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svlad2 [7]
3 years ago
11

Suppose astronomers discover a radio message from a civilization whose planet orbits a star 35 light-years away. Their message e

ncourages us to send a radio answer, which we decide to do. Suppose our governing bodies take 2 years to decide whether and how to answer. When our answer arrives there, their governing bodies also take two of our years to frame an answer to us. How long after we get their first message can we hope to get their reply to ours? Enter your answer in years.
Physics
1 answer:
Grace [21]3 years ago
4 0

Answer:

The duration  is  T  =72 \  years /tex]Explanation:From the question we are told that     The  distance is  [tex]D  =  35 \ light-years = 35 *  9.46 *10^{15} = 3.311 *10^{17} \  m

  Generally the time it would take for the message to get the the other civilization is mathematically represented as

         t =  \frac{D}{c}

Here c  is the speed of light with the value  c =  3.08 *10^{8} \  m/s

=>     t =  \frac{3.311 *10^{17} }{3.08 *10^{8}}

=>     t =  1.075 *10^9 \ s

converting to years

           t =  1.075 *10^9 *  3.17 *10^{-8}

              t =  1.075 *10^9 *  3.17 *10^{-8}

            t =  34 \ years

Now the total time taken is mathematically represented as

      T  =  2*  t  +  2 + 2

=>   T  =  2* 34  +  2 + 2

=>   [tex]T  =72 \  years /tex]

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Light of wavelength 400 nm is incident on a single slit of width 15 microns. If a screen is placed 2.5 m from the slit. How far
olganol [36]

Answer:

0.0667 m

Explanation:

λ = wavelength of light = 400 nm = 400 x 10⁻⁹ m

D = screen distance = 2.5 m

d = slit width = 15 x 10⁻⁶ m

n = order = 1

θ = angle = ?

Using the equation

d Sinθ = n λ

(15 x 10⁻⁶) Sinθ = (1) (400 x 10⁻⁹)

Sinθ = 26.67 x 10⁻³

y = position of first minimum

Using the equation for small angles

tanθ = Sinθ = y/D

26.67 x 10⁻³ = y/2.5

y = 0.0667 m

5 0
3 years ago
A crane raises a crate with a mass of 150 kg to a height of 20 m. Given that
Virty [35]

Answer:

\boxed {\boxed {\sf 29,400 \ Joules}}

Explanation:

Gravitational potential energy is the energy an object possesses due to its position. It is the product of mass, height, and acceleration due to gravity.

E_P= m \times g \times h

The object has a mass of 150 kilograms and is raised to a height of 20 meters. Since this is on Earth, the acceleration due to gravity is 9.8 meters per square second.

  • m= 150 kg
  • g= 9.8 m/s²
  • h= 20 m

Substitute the values into the formula.

E_p= 150 \ kg \times 9.8 \ m/s^2 \times 20 \ m

Multiply the three numbers and their units together.

E_p=1470 \ kg*m/s^2 \times 20 m

E_p=29400 \ kg*m^2/s^2

Convert the units.

1 kilogram meter square per second squared (1 kg *m²/s²) is equal to 1 Joule (J). Our answer of 29,400 kg*m²/s² is equal to 29,400 Joules.

E_p= 29,400 \ J

The crate has <u>29,400 Joules</u> of potential energy.

7 0
3 years ago
Science help answer all
wlad13 [49]
Idk but i hope you figure it out :)
4 0
3 years ago
A boy slides a book across the floor using a force of five and over a distance of 2M what is the kinetic energy of the book afte
Alinara [238K]

Answer: 10Nm or 10J

Explanation:

Given the following :

Force (f) = 5

Distance (d) = 2m

Calculate the kinetic energy assuming no friction

Work done = force × distance

Work done = 5N × 2m = 10Nm

Recall :

Work done = ΔK.E ( change in kinetic energy)

Therefore, kinetic energy of the book after sliding = ΔK. E, which is equal to work done.

Hence, K. E of book after sliding is 10Nm

5 0
3 years ago
How much heat is required to convert 2.55g of water at 28 degrees c to steam?
pantera1 [17]
There are two different processes here:
1) we must add heat in order to bring the temperature of the water from 28^{\circ}C to 100^{\circ}C (the temperature at which the water evaporates)
2) other heat must be added to make the water evaporates

1) The heat needed for process 1) is
Q_1=m C_s \Delta T
where 
m=2.55 g is the water mass
C_s = 4.18 g/J^{\circ}C is the water specific heat
\Delta T=100^{\circ}C-28^{\circ}C=72^{\circ}C is the variation of temperature of the water
If we plug the numbers into the equation, we find
Q_1 = (2.55 g)(4.18 J/g^{\circ}C)(72^{\circ}C)=767.4 J

2) The heat needed for process 2) is
Q=m L_e
where 
m=2.55 g is the water mass
L_e = 2264.7 J/g is the latent heat of evaporation of water
If we plug the numbers into the equation, we find
Q_2=(2.55 g)(2264.7 J/g)=5775.0 J

So, the total heat needed for the whole process is
Q=Q_1+Q_2=767.4 J+5775.0 J=6542.4 J
4 0
3 years ago
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