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tatiyna
2 years ago
11

Water flows through a garden hose which is attached to a nozzle. The water flows through hose with a speed of 1.81 m/s and throu

gh the nozzle with a speed of 18.3 m/s. Calculate the maximum height (in m) to which water could be squirted if it emerges from the nozzle and emerges with the nozzle removed. (a) Emerges from the nozzle. m (b) Emerges with the nozzle removed, assuming the same flow rate. m
Physics
1 answer:
Serggg [28]2 years ago
6 0

Answer:

a) 17.086m

b) 0.1671 m

Explanation:

Given data: speed of water through the hose  = 1.81 m/s

through the nozzle = 18.3 m/s

We know that maximum height of an object with upward velocity v is given by,

a) H = v^2/2g

where H is the maximum height water emerges  

= 18.3^2/(2×9.8) = 17.086 m answer

b) Again,  

H = v^2/2g

= 1.81^2/(2×9.8) = 0.1671 m

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If you exert 250J of work onto a lever that is 5m long, what amount of force are you applying?
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By calculating its wavelength (in nm), show that the second line in the Lyman series is UV radiation.
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Explanation:

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an atom transition is the jump of an electron from an initial state to a final state of lesser emergy

            ΔE = 13.606 (1 / n_{f}^{2} - 1 / n_{i}^{2})

the so-called Lyman series occurs when the final state nf = 1, so the second line occurs when ni = 3, let's calculate the energy of the emitted photon

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let's reduce the energy to the SI system

            DE = 12.094 eV (1.6 10⁻¹⁹ J / 1 ev) = 10.35 10⁻¹⁹ J

let's find the wavelength is this energy, let's use Planck's equation to find the frequency

            E = h f

             f = E / h

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            f = 2.9186 10¹⁵ Hz

now we can look up the wavelength

           c = λ f

           λ = c / f

           λ = 3 10⁸ / 2.9186 10¹⁵

           λ = 1.0278  10⁻⁷ m

let's reduce to nm

            λ = 102.78  nm

This radiation is in the UV range, which occurs for wavelengths less than 400 nm.

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