Answer:
0.173 m.
Explanation:
The fundamental frequency of a closed pipe is given as
fc = v/4l .................. Equation 1
Where fc = fundamental frequency of a closed pipe, v = speed of sound l = length of the pipe.
Making l the subject of the equation,
l = v/4fc ................ Equation 2
also
v = 331.5×0.6T ................. Equation 3
Where T = temperature in °C, T = 18.0 °c
Substitute into equation 3
v = 331.5+0.6(18)
v = 331.5+10.8
v = 342.3 m/s.
Also given: fc = 494 Hz,
Substitute into equation 2
l = 342.3/(4×494)
l = 342.3/1976
l =0.173 m.
Hence the length of the organ pipe = 0.173 m.
The ball will decelerate as it moves upwards.
The magnitude of the ball's acceleration is 0.3 m/s² and it directed backwards.
The given parameters;
- initial velocity of the ball, u = 1.25 m/s
- time of motion of the ball, t = 4.22 s
As the ball rolls up the inclined plane, the velocity decreases and eventually becomes zero when the ball reaches the highest point of the plane.
Thus, the ball decelerate as it moves upwards.
The acceleration of the ball is calculate as;
<em>at the highest point on the incline plane, the final velocity </em>
<em> is zero</em>
Thus, the magnitude of the ball's acceleration is 0.3 m/s² and it directed backwards.
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Resistance is what Inertia goes to.....
The particles can undergo small oscillations around x₂.
The given parameters;
- <em>initial energy of the particles = E₁</em>
- <em>final energy of the particles, E₂ = 0.33E₁</em>
The movement of the particles depends on the kinetic energy of the particles.
When kinetic energy of the particles is 100%, the particles can oscillate from x₁ to x₅.
However, when the total energy of this particles is reduced to one-third (¹/₃) or 33% of the initial energy of the particle, the oscillation of the particles will be reduced.
- The maximum position the particle can oscillate is x₅
- The half position the particles can oscillate is x₃
Since 33% is less than the half of the energy of the particle, the particle will oscillate between x₁ and x₂.
Thus, we can conclude that the particles can undergo small oscillations around x₂.
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To solve this problem it is necessary to apply the concepts related to heat exchange in the vegetable and water.
By definition the exchange of heat is given by
where,
m = mass
c = specific heat
= Change in temperature
Therefore the total heat exchange is given as
Our values are given as,
Total mass is 
= 200lb ,however the mass of solid vegetable and water is given as,
Replacing at our equation we have,
Therefore the heat removed is 22411.2 Btu
