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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Rubbing both pieces cause each piece to have a negative charge.
When two parts have the same they repel each other, so holding one piece up tot he end of the other piece would push it away.
Because one piece is held in the middle by a string, it would rotate the piece in a circle.
If they held the piece to the other end of the one held by a string it would start to rotate in the opposite direction.
Answer:
speed equals distance over time 50 divided by 5.
Gravitational I think would be the answer, Hope this helps!
The relative density of gold is 19.3 it means the ratio obtained by dividing the density of gold by water at temp of 4 degree celcius is 19.3