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N76 [4]
3 years ago
5

Water emerges straight down from a faucet with a 2.51-cm diameter at a speed of 3.04 m/s. (because of the construction of the fa

ucet, there is no variation in speed across the stream.) what is the diameter of the stream 0.200 m below the faucet? neglect any effects due to surface tension.
Physics
1 answer:
Mkey [24]3 years ago
3 0
This is a question on conservation of energy. That is,
mgh + KE1 = KE2
mgh +1/2mv1^2 = 1/2mv2^2
gh + 1/2v1^2 = 1/2v2^2

Where, h = 0.2 m, v1 =3.04 m/s
Therefore,
v2 = Sqrt [2(gh+1/2v1^2)] = Sqrt [2(9.81*0.2 + 1/2*3.04^2)] = 7.26 m/s

Now, Volumetric flow rate, V/time, t = Surface area, A*velocity, v
Where,
V = Av = πD^2/4*3.04 = π*(2.51/100)^2*1/4*3.04 = 1.504*10^-3 m^3/s

At 0.2 m below,
V = 1.504*10^-3 m^3/s = A*7.26
A = (1.504*10^-3)/7.26 = 2.072*10^-4 m^2

But, A = πr^2
Then,
r = Sqrt (A/π) = Sqrt (2.072*10^-4/π) = 0.121*10^-3 m
Diameter = 2r = 0.0162 m = 1.62 cm
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Answer:

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So

        \lambda_p = \frac{h}{m_p * v_p }

so    \lambda_p = \frac{6.62607015 * 10^{-34}}{1.67 *10^{-27} * 3.09529 *10^{7} }

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