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timofeeve [1]
3 years ago
13

Explain in your own words how the Doppler Effect is also applicable in our study of light.

Physics
1 answer:
arlik [135]3 years ago
8 0

well in my own words, i'd saw the the doppler effect is similar to light because sound has a speed, and light does too.

so my theory is if you go fast enough everything would just become black, or maybe white? idk its hard to explain

but what my point is, is taht the doppler effect works in the same way, like if a car is moving towards you the sound is being emitted from the car and being pushed by the speed of the car making it have a much higher pitch, when the car is going away however it drops to a lower pitch due the the sound waves being DRAGGED by the car.

there hoped this helped I guess

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For heat transfer purposes, an egg can be considered to be a 5.5-cm-diameter sphere having the properties of water. An egg that
Harrizon [31]

Answer:

The time taken is   t = 40007 sec  

Explanation:

From the question we are told that

   The diameter of the egg is d_e = 5.5cm  = \frac{5.5}{100} = 5.5*10^{-2}m

    The initial temperature of egg the T_e = 4.3^{o}C

     The temperature of the boiling water T_b = 100^oC

    The heat transfer coefficient is  H  = 800 W/m^2 \cdot K

    The  final temperature is T_e_f = 74^oC

     The  thermal  conductivity of water is k = 0.607 W/m^oC

     The diffusivity of the egg \alpha = 0.146 * 10^{-6} m^2 /s

Using one term approximation

We have the

            \frac{T_e_f - T_b}{T_e - T_b}  = Ae^{-\lambda ^2 \tau}

The radius is  r = \frac{5.5*10^ {-2}}{2} =2.75*10^{-2}m     Note that this radius is approximation to that of  a real egg

    Now we need to obtain the Biot number which help indicate the value of A  \ and \ \lambda to use in the above equation

     The Biot number is mathematically represented as

               Bi = \frac{H r}{k}

Substituting values  

               Bi = \frac{800 * 2.75 *10^{-2}}{0.607}

                    = 36.24

So for this value  which greater than 0.1 the  coefficient \lambda_1 \ and  \ A_1 is  

        \lambda = 3.06632

        A = 1.9942

Substituting this into equation 1 we have

          \frac{74- 100}{4.3 - 100} = 1.9942 e^{-(3.0632^2) \tau}

          0.2717= 1.9942 e^{-(3.0632^2) \tau}

          0.2717= 1.9942 e^{-9.383 \tau}

           0.13624 =  e^{-9.383 \tau}

Taking natural log of both sides

           -1.993 =  -9.383\  \tau

          \tau =  0.2124

    The time required for the egg to be cooked is  mathematically represented as

          t = \frac{\tau r^2}{\alpha }

substituting value  is  

         = \frac{0.2124 * 2.75 *10^{-2}}{0.146 *10^{-6}}

         t = 40007 sec  

8 0
3 years ago
A river flows at a velocity of 3 km/h relative to the riverbank. A boat moves downstream at a velocity of 15 km/h relative to th
Alexus [3.1K]
15+3=18km/hour
Think about it like this. The boat is going 15 faster than the river, and the river is going 3 faster than the bank, so the boat is going 18 faster than the river bank
3 0
3 years ago
Four forces act on a hot-air balloon, as shown
dolphi86 [110]

The magnitude of the resultant force on the balloon is 374.13 N.

The given forces from the image;

  • <em>Upward force = 514 N</em>
  • <em>Downward force = 267 N</em>
  • <em>Eastward force = 678 N</em>
  • <em>Westward force = 397 N</em>

The net vertical force on the balloon is calculated as follows;

F_y = 514 \ N \ \ - \ \ 267 \ N\\\\F_y = 247 \ N

The net horizontal force on the balloon is calculated as follows;

F_x = 678 \ N \ - \ 397 \ N\\\\F_x = 281 \ N

The magnitude of the resultant force on the balloon is calculated as follows;

F = \sqrt{F_y^2 + F_x^2} \\\\F = \sqrt{(247)^2 + (281)^2} \\\\F= 374.13 \ N

Thus, the magnitude of the resultant force on the balloon is 374.13 N.

Learn more here:brainly.com/question/4404327

5 0
2 years ago
The angular speed of an automobile engine is increased at a constant rate from 1120 rev/min to 2560 rev/min in 13.8 s. (a) What
Neko [114]

Complete Question

The angular speed of an automobile engine is increased at a constant rate from 1120 rev/min to 2560 rev/min in 13.8 s.

(a) What is its angular acceleration in revolutions per minute-squared

(b) How many revolutions does the engine make during this 20 s interval?

rev

Answer:

a

 \alpha = 6261 \  rev/minutes^2

b

 \theta  = 613 \ revolutions

Explanation:

From the question we are told that

   The initial  angular speed is w_i =  1120 \ rev/minutes

    The angular speed after t = 13.8 s = \frac{13.8}{60 }  = 0.23 \ minutes  is w_f = 2560 \ rev/minutes

    The time for revolution considered is t_r =  20 \ s  =  \frac{20}{60} = 0.333 \  minutes  

 Generally the angular acceleration is mathematically represented as

         \alpha = \frac{w_f - w_i }{t}

=>      \alpha = \frac{2560  - 1120 }{0.23}  

=>      \alpha = 6261 \  rev/minutes^2

Generally the number of revolution made is t_r =  20 \ s  =  \frac{20}{60} = 0.333 \  minutes  is mathematically represented as

           \theta  =  \frac{1}{2}  * (w_i + w_f)*  t

=>      \theta  =  \frac{1}{2}  * (1120+ 2560 )*  0.333

=>      \theta  = 613 \ revolutions

5 0
3 years ago
A 2kg object initially going 4m/s to the right is later going 8m/s. whats the change in velocity?
Nezavi [6.7K]

The change in velocity is +4 m/s to the right (or -4 m/s to the left).

The object's mass is irrelevant.

6 0
3 years ago
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