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amm1812
4 years ago
6

You don't have any matches, so you want to focus sunlight to ignite a pile of twigs for a campfire. Which mirror would be best t

o use? A.convex mirror B.concave mirror C.flat mirror D.plane mirror
Physics
2 answers:
Anna35 [415]4 years ago
4 0
Out of that list, the concave mirror is the only item that can concentrate sunlight and heat into a small area. But if you could get ahold of a convex lens, that would be even better.
Lera25 [3.4K]4 years ago
3 0

Answer: Concave mirror would be best to use.

Explanation :

To focus sunlight to ignite a pile of twigs for a campfire, a concave mirror should be used.

When the rays of light coming from infinity parallel to principal axis in front of a concave mirror, all the rays get concentrated at the focus of the mirror. That's why the concave mirror is also known as converging.

In this case, a concave mirror should be used so as to converge all the rays at a point.

So, the correct option is (b).        

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Choose the sources of gamma rays. Check all that apply.
harina [27]

Answer:

a. Stars all warm objects

c. Some unstable atomic nuclei

Explanation:

Gamma rays are photons of very high energy (beyond 100keV) enough to remove an electron from its orbit.

They have a very short wavelength, less than 5 meters from the peak, and can be produced by nuclear decay, especially in the breasts of massive stars at the end of life.

They were discovered by the French chemist Paul Villard (1860 to 1934).

While X-rays are produced by electronic transitions in general caused by the collision of an electron with an atom at high speed, gamma rays are produced by nuclear transitions.

 Gamma rays produce damage similar to those caused by X-rays or ultraviolet rays (burns, cancer and genetic mutations).

The sources of gamma rays that we observe in the universe come from <u>massive stars (hypernovas) or some warm objects on the space</u> that end their lives by a gravitational collapse that leads to the formation of a neutron star or a black hole, as well as <u>unstable radioactive nuclei </u>that emit radiation gamma to reach its steady state.

7 0
4 years ago
You decide it is time to clean your pool since summer is quickly approaching. Your pool maintenance guide specifies that the chl
kotykmax [81]

Answer:

Concentration of Cl2: 1,048 ppm

Explanation:

The unit of measurement molarity (M) represents the same magnitud as mole/L (number of moles of the substance in one liter of solution). The concentration in ppm represents the same as mg/L (number of miligrammes of the substance in one liter of solution). In order to convert from M to ppm we have to consider, then, mass and volume. We can see that in both cases the volume is expressed in liters, so we don't have to do anything to change it. The problem comes when you see that you have to convert moles, from the molatiry, to miligrammes, to get the result in mg/L, meaning ppm.

First of all, notice that the substance is Cl2, so we need to find a relationship between the number of moles and its mass. If you look in the periodic table you'll see that the atomic mass for Cl is 35,4 g/mole (grammes of Cl in one mole). So, there is a way to relate the moles to the mass of the substance and it is represented on the equation below:

mass=mole*atomic mass

We want to find the mass (m) and we know the amount of moles of Cl2 in the solution (moles=2,96x10^-5), so, if we use the values known on the equation above we get that:

m=2,96x10^-5 moles*35,4 g/mole\\ m=1,048x10^-3 g

Remember that these grammes are found in one liter of solution. So, this means we have 1,048x10^-3 g/L. Previously we said that ppm=mg/L, so all that's left to do it to convert grammes to miligrammes:

1 g = 1000 mg

If we multiply both sides by 1,048x10^-3:

1 * 1,048x10^-3 g = 1000 * 1,048x10^-3 mg

1,048x10^-3 g = 1,048 mg

Knowing that this amount of mass was found in one liter, we get that the amount of substance in the solution is:

1,048 mg/L

Knowing that <em>mg/L=ppm</em>, then the concentration of Cl2 in ppm is:

<u>1,048 ppm</u>

4 0
3 years ago
If light of wavelength 700 nm strikes such a photocathode, what will be the maximum kinetic energy, in eV , of the emitted elect
Oksana_A [137]

If the light of wavelength 700 nm strikes such a photocathode the maximum kinetic energy, in eV, of the emitted electrons is 0.558 eV.

so - $KE_{max} = hc/lembda}  work

threshold when KE = 0

hc/lambda = work = 1240/900=1.38 eV

b) Kemax = hc/lambda - work = 1240/640 -1.38=0.558 eV

What is photocathode?

  • A photocathode electrolyte interface can be used in a photoelectrolysis cell as the primary light-harvesting junction (in conjunction with an appropriate electrochemical anode) or as an optically complementary photoactive half-cell in a tandem photoelectrode photoelectrolysis cell (Hamnett, 1982; Kocha et al, 1994).
  • In the case of the former, the electrode should ideally harvest photon energy across the majority of the solar spectrum in order to achieve the highest energy conversion efficiency possible.
  • In the latter case, however, the photocathode may only be active in a specific band of the solar spectrum in order to generate a cathodic photocurrent sufficient to match the current generated in the photoanodic half-cell.

To learn more about Photocathode from the given link:

brainly.com/question/9861585

#SPJ4

3 0
2 years ago
(Select all that apply.)
krok68 [10]

Answer:

boulder perched...

compressed spring

6 0
3 years ago
A soccer player takes a free kick from a spot that is 20 m from the goal. The ball leaves his foot at an angle of 38 ∘, and it e
Sergio039 [100]
This dude can kick the ball
7 0
3 years ago
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