<span> If you make 46 pushes then your sister swings through 46 cycles and she does so in 90 seconds. The frequency would be 46/90 = 0.511 Hz.</span>
The lowest constant acceleration needed for takeoff from a 1.80 km runway is 2.8 m/s².
To find the answer, we need to know about the Newton's equation of motion.
<h3>What's the Newton's equation of motion to find the acceleration in term of initial velocity, final velocity and distance?</h3>
- The Newton's equation of motion that connects velocity, distance and acceleration is V² - U²= 2aS
- V= final velocity, U= initial velocity, S= distance and a= acceleration
<h3>What's the acceleration, if the initial velocity, final velocity and distance are 0 m/s, 360km/h and 1.8 km respectively?</h3>
- Here, S= 1.8 km or 1800 m, V= 360km/h or 100m/s , U= 0 m/s
- So, 100²-0= 2×a×1800
=> 10000= 3600a
=> a= 10000/3600 = 2.8 m/s²
Thus, we can conclude that the lowest constant acceleration needed for takeoff from a 1.80 km runway is 2.8 m/s².
Learn more about the Newton's equation of motion here:
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Answer:Decreases
Explanation:
Given
Volume is held constant that is it is a isochoric process.
We know that
PV=nRT
as n,V& R are constant therefore only variables are
P & T
so 
As 
is decreasing therefore Pressure must also decrease so that ratio remains constant.
I would say D. all of the above
