The time after being ejected is the boulder moving at a speed 20.7 m/s upward is 2.0204 s.
<h3>What is the time after being ejected is the boulder moving at a speed 20.7 m/s upward?</h3>
The motion of the boulder is a uniformly accelerated motion, with constant acceleration
a = g = -9.8 
downward (acceleration due to gravity).
By using Suvat equation:
v = u + at
where: v is the velocity at time t
u = 40.0 m/s is the initial velocity
a = g = -9.8 
is the acceleration
To find the time t at which the velocity is v = 20.7 m/s
Therefore,
The time after being ejected is the boulder moving at a speed 20.7 m/s upward is 2.0204 s.
The complete question is:
A large boulder is ejected vertically upward from a volcano with an initial speed of 40.0 m/s. Ignore air resistance. At what time after being ejected is the boulder moving at 20.7 m/s upward?
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The average acceleration between t = 5.6 s and t = 8.5 s is 2.31 m/s²
<h3>What is acceleration?</h3>
Acceleration is defined as the rate change of velocity with time.
acceleration a = (Δv) / (Δt)
An object is moving with initial velocity u =5.7 m/s and its final velocity v= -1.0 m/s.
Time taken for the change in speed, t= 8.5 - 5.6 = 2.9 seconds
The acceleration is given by
a = (-1 - 5.7)/ 2.9
a = - 2.31 m/s²
|a | = 2.31 m/s²
Thus, the object's acceleration is 2.31 m/s²
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Answer:
dV/dt = 9 cubic inches per second
Explanation:
Let the height of the cylinder is h
Diameter of cylinder = height of the cylinder = h
Radius of cylinder, r = h/2
dh/dt = 3 inches /s
Volume of cylinder is given by
put r = h/2 so,
Differentiate both sides with respect to t.
Substitute the values, h = 2 inches, dh/dt = 3 inches / s
dV/dt = 9 cubic inches per second
Thus, the volume of cylinder increases by the rate of 9 cubic inches per second.
