Answer:
h=17357.9m
Explanation:
The atmospheric pressure is just related to the weight of an arbitrary column of gas in the atmosphere above a given area. So, if you are higher in the atmosphere less gass will be over you, which means you are bearing less gas and the pressure is less.
To calculate this, you need to use the barometric formula:

Where R is the gas constant, M the molar mass of the gas, g the acceleration of gravity, T the temperature and h the height.
Furthermore, the specific gas constant is defined by:

Therefore yo can write the barometric formula as:

at the surface of the planet (h =0) the pressure is ![P_0[\tex]. The pressure at the height requested is half of that:[tex]P=\frac{P_0}{2}](https://tex.z-dn.net/?f=P_0%5B%5Ctex%5D.%20The%20pressure%20at%20the%20height%20requested%20is%20half%20of%20that%3A%3C%2Fp%3E%3Cp%3E%5Btex%5DP%3D%5Cfrac%7BP_0%7D%7B2%7D)
applying to the previuos equation:

solving for h:
h=17357.9m
Looks like something went wrong while editing this question
The question needs to be between 20 and 5000 characters
Answer:
___________________________________
<h3>a. Let
us assume a body has initial velocity 'u' and it is subjected to a uniform acceleration 'a' so that the final velocity 'v' after a time interval 't'. Now, By the definition of acceleration, we have:</h3>

It is first equation of motion.
___________________________________
<h3>
b. Let us assume a body moving with an initial velocity 'u'. Let it's final body 'v' after a time interval 't' and the distance travelled by the body becomes 's' then we already have,</h3>

Putting the value of v from the equation (i) in equation (ii), we have,

It is third equation of motion.
________________________________
<h3>
c. Let us assume a body moving with an initial velocity 'u'. Let it's final velocity be 'v' after a time and the distance travelled by the body be 's'. We already have,</h3>


Putting the value of t from (i) in the equation (ii)

It is forth equation of motion.
________________________________
Hope this helps...
Good luck on your assignment..
Machines capable of manufacturing exactly the same component time after time,
with exactly the resistance you want, would be very expensive, and so would the
products they turn out. A resistor would cost a dollar instead of a few pennies.
The machine itself, and its output, work within tolerances.
The cheapest mass-produced resistors are guaranteed to be within 20% above
or below the resistance marked on them. And you know what ? For most bench-
work and prototyping, that's usually close enough.
Answer:
2.0 Hz
Explanation:
2.0 Hz is the most closest to the middle number in between 2.8 and 1.5 Hz. 1.6 Hz would not be suitable as it is too close to 1.5 Hz and 2.7 would also not be suitable as it is too close to 2.8 Hz.