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
#SPJ4
Answer:
0.2 m/s
Explanation:
given,
mass of astronaut, M = 85 Kg
mass of hammer, m = 1 Kg
velocity of hammer , v =17 m/s
speed of astronaut, v' = ?
initial speed of the astronaut and the hammer be equal to zero = ?
Using conservation of momentum
(M + m) V = M v' + m v
(M + m) x 0 = 85 x v' + 1 x 17
85 v' = -17
v' = -0.2 m/s
negative sign represent the astronaut is moving in opposite direction of hammer.
Hence, the speed of the astronaut is equal to 0.2 m/s
The football and air resistance between contact
What models are you talking about
Answer:
Diamagnetic
Explanation:
Hunds rule states that electrons occupy each orbital singly first before pairing takes place in degenerate orbitals. This implies that the most stable arrangement of electrons in an orbital is one in which there is the greatest number of parallel spins(unpaired electrons).
For vanadium V ion, there are 18 electrons which will be arranged as follows;
1s2 2s2 2p6 3s2 3p6.
All the electrons present are spin paired hence the ion is expected to be diamagnetic.
