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Alex
3 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]3 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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A particle, whose acceleration is constant, is moving in the negative x direction at a speed of 4.91 m/s, and 12.9 s later the p
Zinaida [17]

Answer:

The particle’s velocity is -16.9 m/s.

Explanation:

Given that,

Initial velocity of particle in negative x direction= 4.91 m/s

Time = 12.9 s

Final velocity of particle in positive x direction= 7.12 m/s

Before 12.4 sec,

Velocity of particle in negative x direction= 5.32 m/s

We need to calculate the acceleration

Using equation of motion

v = u+at

a=\dfrac{v-u}{t}

Where, v = final velocity

u = initial velocity

t = time

Put the value into the equation

a=\dfrac{7.12-(-4.91)}{12.9}

a=0.933\ m/s^2

We need to calculate the initial speed of the particle

Using equation of motion again

v=u+at

u=v-at

Put the value into the formula

u=-5.321-0.933\times12.4

u=-16.9\ m/s

Hence, The particle’s velocity is -16.9 m/s.

4 0
3 years ago
What kind of light results when the total spectrum of refracted light is recombined? violet green white red?
ElenaW [278]
Hello
The final light will be white. In fact, each color of the visible spectrum is an electromagnetic wave with its own specific frequency and wavelength. White, instead, does not have a specific frequency: it is the sum of all the different wavelengths of the visible spectrum. Therefore, when recombining the spectrum of the refracted light all the different frequencies recombine together, and their sum gives white light.

(edited)
4 0
3 years ago
An electron and a proton each have a thermal kinetic energy of 3kBT/2. Calculate the de Broglie wavelength of each particle at a
S_A_V [24]

Answer:

Given:

Thermal Kinetic Energy of an electron, KE_{t} = \frac{3}{2}k_{b}T

k_{b} = 1.38\times 10^{- 23} J/k = Boltzmann's constant

Temperature, T = 1800 K

Solution:

Now, to calculate the de-Broglie wavelength of the electron, \lambda_{e}:

\lambda_{e} = \frac{h}{p_{e}}

\lambda_{e} = \frac{h}{m_{e}{v_{e}}              (1)

where

h = Planck's constant = 6.626\times 10^{- 34}m^{2}kg/s

p_{e} = momentum of an electron

v_{e} = velocity of an electron

m_{e} = 9.1\times 10_{- 31} kg = mass of electon

Now,

Kinetic energy of an electron = thermal kinetic energy

\frac{1}{2}m_{e}v_{e}^{2} = \frac{3}{2}k_{b}T

}v_{e} = \sqrt{2\frac{\frac{3}{2}k_{b}T}{m_{e}}}

}v_{e} = \sqrt{\frac{3\times 1.38\times 10^{- 23}\times 1800}{9.1\times 10_{- 31}}}

v_{e} = 2.86\times 10^{5} m/s                    (2)

Using eqn (2) in (1):

\lambda_{e} = \frac{6.626\times 10^{- 34}}{9.1\times 10_{- 31}\times 2.86\times 10^{5}} = 2.55 nm

Now, to calculate the de-Broglie wavelength of proton, \lambda_{e}:

\lambda_{p} = \frac{h}{p_{p}}

\lambda_{p} = \frac{h}{m_{p}{v_{p}}                             (3)

where

m_{p} = 1.6726\times 10_{- 27} kg = mass of proton

v_{p} = velocity of an proton

Now,

Kinetic energy of a proton = thermal kinetic energy

\frac{1}{2}m_{p}v_{p}^{2} = \frac{3}{2}k_{b}T

}v_{p} = \sqrt{2\frac{\frac{3}{2}k_{b}T}{m_{p}}}

}v_{p} = \sqrt{\frac{3\times 1.38\times 10^{- 23}\times 1800}{1.6726\times 10_{- 27}}}

v_{p} = 6.674\times 10^{3} m/s                               (4)                    

Using eqn (4) in (3):

\lambda_{p} = \frac{6.626\times 10^{- 34}}{1.6726\times 10_{- 27}\times 6.674\times 10^{3}} = 5.94\times 10^{- 11} m = 0.0594 nm

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3 years ago
Question 15 of 25
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Combustion reaction
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2 years ago
Waves combine to produce a smaller or zero-amplitude wave in a process called
vampirchik [111]
ANSWER: Destructive Interference

-This is the exact definition of the question you have provided, look this term up if you do not believe lol.

Hope this Helps!
6 0
3 years ago
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