What happens to the period of a wave as the frequency increases?.
1 answer:
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Answer:
a)
b)
Explanation:
Given:
height of water in one arm of the u-tube,
a)
Gauge pressure at the water-mercury interface,:
we've the density of the water
b)
Now the same pressure is balanced by the mercury column in the other arm of the tube:
<u>Now the difference in the column is :</u>
Answer:
Explanation:
For answer this we will use the law of the conservation of the angular momentum.
so:
where is the moment of inertia of the merry-go-round,
is the initial angular velocity of the merry-go-round,
is the moment of inertia of the merry-go-round and the child together and
is the final angular velocity.
First, we will find the moment of inertia of the merry-go-round using:
I =
I =
I = 359.375 kg*m^2
Where is the mass and R is the radio of the merry-go-round
Second, we will change the initial angular velocity to rad/s as:
W = 0.520*2 rad/s
W = 3.2672 rad/s
Third, we will find the moment of inertia of both after the collision:
Finally we replace all the data:
Solving for :
Answer:
the two gliders collide, the mobile glider will transfer a bit of time to the fixed glider, which is why it comes out with a speed that is smaller than that of the bullet glider.
Explanation:
When the two gliders collide, the mobile glider will transfer a bit of time to the fixed glider, which is why it comes out with a speed that is smaller than that of the bullet glider.
Changes can occur that the gliders unite and move with a cosecant speed less than the initial one.
The whole process must be analyzed using conservation of the moment.
p₀ = m v₀
celestines que clash case
p_f = (m + M) v
po = pf
m v₀ = (n + M) v
v =
calculemos
v=
v= 0.09 m/s
elastic shock case
p₀ = m v₀
p_f = m v₁ +M v₂
p₀ = p_f
m v₀ = m v₁ + m v₂