Concentration involves the relative amounts of solvent and solute in a solution, when strength is related to the extent to which a substance dissociates :))
i hope this is helpful
have a nice day
Within an atom, there are three elementary particles: the proton, neutron, and electron. Most of the mass of an atom is situated within the nucleus, which is the central part of the atom. It is made up of protons and neutrons, which are the heaviest subatomic particles. The electrons within the atom, orbit around the nucleus at a very far distance. Electrons are also a part of the lightest group of subatomic particles called leptons. That is why these electrons don't contribute much to the majority of an atoms mass. They are very light and they orbit at very far distances.
The moment of inertia of the flywheel is 2.63 kg-
It is given that,
The maximum energy stored on the flywheel is given as
E=3.7MJ= 3.7×
J
Angular velocity of the flywheel is 16000
= 1675.51
So to find the moment of inertia of the flywheel. The energy of a flywheel in rotational kinematics is given by :
E = 
By rearranging the equation:
I = 
I = 2.63 kg-
Thus the moment of inertia of the flywheel is 2.63 kg-.
Learn more about moment of inertia here;
brainly.com/question/13449336
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I'm not sure what "60 degree horizontal" means.
I'm going to assume that it means a direction aimed 60 degrees
above the horizon and 30 degrees below the zenith.
Now, I'll answer the question that I have invented.
When the shot is fired with speed of 'S' in that direction,
the horizontal component of its velocity is S cos(60) = 0.5 S ,
and the vertical component is S sin(60) = S√3/2 = 0.866 S . (rounded)
-- 0.75 of its kinetic energy is due to its vertical velocity.
That much of its KE gets used up by climbing against gravity.
-- 0.25 of its kinetic energy is due to its horizontal velocity.
That doesn't change.
-- So at the top of its trajectory, its KE is 0.25 of what it had originally.
That's E/4 .
Given:
u(initial velocity)=0
v(final velocity)= 10 m/s
t= 4 sec
Now we know that
v= u + at
Where v is the final velocity
u is the initial velocity
a is the acceleration measured in m/s^2
t is the time measured in sec
10=0+ax4
a=10/4
a=2.5 m/s^2
