary to use the equations of description of the simple harmonic movement in which the acceleration and angular velocity are expressed as a function of the Amplitude.
Our values are given as


The angular velocity of a body can be described as a function of frequency as



PART A) The expression for the maximum angular velocity is given by the amplitude so that



PART B) The maximum acceleration on your part would be given by the expression



We are given an electromagnetic wave with a frequency of 5.09 x 10^14 Hz and travelling through a transparent medium. If the medium was vacuum, the speed of the wave would be equal to the speed of light. Otherwise, the main factor that would determine the speed of the wave is its wavelength.
So results can be shared and used by other scientists that want to use or replicate your experiment.
Answer:
(a). The spring compressed is
.
(b). The acceleration is 1.5 g.
Explanation:
Given that,
Acceleration = a
mass = m
spring constant = k
(a). We need to calculate the spring compressed
Using balance equation

....(I)
The spring compressed is
.
(b). If the compression is 2.5 times larger than it is when the mass sits in a still elevator,
The compression is given by

Here, acceleration is zero
So, 
We need to calculate the acceleration
Put the value of x in equation (I)




Hence, (a). The spring compressed is
.
(b). The acceleration is 1.5 g.
The equivalent resistance of several devices connected in parallel is given by

where

are the resistances of the various devices. We can see that every time we add a new device in parallel, the term

increases, therefore the equivalent resistance of the circuit

decreases.
But Ohm's law:

tells us that if the equivalent resistance decreases, the total current in the circuit increases. The power dissipated through the circuit (and so, the heat produced) depends on the square of the current:

therefore if there are too many devices connected in parallel, this can be a problem because there could be too much power dissipated (and too much heat) through the circuit.