Answer:
L = 0.7 m
Explanation:
This is a resonance exercise, in this case the air-filled pipe is open at both ends, therefore we have bellies at these points.
λ / 2 = L 1st harmonic
λ = L 2nd harmonic
λ = 2L / 3 3rd harmonic
λ = 2L / n n -th harmonic
the speed of sound is related to wavelength and frequencies
v =λ f
f = v /λ
we substitute
f = v n / 2L
the speed of sound in air is v = 343 m / s
suppose that the frequency of f = 980Hz occurs in harmonic n
f₁ = v n / 2L
f₂ = v (n + 1) / 2L
f₃ = v (n + 2) / 2L
we substitute the values
2 980/343 = n / L
2 1260/343 = (n + 1) / L
2 1540/343 = (n + 2) / L
we have three equations, let's use the first two
5.714 = n / L
7.347 = (n + 1) / L
we solve for L and match the expressions
n / 5,714 = (n + 1) / 7,347
7,347 n = 5,714 (n + 1)
n (7,347 -5,714) = 5,714
n = 5,714 / 1,633
n = 3.5
as the number n must be integers n = 4 we substitute in the first equation
L = n / 5,714
L = 0.7 m
Answer:
Explanation:
To find the angular velocity of the tank at which the bottom of the tank is exposed
From the information given:
At rest, the initial volume of the tank is:
where;
height h which is the height for the free surface in a rotating tank is expressed as:
at the bottom surface of the tank;
r = 0, h = 0
∴
0 = 0 + C
C = 0
Thus; the free surface height in a rotating tank is:
Now; the volume of the water when the tank is rotating is:
dV = 2π × r × h × dr
Taking the integral on both sides;
replacing the value of h in equation (2); we have:
![V_f = \dfrac{ \pi \omega ^2}{g} \Big [ \dfrac{r^4}{4} \Big]^R_0](https://tex.z-dn.net/?f=V_f%20%3D%20%5Cdfrac%7B%20%5Cpi%20%5Comega%20%5E2%7D%7Bg%7D%20%5CBig%20%5B%20%20%5Cdfrac%7Br%5E4%7D%7B4%7D%20%5CBig%5D%5ER_0)
![V_f = \dfrac{ \pi \omega ^2}{g} \Big [ \dfrac{R^4}{4} \Big] --- (3)](https://tex.z-dn.net/?f=V_f%20%3D%20%5Cdfrac%7B%20%5Cpi%20%5Comega%20%5E2%7D%7Bg%7D%20%5CBig%20%5B%20%20%5Cdfrac%7BR%5E4%7D%7B4%7D%20%5CBig%5D%20---%20%283%29)
Since the volume of the water when it is at rest and when the angular speed rotates at an angular speed is equal.
Then 
Replacing equation (1) and (3)
Finally, the angular velocity of the tank at which the bottom of the tank is exposed = 10.48 rad/s
Answer:
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Answer:
Fx= 50.0 Pounds : Components of the force along the x-axis
Fy= 86.6 Pounds : Component of the force along the y-axis
Explanation:
Conceptual Analysis
To find the components (Fx, Fy) of the total force (F), we apply the trigonometric concepts for a right triangle, where the perpendicular sides of the triangle are the components (Fx, Fy) of the force (F), the hypotenuse (h) is the magnitude of the total force F and β is the angle that forms the horizontal component with the hypotenuse.
Formulas
cos β : x/h : x: side adjacent to the β angle h: hypotenuse (1)
sin β = y/h : y: side opposite to the β angle h: hypotenuse (2)
Known Data
Known data
F= 1.00 * 10² pounds = 100 pounds : magnitude of total force
β = 60.0° to the x-axis. : Angle that forms the force with the x-axis
Problem Development
We apply the formula 1 to calculate horizontal component (Fx)
cos β :Fx/F
Fx= F cosβ = 100*cos 60° = 50.0 Pounds
We apply the formula 2 to calculate vertical component (Fy)
sin β = Fy/F
Fy= F sinβ = 100*sin 60° = 86.6 Pounds
