One of the harmonics of a column of air in a tube that is open at one end and closed at the other has a frequency of 448 hz, and
1 answer:
The general formula for the frequency of the nth-harmonic of the column of air in the tube is given by
where f1 is the fundamental frequency.
In our problem, we have two harmonics, one of order n and the other one of order (n+1) (because it is the next higher harmonic), so their frequencies are
so their difference is
So, the difference between the frequencies of the two harmonics is just the fundamental frequency of the column of air in the tube, which is:
where f1 is the fundamental frequency.
In our problem, we have two harmonics, one of order n and the other one of order (n+1) (because it is the next higher harmonic), so their frequencies are
so their difference is
So, the difference between the frequencies of the two harmonics is just the fundamental frequency of the column of air in the tube, which is:
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Answer:
The initial temperature of the hot water is (assuming that no heat was lost to the surroundings.)
Explanation:
Let denote the mass of the hot water.
The question states that the mass of the water at is three times the mass of the hot water. If the mass of the hot water is
, the mass of the cold water would be
.
Let denote the specific heat capacity of water. Let
denote the mass of some water. The energy required to change the temperature of that much water by
(without state change) would be:
.
The temperature change for the cold water was:
.
Energy required to raise the temperature of water with mass from
to
:
.
On the other hand, if the initial temperature of the hot water is (where
,) the temperature change would be:
.
Calculate the energy change involved:
.
If no energy was lost to the surroundings, should be equal to
. That is:
.
Simplify and solve for :
.
.
Therefore, the initial temperature of the hot water would be .
Answer:
0.00198 secs
Explanation:
Parameters given:
Mass of baseball, m = 0.14 kg
Initial velocity of baseball, u = 33.6 m/s
Force applied to baseball, F = -5000 N
(The force is applied in an opposite direction to the initial velocity)
Final velocity, v = -37 m/s
Using the impulse-momentum theory, we have that the impulse applied to the baseball is equal to the change in momentum of the baseball:
FΔt = m(v - u)
Time interval, Δt, will be given as:
Δt =
Δt =
Δt =
Δt =
Δt =
The bat and the baseball were in contact for 0.00198 secs.