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Monica [59]
4 years ago
10

Which of the following is a possible use for nuclear energy?

Physics
2 answers:
belka [17]4 years ago
5 0
<span>Nuclear energy can be used to power all of the above choices. Nuclear power plants produce radioactive waste that must be stored properly. It is very impossible for a nuclear power plant to have no waste at all since lots of chemicals are used to create the process as it gives energy to other machines, weapons such as bombs and  powering submarines.  Radioactive waste can not be released into local water supplies since the wastes are very radioactive and may cause mutation to the fishes and bioaccumulation which will affect humans as well. It will also cause air pollution if the chemicals are not stored properly.</span>
Bess [88]4 years ago
4 0

Answer:

D. All of the above. you're welcome

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Escape speed from the surface of earth is about
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3 years ago
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When charges qa, qb, and qc are placed respectively at the corners a, b, and c of a right triangle, the potential at the midpoin
earnstyle [38]

Answer:

8v

Explanation:

First we apply super position principle

Vt= v1 + v2+ v3

Remove qa

But vt= 20v

So V = v2+v3

V1= 20-15

= 5v

Remove qb

V= v1+v3

V=8v

So the potential when qa and qc are remove is the potential due to qb

Which is 8v

7 0
3 years ago
Two strong magnets were brought close to each other. They were repelling each other. Explain what must have happened.​
r-ruslan [8.4K]

When two magnets are brought together, the opposite poles will attract one another, but the like poles will repel one another. This is similar to electric charges. Like charges repel, and unlike charges attract.

pls. mark brainliest am. dyning for it

8 0
3 years ago
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What is the wavelength, in nm, of a photon with energy
lara31 [8.8K]

Answer:

(a)  λ = 4136 nm → infrared

(b) λ = 413.6 nm → visible light

(c) λ = 41.36 nm → ultraviolet

Explanation:

The wavelength of infrared is on the range of 700 nm to 1000000 nm

The wavelength of visible light is between 400 nm and 700 nm

The wavelength of ultraviolet ray on the range of 10 nm to 400 nm

The wavelength of photon is given by;

E = hf

f is the frequency of the wave = c / λ

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

Where;

c is the speed of light = 3 x 10⁸ m/s

h is Planck's constant = 6.626 x 10⁻³⁴ J/s

(a) 0.3 eV = 0.3 x 1.602 x 10⁻¹⁹ J

\lambda = \frac{(6.626 * 10^{-34})(3*10^8)}{(0.3)*(1.602*10^{-19})}\\\\\lambda = 4.136 *10^{-6} \ m

λ = 4136 x 10⁻⁹ m

λ = 4136 nm → infrared

(b) 3.0 eV

\lambda = \frac{(6.626 * 10^{-34})(3*10^8)}{(3)*(1.602*10^{-19})}\\\\\lambda = 4.136 *10^{-7} \ m

λ = 413.6 x 10⁻⁹ m

λ = 413.6 nm →visible light

(c) 30 eV

\lambda = \frac{(6.626 * 10^{-34})(3*10^8)}{(30)*(1.602*10^{-19})}\\\\\lambda = 4.136 *10^{-8} \ m

λ = 41.36 x 10⁻⁹ m

λ = 41.36 nm →ultraviolet

5 0
3 years ago
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