Answer:
Yes, it is reasonable to neglect it.
Explanation:
Hello,
In this case, a single molecule of oxygen weights 32 g (diatomic oxygen) thus, the mass of kilograms is (consider Avogadro's number):
After that, we compute the potential energy 1.00 m above the reference point:
Then, we compute the average kinetic energy at the specified temperature:
Whereas 
stands for the Avogadro's number for which we have:
In such a way, since the average kinetic energy energy is about 12000 times higher than the potential energy, it turns out reasonable to neglect the potential energy.
Regards.
According to the conservation of mechanical energy, the kinetic energy just before the ball strikes the ground is equal to the potential energy just before it fell.
Therefore, we can say KE = PE
We know that PE = m·g·h
Which means KE = m·g·h
We can solve for h:
h = KE / m·g
= 20 / (0.15 · 9.8)
= 13.6m
The correct answer is: the ball has fallen from a height of 13.6m.
The body system on the chart
a,b,c is your answer light and sound are not considered matter and heat is energy created from matter and electricity is particles moving basically therefore electricity is matter hope this helps
We will apply the conservation of linear momentum to answer this question.
Whenever there is an interaction between any number of objects, the total momentum before is the same as the total momentum after. For simplicity's sake we mostly use this equation to keep track of the momenta of two objects before and after a collision:
m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'
Note that v₁ and v₁' is the velocity of m₁ before and after the collision.
Let's choose m₁ and v₁ to represent the bullet's mass and velocity.
m₂ and v₂ represents the wood block's mass and velocity.
The bullet and wood will stick together after the collision, so their final velocities will be the same. v₁' = v₂'. We can simplify the equation by replacing these terms with a single term v'
m₁v₁ + m₂v₂ = m₁v' + m₂v'
m₁v₁ + m₂v₂ = (m₁+m₂)v'
Let's assume the wood block is initially at rest, so v₂ is 0. We can use this to further simplify the equation.
m₁v₁ = (m₁+m₂)v'
Here are the given values:
m₁ = 0.005kg
v₁ = 500m/s
m₂ = 5kg
Plug in the values and solve for v'
0.005×500 = (0.005+5)v'
v' = 0.4995m/s
v' ≅ 0.5m/s
