Answer:
Let the mass of the book be "m", acceleration due to gravity be "g", velocity be "v" and height be "h".
Now if we are holding a book at a certain height (h), <em><u>the potential energy will be maximum which is equal to mass× acceleration due to gravity× height (= mgh)</u>.</em>
(Remember: kinetic energy =0)
Now we consider that the book is dropped, in this case a force will act downward towards the centre of the earth, <em><u>Force= mass× acceleration due to gravity (F=mg)</u></em>. It is equal to the weight of the book.
While the book is falling, the potential energy stored in the book converts into kinetic energy and strikes the floor with <em><u>the maximum kinetic energy= (1/2)×mass×velocity² (=1/2mv²)</u>.</em>
(Remember: kinetic energy=0)
Due to this process the whole energy is conserved.
As the potential energy decreases kinetic energy increases.
Answer:
the mass of the air in the classroom = 2322 kg
Explanation:
given:
A classroom is about 3 meters high, 20 meters wide and 30 meters long.
If the density of air is 1.29 kg/m3
find:
what is the mass of the air in the classroom?
density = mass / volume
where mass (m) = 1.29 kg/m³
volume = 3m x 20m x 30m = 1800 m³
plugin values into the formula
1.29 kg/m³ = <u> mass </u>
1800 m³
mass = 1.29 kg/m³ ( 1800 m³ )
mass = 2322 kg
therefore,
the mass of the air in the classroom = 2322 kg
The alpha particle is emitted at 4235 m/s
Explanation:
We can use the law of conservation of momentum to solve the problem: the total momentum of the original nucleus must be equal to the total momentum after the alpha particle has been emitted. Therefore:
where:
is the mass of the original nucleus
is the initial velocity of the nucleus
is the mass of the alpha particle
is the final velocity of the alpha particle
is the mass of the daughter nucleus
is the final velocity of the nucleus
Solving for
, we find the final velocity of the alpha particle:

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600/3 = 200
the slope is 200m/min
OR
600/ (3/60) =
600 x 60/3 =
600 x 20 = 12,000 meters per hour
The initial momentum of the system can be expressed as,

The final momentum of the system can be given as,

According to conservation of momentum,

Plug in the known expressions,

Initially, the second mass move towards the first mass therefore the initial speed of second mass will be taken as negative and the recoil velocity of first mass is also taken as negative.
Plug in the known values,

Thus, the final velocity of second mass is 2.99 m/s.