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GREYUIT [131]
3 years ago
8

Which of the following is NOT one of the three keys to keeping water clean? Reduction of pollutants Effective cleanup of oil and

gasoline spills Encouraging algae growth Proper sewage treatment
Physics
2 answers:
Irina18 [472]3 years ago
6 0
"Reduction of Pollutants"

Andreyy893 years ago
6 0

Answer: Encouraging algae growth

Explanation:

Algae are tiny single celled aquatic plants which are found in colonies over the surface of water body and on the shorelines of the water body. The algal growth restricts the natural flow and circulation in the water body. If the water body consists of high concentration of phosphates and nitrogen this will lead to abundance in the algal growth. The algal bloom is likely to deteriorate the quality of freshwater as it will lead to the depletion of oxygen in the water body. This will likely to affect the aquatic food chain in the water body.

Hence, encouraging algae growth is not one of the three keys to keep water clean.

The following are the three keys which can be used to keep water clean.

Reduction of pollutants.

Effective cleanup of oil and gasoline spills.

Proper sewage treatment.

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When light of wavelength 240 nm falls on a cobalt surface, electrons having a maximum kinetic energy of 0.17 eV are emitted. Fin
dusya [7]

Answer:

(a) 5.04 eV (B) 248.14 nm (c) 1.21\times 10^{15}Hz

Explanation:

We have given Wavelength of the light  \lambda = 240 nm

According to plank's rule ,energy of light

E = h\nu = \frac{hc}{}\lambda

E = h\nu = \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{ 240\times 10^{-9} m\times 1.6\times 10^{-19}J/eV}= 5.21 eV

Maximum KE of emitted electron i= 0.17 eV

Part( A) Using Einstien's equation

E = KE_{max}+\Phi _{0}, here \Phi _0 is work function.

\Phi _{0}=E - KE_{max}= 5.21 eV-0.17 eV = 5.04 eV

Part( B) We have to find cutoff wavelength

\Phi _{0} = \frac{hc}{\lambda_{cuttoff}}

\lambda_{cuttoff}= \frac{hc}{\Phi _{0} }

\lambda_{cuttoff}= \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{5.04 eV\times 1.6\times 10^{-19}J/eV }=248.14 nm

Part (C) In this part we have to find the cutoff frequency

\nu = \frac{c}{\lambda_{cuttoff}}= \frac{3\times 10^{8}m/s}{248.14 \times 10^{-19} m }= 1.21\times 10^{15} Hz

5 0
3 years ago
Is this a testable question?
faltersainse [42]
This is a non testable question because it cannot be answered by doing an experiment. But it could be modified for example Dogs are more obedient then cats.
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levacccp [35]

0.77 m/s2 directed 35° south of west

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