Answer:
d = 4.9 m
Explanation:
It is mentioned that a stone is dropped from a certain height. It is required to find the distance covered by it in one second.
The initial speed of the stone is equal to 0 as it was at rest. Let d is the distance covered by the stone.
Using second equation of motion :
Put u = 0 and a = g
So, the distance covered by it in one second is 4.9 m.
Low pressure has a bit less of a function than high pressure, high pressure is more useful in certain terms
Ω₀ = the initial angular velocity (from rest)
t = 0.9 s, time for a revolution
θ = 2π rad, the angular distance traveled
Let
α = the angular acceleration
ω = the final angular velocity
The angular rotation obeys the equation
(1/2)*(α rad/s²)*(0.9 s)² = (2π rad)
α = 15.514 rad/s²
The final angular velocity is
ω = (15.514 rad/s²)*(0.9 s) = 13.963 rad/s
If the thrower's arm is r meters long, the tangential velocity of release will be
v = 13.963r m/s
Answer: 13.963 rad/s
Answer:
you have probably missed some details in the question.
Answer:
The correct answer is a rarefaction.
Explanation:
Sound waves are longitudinal waves that propagate in a medium, such as air. As the vibration continues, a series of successive condensations and rarefactions form and propagate from it. The pattern created in the air is something like a sinusoidal curve to represent a sound wave.
There are peaks in the sine wave at the points where the sound wave has condensations and valleys where it has rarefactions.
Have a nice day!
