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viva [34]
3 years ago
9

Which of the following is a safety precaution to take in order to avoid wildlife

Physics
2 answers:
Luden [163]3 years ago
6 0

Answer:

C. Stay away from food and water sources

e-lub [12.9K]3 years ago
4 0

Answer:

c

Explanation:

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When two hydrogen atoms bond, the positive nucleus of one atom attracts the
mafiozo [28]
<span>When two hydrogen atoms bond, the positive nucleus of one atom attracts the electrons of both the atoms since the electrons are shared between the two atoms now.</span>
6 0
3 years ago
An LED operation at 850 nm center wavelength has a spectral width of 45 nm. What is the pulse spreading in ns/km
Papessa [141]

Answer:

\mathbf{\dfrac{\sigma_{mat}}{L} = 3.6 \ ns/ km}

Explanation:

From the given information, the LED is operating with a given wavelength of 850 nm or 0.85 μm.

Hence, the material dispersion is \dfrac {d \tau _{mat}}{d \lambda } \simeq (80 \ ps / (nm.km) \ )

Now, using the pulse spread formula:

\dfrac{\sigma_{mat}}{L} = \dfrac{d \tau _{mat} }{d \lambda} \sigma \lambda

\dfrac{\sigma_{mat}}{L} = (80 \ ps/ ( m.km) \ )  \times (45 \ nm)

Thus, the pulse spreading as a result of  material dispersion is:\mathbf{\dfrac{\sigma_{mat}}{L} = 3.6 \ ns/ km}

3 0
3 years ago
Why were quasars and active galaxies not initially recognized as being “special” in some way?
djyliett [7]

Answer:

In the past, astronomers look into the sky and the universe as a whole with an idea that it is a place where stars are born, transition through their life stages and ultimately die and this is because they couldnt differentiate between stars, quasars and active galaxies because with advanced equipment, they all look similar but as they technological ages arrived, they were able see that they are not the same.

8 0
3 years ago
Neutral hydrogen can be modeled as a positive point charge +1.6×10^−19C surrounded by a distribution of negative charge with vol
Archy [21]

Answer:

a) 1.082 × 10⁻¹⁹C  ( e = 1.6 × 10⁻¹⁹C)  

b) 3.466 × 10¹¹ N/C

Explanation:

a)

p(r) = -A exp ( - 2r/a₀)

Q = ₀∫^∞ ₀∫^π ₀∫^2xπ p(r)dV  =  -A  ₀∫^∞ ₀∫^π ₀∫^2π   exp ( - 2r/a₀)r² sinθdrdθd∅

Q = -4πA ₀∫^∞ exp ( - 2r/a₀)r²dr = -e

now using integration by parts;

A = e / πa₀³

p(r) =  - (e / πa₀³) exp (-2r/a₀)

Now Net charge inside a sphere of radius a₀ i.e Qnet is;

= e - (e / πa₀³)  ₀∫^a₀ ₀∫^π ₀∫^2π  r² exp (-2r/a₀)dr

= e - e + 5e exp (-2) = 1.082 × 10⁻¹⁹C  ( e = 1.6 × 10⁻¹⁹C)  

b)

Using Gauss's law,

E × 4πa₀ ² = Qnet / ∈₀

E = 4πa₀ ² × Qnet × 1/a₀²

E = 3.466 × 10¹¹ N/C

4 0
3 years ago
What is the answer to these questions?
Oksana_A [137]

Answer:

THES IS NOT

Explanation:

THIS PAPAER IS A FAKE PAPAER BEACISE POGI TALAGA AKO

7 0
3 years ago
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