Air pressure changes with altitude because of issues related to gravity. Molecules have more weight the closer they are to the Earth and more of them move to lower elevations as a result; this causes increased pressure because there are more molecules in number and proximity. Conversely, air at higher elevations has less weight, but also forces pressure on those layers below it, resulting in the molecules closer to the Earth supporting more weight, increasing the pressure
To solve this problem we will apply the concepts related to equilibrium, for this specific case, through the sum of torques.

If the distance in which the 600lb are applied is 6in, we will have to add the unknown Force sum, at a distance of 27in - 6in will be equivalent to that required to move the object. So,



So, Force that must be applied at the long end in order to lift a 600lb object to the short end is 171.42lb
Answer:
Microwave towers are telecommunications towers that use microwaves to transmit telephone and television signals to other microwave towers.
Explanation:
This technology replaced existing transmission wires, but it is almost entirely obsolete as of 2015 due to the advent of fiber optics and other modern methods of telecommunication.
Answer:
Your pinball machine was built using two kinds of simple machines: a lever and an inclined plane. The lever shot the marble to the top of the box with lots of force. The inclined planes made the marble wind its way down to the bottom.
Answer:

Explanation:
A parallel-plate capacitors consist of two parallel plates charged with opposite charge.
Since the distance between the plates (1 cm) is very small compared to the side of the plates (19 cm), we can consider these two plates as two infinite sheets of charge.
The electric field between two infinite sheets with opposite charge is:

where
is the surface charge density, where
Q is the charge on the plate
A is the area of the plate
is the vacuum permittivity
In this problem:
- The side of one plate is
L = 19 cm = 0.19 m
So the area is

Here we want to find the maximum charge that can be stored on the plates such that the value of the electric field does not overcome:

Substituting this value into the previous formula and re-arranging it for Q, we find the charge:
