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SashulF [63]
3 years ago
5

What is the acceptable level for chlorine

Physics
1 answer:
Katena32 [7]3 years ago
8 0
The EPA requires treated tap water<span> to have a detectable </span>level of chlorine<span> to help prevent contamination. The allowable </span>chlorine <span>in </span>drinking water<span> (up to 4 parts per million) pose “no known or expected health risk [including] an adequate margin of safety.”</span>
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An alkane has molecular weight 86. Write its molecular formula. What will be its physical state?
Brums [2.3K]

C6H14

Gaseous state

it's unsaturated hence gaseous

8 0
3 years ago
A rod of very small diameter with a mass 2m and length 3L is placed along the xaxis with one end at the origin. An identical rod
rewona [7]

Answer:

coordinates of the center of mass for these two rods

(x_{cm}, y_{cm})= (\frac{3L}{4},  \frac{3L}{4})cm

Explanation:

given

mass of a rod = 2m

length of the rod = 3L

mass of two rods = 2(2m) = 4m

radius = diameter/2 = \frac{3L}{2}

attached is the diagram and solution to the question

5 0
3 years ago
A helicopter has blades of length 4.0 m rotating at 3.0 rev/s in a horizontal plane.If the vertical component of the Earth’s mag
VARVARA [1.3K]

Answer:

Induced emf, \epsilon=9.79\times 10^7\ volts

Explanation:

Given that,

Length of the helicopter, l = 4 m

Angular speed of the helicopter, \omega=3\ rev/s=18.84\ rad/s

The vertical component of the Earth’s magnetic field is, B=6.5\times 10^5\ T

We need to find the induced emf between the tip of a blade and the hub. The induced emf in terms of angular velocity of an rotating object is given by :

\epsilon=\dfrac{1}{2}B\omega l^2

\epsilon=\dfrac{1}{2}\times 6.5\times 10^5\times 18.84\times (4)^2

\epsilon=9.79\times 10^7\ volts

So, the induced emf between the tip of a blade and the hub is \epsilon=9.79\times 10^7\ volts. Hence, this is the required solution.

5 0
3 years ago
Ultraviolet light of wavelength 270 nm strikes a metal whose work function is 2.3 eV.What is the shortest de Broglie wavelength
soldier1979 [14.2K]

Answer:

The shortest de Broglie wavelength for the electrons that are produced as photoelectrons is 0.81 nm

Explanation:

Given;

wavelength of ultraviolet light, λ = 270 nm

work function of the metal, φ = 2.3 eV = 2.3 x 1.602 x 10⁻¹⁹ J = 3.685 x 10⁻¹⁹ J

The energy of the ultraviolet light is given by;

E = \frac{hc}{\lambda}\\\\E = \frac{(6.626*10^{-34} )(3*10^{8}) }{270*10^{-9} }\\\\E = 7.362 * 10^{-19} \ J

The energy of the incident light is related to kinetic energy of the electron and work function of the metal by the following equation;

E = φ  + K.E

K.E = E - φ

K.E = (7.362 x 10⁻¹⁹ J) - (3.685 x 10⁻¹⁹ J )

K.E = 3.677 x 10⁻¹⁹ J

K.E = ¹/₂mv²

mv² = 2K.E

velocity of the electron is given by;

V = \sqrt{\frac{2K.E}{m} }\\\\V =  \sqrt{\frac{2(3.677*10^{-19}) }{9.1*10^{-31} } }\\\\V = 8.99*10^{5}  \ m/s

the shortest de Broglie wavelength for the electrons is given by;

\lambda = \frac{h}{mv}\\ \\\lambda = \frac{6.626*10^{-34} }{(9.1*10^{-31})( 8.99*10^{5} )}\\\\\lambda = 8.10*10^{-10} \ m\\\\\lambda = 0.81 \ nm

Therefore, the shortest de Broglie wavelength for the electrons that are produced as photoelectrons is 0.81 nm

7 0
3 years ago
Please help! I haven't really been able to understand this topic :/
Anettt [7]

Ok but y I thought it was upside down tho...

4 0
3 years ago
Read 2 more answers
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