KE = ½mv² = ½(4.00 kg)(16.0 m/s)² = 512 J
Answer:
the velocity of the bullet-wood system after the collision is 2.48 m/s
Explanation:
Given;
mass of the bullet, m₀ = 20 g = 0.02 kg
velocity of the bullet, v₀ = 250 m/s
mass of the wood, m₁ = 2 kg
velocity of the wood, v₁ = 0
Let the velocity of the bullet-wood system after collision = v
Apply the principle of conservation of linear momentum to calculate the final velocity of the system;
Initial momentum = final momentum
m₀v₀ + m₁v₁ = v(m₀ + m₁)
0.02 x 250 + 2 x 0 = v(2 + 0.02)
5 + 0 = v(2.02)
5 = 2.02v
v = 5/2.02
v = 2.48 m/s
Therefore, the velocity of the bullet-wood system after the collision is 2.48 m/s
<span>A: put an atom on a poster in the exhibit
Good luck. The poster itself is made of trillions of trillions of trillions
of atoms. You could not see the extra one any easier than you could
see the ones that are already there, and even if you could, it would be
lost in the crowd.
B: use a life size drawing of an atom
Good luck. Nobody has ever seen an atom. Atoms are too small
to see. That's a big part of the reason that nobody knew they exist
until less than 200 years ago.
D: set up a microscope so that visitors can view atoms
Good luck. Atoms are way too small to see with a microscope.
</span><span><span>C: Display a large three dimensional model of an atom.
</span> </span>Finally ! A suggestion that makes sense.
If something is too big or too small to see, show a model of it
that's just the right size to see.
If he's falling in a straight line and his speed is not changing, that tells you that his acceleration is zero.
And THAT tells you that the forces on him are balanced, the net force acting on him is zero, and his motion is the same as it would be if there were NO force acting on him.
