a block of mass 5kg slides down an inclined plane that has an angle of 10 degrees if the inclined plane has no friction and the
2 answers:
In this case, the inclined plane is frictionless and therefore energy is perfectly conserved with no losses. The block originally starts off with some gravitational potential energy, which is then completely converted into kinetic energy. So the initial gravitational potential energy is equal to the final kinetic energy.
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Answer:
The value is
Explanation:
From the question we are told that
The initial velocity of the proton is
At a distance R from the nucleus the velocity is
The velocity considered is
Generally considering from initial position to a position of distance R from the nucleus
Generally from the law of energy conservation we have that
Here is the change in kinetic energy from initial position to a position of distance R from the nucleus , this is mathematically represented as
=>
=>
=>
And is the change in electric potential energy from initial position to a position of distance R from the nucleus , this is mathematically represented as
Here is zero because the electric potential energy at the initial stage is zero so
So
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=>
Generally considering from initial position to a position of distance from the nucleus
Here represented the distance of the proton from the nucleus where the velocity is
Generally from the law of energy conservation we have that
Here is the change in kinetic energy from initial position to a position of distance R from the nucleus , this is mathematically represented as
=>
=>
=>
And is the change in electric potential energy from initial position to a position of distance
from the nucleus , this is mathematically represented as
Here is zero because the electric potential energy at the initial stage is zero so
So
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=>
Divide equation 2 by equation 1
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Answer:68.15m/s
Explanation:
<u><em>Given: </em></u>
v₁=15m/s
a=6.5m/s²
v₁=?
x=340m
<u><em>Formula:</em></u>
v₁²=v₁²+2a (x)
<u>Set up:</u>
=
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