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GenaCL600 [577]

3 years ago

15

D. Fill in the blanks in the following paragraph to correctly identify the conductive properties of materials. (1 point)

1 answer:

Salsk061 [2.6K]3 years ago

4 0

Answer:

conductor, insulator, resists, semiconductor, 3 and 5 group

Explanation:

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At noon on a clear day, sunlight reaches the earth\'s surface at Madison, Wisconsin, with an average intensity of approximately

Dmitrij [34]

Intensity of sunlight at given position is defined as power received per unit area

so here we can say

$I = 2 kJ/s*m^2$

area on which photons are received is given as

$A = 4.80 cm^2 = 4.80 * 10^-4 m^2$

now we can find the power received due to sunlight

$P = I*A$

$P = 2* 10^3 * 4.80 * 10^-4$

$P = 0.96 Watt$

now we can say this power is due to photons that strikes on surface of earth

so here we can say

$P = N\frac{hc}{\lambda}$

given here that

$\lambda = 510 nm$

$0.96 = N\frac{6.6 * 10^{-34}* 3 * 10^8}{510*10^{-9}}$

$0.96 = N * 3.88 * 10^{-19}$

$N = \frac{0.96}{3.88*10^{-19}}$

$N = 2.47 * 10^{18}$

so it will strike 2.47 * 10^18 photons on given area per second

3 0

3 years ago

List the 4 things technology can be.

inessss [21]

Technology can be the four things listed below...

Phones
Computers
TVs
Playstation or Xbox

6 0

3 years ago

Read 2 more answers

Sophie [7]

The correct answer is heterozygous.

7 0

3 years ago

Read 2 more answers

eimsori [14]

Explanation:

$d = \frac{1}{2} a {t}^{2} + vt \\ d = 75 \\ t = 5 \\ v = 0 \\ \\ 75 = \frac{1}{2} \times 25a + 0 \\ a = 6 \frac{m}{ {s}^{2} } \\$

7 0

3 years ago

In a photoelectric effect experiment, electrons emerge from a copper surface with a maximum kinetic energy of 1.10 eV when light

Sladkaya [172]

Answer:A) 220 nm

Explanation:

Given

Maximum Kinetic Energy $K.E.=1.10 eV$

Work Function $W=4.65 eV$

from Einstein Equation

$h\mu =W+K.E.$

$h\cdot \frac{c}{\lambda }=W+K.E.$

$h\cdot \frac{c}{\lambda }=4.65+1.10$

$6.626\times 10^{-34}\cdot \frac{3\times 10^8}{\lambda }=5.75$

$1 eV=1.6\times 10^{-19} J$

thus $5.75 eV=9.2\times 10^{-19} J$

$\lambda =\frac{6.626\times 10^{-34}\time 3\times 10^8}{9.2\times 10^{-19}}$

$\lambda =216.06 nm\approx 220 nm$

6 0

3 years ago