Answer:

Explanation:
The equation of equlibrium for the box is:

The formula for the acceleration, given in
, is:

Velocity can be derived from the following definition of acceleration:





![v =\sqrt{2\cdot[(2.278\,\frac{m}{s^{2}})\cdot x |_{0\,m}^{27\,m}-(0.034\,\frac{1}{s^{2}})\cdot x^{2}|_{0\,m}^{27\,m}] }](https://tex.z-dn.net/?f=v%20%3D%5Csqrt%7B2%5Ccdot%5B%282.278%5C%2C%5Cfrac%7Bm%7D%7Bs%5E%7B2%7D%7D%29%5Ccdot%20x%20%7C_%7B0%5C%2Cm%7D%5E%7B27%5C%2Cm%7D-%280.034%5C%2C%5Cfrac%7B1%7D%7Bs%5E%7B2%7D%7D%29%5Ccdot%20x%5E%7B2%7D%7C_%7B0%5C%2Cm%7D%5E%7B27%5C%2Cm%7D%5D%20%20%7D)
The speed after the box has travelled 17 meters is:

Decreasing the distance between two objects having a considerable mass would increase the attraction on gravitation. The reverse is true that if you separate or inrease the objects distance would substantially decrease their gravitational attraction. Most object in our planet is held by its gravitational force towards it's center.
The definition of a scale of 1: 166 will mean that the scale of 1 in the model will be equivalent to 166 times the measurement in the real model, therefore we will have that the height would be 166 times smaller than the 179m given:

The same for the diameter,

The volume of a cylinder is given as




Therefore the volume would be 
Answer: Air
Explanation: Of the three mediums (gas, liquid, and solid) sound waves travel the slowest through gases, faster through liquids, and fastest through solids. Air and hydrogen have nearly the same elastic properties, but the density of hydrogen is less than that of air. Sound thus travels faster (about 4 times as fast) in hydrogen than in air.
Answer:


Explanation:

For transitions:


Thus solving it, we get:




Also, 
Where,
h is Plank's constant having value 
c is the speed of light having value 
So,


So,

Also, 
So,


