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?
To learn more about uniformly accelerated motion refer to:
brainly.com/question/14669575
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6 J is the impulse caused by the change in velocity of 2 kg box from 2 m/s to 5 m/s.
Answer:
The magnitude of impulse is 6 J.
Explanation:
Impulse is the force acting on any object for a given time interval. As force is equal to the product of mass and acceleration and acceleration is the rate of change of velocity with time. Then the product of force with time interval will be equal to the product of mass with change in velocity.
F = m a = 
FΔt = mΔv
Impulse = FΔt=mΔv
As the mass of box is given as 2 kg and the velocity changes from 2 m/s to 5 m/s, then the impulse = 2 × (5-2) = 2 ×3 =6 J
So 6 J is the impulse caused by the change in velocity of 2 kg box from 2 m/s to 5 m/s.
If the object is moving in a straight line at a constant speed, then that's
the definition of zero acceleration. It can only happen when the sum of
all forces (the 'net' force) on the object is zero.
And it doesn't matter what the object's mass is. That argument is true
for specks of dust, battleships, rocks, stars, rock-stars, planets, and
everything in between.
<span>The number in front is the number of molecules (or atoms) taking part in the (balanced) chemical reaction equation.</span>
Here we can say that there is no external torque on this system
So here we can say that angular momentum is conserved
so here we will have
now we have
similarly let the final distance is "r"
so now we have
now from above equation we have
so final distance is 0.04 m between them
