Answer:
a) 24.692 m/s
b) 19.4 m
Explanation:
To calculate the velocity at the nozzle outflow (V2) we use the Bernoulli equation:
We know that the velocity above the oil surface (V1) and the pressure at the nozzle outflow (P2) are negligible, the height in the exit is zero (Z2) then:
a) The velocity (V2) is:
![V2=[(\frac{P1}{pg}+Z1)(2g)]^{1/2}](https://tex.z-dn.net/?f=V2%3D%5B%28%5Cfrac%7BP1%7D%7Bpg%7D%2BZ1%29%282g%29%5D%5E%7B1%2F2%7D)
Substituting the known values we can get the velocity at the out:
Atmospheric pressure= 101000 Pa
Oil density= 0.88x(Water density)=0.88(1000kg/m3)=880kg/m3
![V2=[(\frac{150000Pa+101000 Pa}{(880 kg/m3)(9.81m/s)}+2m)(2(9.81m/s2))]^{1/2}](https://tex.z-dn.net/?f=V2%3D%5B%28%5Cfrac%7B150000Pa%2B101000%20Pa%7D%7B%28880%20kg%2Fm3%29%289.81m%2Fs%29%7D%2B2m%29%282%289.81m%2Fs2%29%29%5D%5E%7B1%2F2%7D)
b) To calculate the height we have to apply the Bernoulli equation between the outflow and the maximum height (Z3), so:
We know that the velocity above the stream (V3) and the pressure at the nozzle outflow (P2) are negligible, the pressure at the top of the stream (P3) is the atmospheric pressure, then:
Substituting the known values, the height (Z3) is:
Z3=Maximum Height=19.376=19.4 m
Answer:
(A) Fo = 72 Hz
(B) The pipe is open at both ends
(C) The length of the pipe is 2.38m
This problem involves the application of the knowledge of standing waves in pipes.
Explanation:
The full solution can be found in the attachment below.
For pipes open at both ends the frequency of the pipe is given by
F = nFo = nv/2L where n = 1, 2, 3, 4.....
For pipes closed at one end the frequency of the pipe is given by
F = nFo = nv/4L where n = 1, 3, 5, 7...
The full solution can be found in the attachment below.
Answer:
4°C
Explanation:
Water is densest at 4°C. Since dense water sinks, the bottom of the lake will be 4°C.
Immersed => totally submerged
a) Volume occupied by the ball, V = area of the base of the cylynder * water level rise
V =π(r^2)(4cm) = π*100cm^2*4cm = 400 π cm^3
Volume of a sphere = [4/3]π(r^3) = 400 π cm^3 =>
r^3 = 300 cm^3 => r = ∛(300 cm^3) = 6.69 cm
b) V = π(100cm)^2 (8cm) = 80,000 π cm^3
[4/3]πr^3 = 80,000π cm^3 => r^3 = 60,000 cm^3 => r = 39.15 cm
