Answer:
Explanation:
solution is in the attachment below
Answer:
1.23 m/s
Explanation:
p=mv
57.2 = 46.5v
v= 57.2/46.5
v= 1.23
If you want to verify your answer, just insert the value of v in the equation.
Answer:
45000 K .
Explanation:
Given :
A liter of a gas weigh 2 gram at 300 kelvin temperature and 1 atm pressure
We need to find the temperature in which 1 litre of the same gas weigh 1 gram
in pressure 75 atm.
We know, by ideal gas equation :

Here , n is no of moles , 
Putting initial and final values and dividing them :


Hence , this is the required solution.
Answer:
The correct answer to the following question will be Option A (moment arm; pivot point).
Explanation:
- The moment arm seems to be the duration seen between joint as well as the force section trying to act mostly on the joint. Each joint that is already implicated in the workout seems to have a momentary arm.
- The moment arm extends this same distance from either the pivot point to just the position of that same pressure exerted.
- The pivotal point seems to be the technical indicators required to fully measure the appropriate demand trends alongside different time-frames.
The other three choices are not related to the given situation. So that option A is the appropriate choice.
<span>by placing dams in tidal areas. ITS the only one that has anything to do with water</span>