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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In electricity, the most famous and basic equation is the Ohm's Law which relates the parameters voltage, current and resistance. One form of this law as written in equation is V = IR, where V is the voltage in volts, I is the current in amperes and R is the resistance in ohms. These parameters depends in the arrangements, whether it's series or parallel.
In a series connection, the voltage is greater across a high-resistance resistor. Therefore, the voltage is much greater for the 20-ohm resistor. However,if it is a parallel circuit, the voltage is just the same for both resistors.
Answer:

Explanation:
Given that,
Radius, r = 2 m
Velocity, v = 1 m/s
We need to find the magnitude of the centripetal acceleration. The formula for the centripetal acceleration is given by :

So, the magnitude of centripetal acceleration is
.