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Alex
2 years ago
6

If the Sun suddenly went dark, we would not know it until its light stopped arriving on Earth. How long would that be, in second

s, given that the Sun is 1.50 × 1011 m away?
Physics
1 answer:
Gre4nikov [31]2 years ago
5 0

Answer: 500 s

Explanation:

Speed v is defined as a relation between the distance d and time t:

v=\frac{d}{t}

Where:

v=3(10)^{8}m/s is the speed of light in vacuum

d=1.5(10)^{11}m is the distance between the Earth and Sun

t is the time it takes to the light to travel the distance d

Isolating t:

t=\frac{d}{v}

t=\frac{1.5(10)^{11}m}{3(10)^{8}m/s}

Finally:

t=500 s

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The land form is formed by lava flows of low viscosity

Hope this helps! :)

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3 years ago
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An atomic scientist was studying an atom and through experimentation found that the total negative charge of the area surroundin
Mars2501 [29]

Answer:

c  hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh

Explanation:

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4 0
3 years ago
When a 20-V emf is placed across two resistors in series, a current of 2.0 A is present in each of the resistors. When the same
ehidna [41]

Answer:

7.24 ohm

Explanation:

Let R1 and R2 are resistance of two resistors.

Emf=E=20 V

Current,I=2 A

Current,I'=10 A

We have to find the magnitude of the greater of the two resistances.

In series

R=R_1+R_2

V=IR

By using the formula

20=2(R_1+R_2)

R_1+R_2=\frac{20}{2}=10...(1)

In parallel

\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}

\frac{1}{R}=\frac{R_2+R_1}{R_1R_2}

R=\frac{R_1R_2}{R_1+R_2}

20=10(\frac{R_1R_2}{R_1+R_2}

2=\frac{R_1R_2}{10}

R_1R_2=20

R_2=\frac{20}{R_1}

Substitute the value

\frac{20}{R_1}+R_1=10

R^2_1+20=10R_1

R^2_1-10R_1+20=0

R_1=\frac{10\pm\sqrt{(-10)^2-4(20)}}{2}

By using quadratic formula

x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

R_1=\frac{10\pm 2\sqrt 5}{2}

R_1=\frac{10+2\sqrt 5}{2}=5+\sqrt 5=7.24 ohm

R_1=\frac{10-2\sqrt 5}{2}=2.76 ohm

Substitute the value

R_2=\frac{20}{7.24}=2.76 ohm

R_2=\frac{20}{2.76}=7.24 ohm

Hence, the magnitude of the greater of the two resistance=7.24 ohm

8 0
3 years ago
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Find the voltage change when: a. An electric field does 12 J of work on a 0.0001-C charge. b. The same electric field does 24 J
kondaur [170]

Explanation:

Given that,

(a) Work done by the electric field is 12 J on a 0.0001 C of charge. The electric potential is defined as the work done per unit charged particles. It is given by :

V=\dfrac{W}{q}

V=\dfrac{12}{0.0001}

V=12\times 10^4\ Volt

(b) Similarly, same electric field does 24 J of work on a 0.0002-C charge. The electric potential difference is given by :

V=\dfrac{W}{q}

V=\dfrac{24}{0.0002}

V=12\times 10^4\ Volt

Therefore, this is the required solution.

7 0
3 years ago
A LED light source contains a 0.5-Watts GaAs (Eg =1.43 eV) LED. Assuming that 0.12% of the electric energy is converted to emiss
Ray Of Light [21]

Answer:

Explanation:

energy emitted by  source per second  = .5 J

Eg = 1.43 eV .

Energy converted into radiation = .5 x .12 = .06 J

energy of one photon = 1.43 eV

= 1.43 x 1.6 x 10⁻¹⁹ J

= 2.288 x 10⁻¹⁹ J .

no of photons generated = .06 / 2.288 x 10⁻¹⁹

= 2.6223 x 10¹⁷

wavelength of photon λ = 1275 / 1.43 nm

= 891.6 nm .

momentum of photon = h / λ  ;  h is plank's constant

= 6.6 x 10⁻³⁴ / 891.6 x 10⁻⁹

= .0074 x 10⁻²⁵ J.s

Total momentum of all the photons generated

= .0074 x 10⁻²⁵  x 2.6223 x 10¹⁷

= .0194 x 10⁻⁸ Js

b ) spectral width in terms of wavelength = 30 nm

frequency width = ?

n = c / λ  , n is frequency , c is velocity of light and λ is wavelength

differentiating both sides

dn = c x dλ / λ²

given dλ = 30 nm

λ = 891.6 nm

dn = 3 x 10⁸ x 30 x 10⁻⁹ / ( 891.6  x 10⁻⁹ )²

= 11.3 x 10¹² Hz .

c )

10 nW = 10  x 10⁻⁹ W

= 10⁻⁸ W .

energy of 50 dB

50 dB = 5 B

I / I₀ = 10⁵   ;   decibel scale is logarithmic , I is energy of sound having dB = 50 and  I₀ = 10⁻¹² W /s

I = I₀ x 10⁵

= 10⁻¹² x 10⁵

= 10⁻⁷ W

= 10 x 10⁻⁸ W

power required

= 10⁻⁸ + 10 x 10⁻⁸ W

= 11  x 10⁻⁸ W.

5 0
3 years ago
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