The acceleration of gravity on or near the Earth's surface is 9.8 m/s² downward.
Is that right ? I don't hear any objection, so I'll assume that it is.
That means that during every second that gravity is the only force on an object,
the object either gains 9.8m/s of downward speed, or it loses 9.8m/s of upward
speed. (The same thing.)
If the rock starts out going up at 14.2 m/s, and loses 9.8 m/s of upward speed
every second, it runs out of upward gas in (14.2/9.8) = <em>1.449 seconds</em> (rounded)
At that point, since it has no more upward speed, it can't go any higher. Right ?
(crickets . . .)
Answer:
What do you need help with?
Explanation:
At stp (standard temperature and pressure), the temperature is T=0 C=273 K and the pressure is p=1.00 atm. So we can use the ideal gas law to find the number of moles of helium:
where p is the pressure (1.00 atm), V the volume (20.0 L), n the number of moles, T the temperature (273 K) and
the gas constant. Using the numbers and re-arranging the formula, we can calculate n:
-- Class I lever
The fulcrum is between the effort and the load.
The Mechanical Advantage can be anything, more or less than 1 .
Example: a see-saw
-- Class II lever
The load is between the fulcrum and the effort.
The Mechanical Advantage is always greater than 1 .
Example: a nut-cracker, a garlic press
-- Class III lever
The effort is between the fulcrum and the load.
The Mechanical Advantage is always less than 1 .
I can't think of an example right now.
