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8090 [49]
3 years ago
11

What is the equivalent resistance between points a and b in the figure?

Physics
1 answer:
madam [21]3 years ago
3 0
20.o hope this helps
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A laser pulse of duration 25 ms has a total energy of 1.4 J. The wavelength of this radiation is
SpyIntel [72]

Answer:

n = 4 x 10¹⁸ photons

Explanation:

First, we will calculate the energy of one photon in the radiation:

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

where,

E = Energy of one photon = ?

h = Plank's Constant = 6.625 x 10⁻³⁴ J.s

c = speed of light = 3 x 10⁸ m/s

λ = wavelength of radiation = 567 nm = 5.67 x 10⁻⁷ m

Therefore,

E = \frac{(6.625\ x\ 10^{-34}\ J.s)(3\ x\ 10^8\ m/s)}{5.67\ x\ 10^{-7}\ m}

E = 3.505 x 10⁻¹⁹ J

Now, the number of photons to make up the total energy can be calculated as follows:

Total\ Energy = nE\\1.4\ J = n(3.505\ x\ 10^{-19}\ J)\\n = \frac{1.4\ J}{3.505\ x\ 10^{-19}\ J}\\

<u>n = 4 x 10¹⁸ photons</u>

8 0
3 years ago
How does sound travel through a medium PLZ HURRY WILL GIVE BRAINLY
ZanzabumX [31]

sound travels through a medium in mechanical waves a mechanical wave is a disturbance that moves and transports energy from one place to another through a medium

5 0
3 years ago
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An object traveling around another object in space is in
drek231 [11]
It is orbiting the object.
8 0
4 years ago
Gravity affects atmospheric pressure true or false
Burka [1]

Answer:

false

Explanation:

5 0
3 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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