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Keith_Richards [23]
3 years ago
9

25. Which type of skin cancer develops from the youngest epidermal cells

Physics
1 answer:
Len [333]3 years ago
4 0

Answer:

Basal cell carcinoma

Explanation:

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What is physics? Why is it important in our life​
-Dominant- [34]

Answer:

It helps us to know or be aware of some things that happen regular and teaches us about laws that guide us

8 0
4 years ago
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3. What type of current is produced by a battery? (1 point
ladessa [460]

Answer:

direct current hope this helped : )

6 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

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3 0
2 years ago
Q|C A string with linear density 0.500 g/m is held under tension 20.0 N. As a transverse sinusoidal wave propagates on the strin
julia-pushkina [17]

A string with linear density 0.500 g/m.

Tension 20.0 N.

The maximum speed  v_{y, max}

The energy contained in a section of string 3.00 m long as a function of v_{y, max}.

We are given following data for string with linear density held under tension :

μ = 0.5 \frac{g}{m}

  = 0.5 x 10⁻³ \frac{kg}{m}

T = 20 N

If string is L = 3m long, total energy as a function of v_{y, max} is given by:

E = 1/2 x μ x L x ω² x A²

  = 1/2 x μ x L x v^{2} _{y, max}

  = 7.5 x 10⁻⁴ v^{2} _{y, max}

So, The total energy as a function of  v^{2} _{y, max} = 7.5 x 10⁻⁴ v^{2} _{y, max}

Learn more about linear density problem here:

brainly.com/question/17190616

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3 0
2 years ago
A 1,250 W electric motor is connected to a 220 Vrms, 60 Hz source. The power factor is lagging by 0.65. To correct the pf to 0.9
Black_prince [1.1K]

Answer:

C = 46.891 \mu F

Explanation:

Given data:

v = 220 rms

power factor = 0.65

P = 1250 W

New power factor is 0.9 lag

we knwo that

s = \frac{P}{P.F} < COS^{-1} 0.65

S = \frac{1250}{0.65} < 49.45

s = 1923.09 < 49.65^o

s = [1250 + 1461 j] vA

P.F new = cos [tan^{-1} \frac{Q_{new}}{P}]

solving for Q_{new}

Q_{new} = P tan [cos^{-1} P.F new]

Q_{new} = 1250 [tan[cos^{-1}0.9]]

Q_{new} = 605.40 VARS

Q_C = Q - Q_{new}

Q_C = 1461 - 605.4 = 855.6 vars

Q_C = \frac[v_{rms}^2}{xc} =v_{rms}^2 \omega C

C = \frac{Q_C}{ v_{rms}^2 \omega}

C = \frac{855.6}{220^2 \times 2\pi \times 60}

C = 4..689 \times 10^{-5} Faraday

C = 46.891 \mu F

6 0
4 years ago
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