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svlad2 [7]
2 years ago
4

Fizeau’s method for measuring the speed of light using a rotating toothed wheel. The speed of rotation of the wheel controls wha

t an observer sees. For example, if the light passing the opening at point A should return at the instant that tooth B had rotated into position to cover the return path, the light would not reach the observer. At a faster rate of rotation, the opening at point C could move into position to allow the reflected beam to reach the observer. Toothed Wheel Mirror A B C d Calculate the minimum angular speed of the wheel for light that passed through opening A to return through opening C to reach the observer. In an experiment to measure the speed of light using the apparatus of Fizeau, the distance between the toothed wheel and mirror was 10.72 km and the wheel had 671 notches. The experimentally determined value of c was 2.908 × 108 m/s . Answer in units of rad/s.
Engineering
1 answer:
Mashutka [201]2 years ago
8 0

Answer:

The minimum angular speed, w is 126.94 rad/s

Explanation:

Given:

C = 2.908×10⁸ m/s

d = 10.72 km ⇒ 10.72×10³ m

There are 671 notches

⇒ Δθ = \frac{2\pi}{671} ==> 9.359×10⁻³ rad

C = 2d / Δt    ⇒    Δt = 2d/C

⇒

w = Δθ / Δt = CΔθ / 2d

substitute for given parameters

w = [2.908×10⁸×9.359×10⁻³] / [2×10.72×10³]

   = 27.215972×10⁵ / 21.44×10³

   = 1.2694×10²

w ⇒ 126.94 rad/s

You might be interested in
Find the largest number. The process of finding the maximum value (i.e., the largest of a group of values) is used frequently in
salantis [7]

Answer:

See Explanation

Explanation:

Required

- Pseudocode to determine the largest of 10 numbers

- C# program to determine the largest of 10 numbers

The pseudocode and program makes use of a 1 dimensional array to accept input for the 10 numbers;

The largest of the 10 numbers is then saved in variable Largest and printed afterwards.

Pseudocode (Number lines are used for indentation to illustrate the program flow)

1. Start:

2. Declare Number as 1 dimensional array of 10 integers

3. Initialize: counter = 0

4. Do:

4.1 Display “Enter Number ”+(counter + 1)

4.2 Accept input for Number[counter]

4.3 While counter < 10

5. Initialize: Largest = Number[0]

6. Loop: i = 0 to 10

6.1 if Largest < Number[i] Then

6.2 Largest = Number[i]

6.3 End Loop:

7. Display “The largest input is “+Largest

8. Stop

C# Program (Console)

Comments are used for explanatory purpose

using System;

namespace ConsoleApplication1

{

   class Program

   {

       static void Main(string[] args)

       {

           int[] Number = new int[10];  // Declare array of 10 elements

           //Accept Input

           int counter = 0;

           while(counter<10)

           {

               Console.WriteLine("Enter Number " + (counter + 1)+": ");

               string var = Console.ReadLine();

               Number[counter] = Convert.ToInt32(var);

               counter++;                  

           }

           //Initialize largest to first element of the array

           int Largest = Number[0];

           //Determine Largest

           for(int i=0;i<10;i++)

           {

               if(Largest < Number[i])

               {

                   Largest = Number[i];

               }

           }

           //Print Largest

           Console.WriteLine("The largest input is "+ Largest);

           Console.ReadLine();

       }

   }

}

8 0
2 years ago
air at 600 kPa, 330 K enters a well-insulated, horizontal pipe having a diameter of 1.2 cm and exits at 120 kPa, 300 K. Applying
notsponge [240]

Answer:

a) 251.31 m/s and 55.29 m/s

b) The mass flow rate is 0.0396 kg/s

c) The rate of entropy production is 0.0144 kW/K

Explanation:

a) The steady state is:

mi = mo

\frac{A_{i}V_{i}  }{v_{i} } =\frac{A_{o}V_{o}  }{v_{o} } \\V_{i} =V_{o}(\frac{v_{i}}{v_{o}} )=V_{o}(\frac{T_{i}P_{o} }{T_{o}P_{i} } )=V_{o}(\frac{330*120}{300*600}) =0.22V_{o}

The energy balance is:

h_{i}+\frac{V_{i}^{2} }{2} =h_{2}+\frac{V_{o}^{2} }{2} \\h_{i}-h_{o}+(\frac{V_{i}^{2}-V_{o}^{2}  }{2} )=0\\V_{o}=\sqrt{\frac{2(h_{i}-h_{o})}{0.9516} } =\sqrt{\frac{2*(330.24-300.19)x10^{3} }{0.9516} } =251.31m/s

Vi = 0.22 * 251.31 = 55.29 m/s

b) The mass flow rate is:

m=\frac{A_{o}V_{o}}{v_{o}} =\frac{\pi d^{2}V_{o}P_{o} }{4RT_{o}} =\frac{\pi *(0.012^{2})*251.31*120x10^{3}  }{4*287*300} =0.0396kg/s

c) The entropy produced is equal to:

\frac{Q}{T} +S_{gen} =m(s_{2} -s_{1} )\\0+S_{gen} =m(s_{2} -s_{1} )\\S_{gen} =m(s_{2} -s_{1} )\\S_{gen}=0.0396*(c_{p} ln\frac{T_{o}}{T_{i}} -Rln\frac{P_{o}}{P_{i}} )=0.0396*(1.004ln\frac{300}{330} -0.287ln\frac{120}{600} )=0.0144kW/K

3 0
2 years ago
Evan notices a small fire in his workplace. Since the fire is small and the atmosphere is not smoky he decides to fight the fire
Norma-Jean [14]

Answer:

not calling the firemean

Explanation:

7 0
2 years ago
Explain why failure of this garden hose occurred near its end and why the tear occurred along its length. Use numerical values t
alukav5142 [94]

Answer:

  • hoop stress
  • longitudinal stress
  • material used

all this could led to the failure of the garden hose and the tear along the length

Explanation:

For the flow of water to occur in any equipment, water has to flow from a high pressure to a low pressure. considering the pipe, water is flowing at a constant pressure of 30 psi inside the pipe which is assumed to be higher than the allowable operating pressure of the pipe. but the greatest change in pressure will occur at the end of the hose because at that point the water is trying to leave the hose into the atmosphere, therefore the great change in pressure along the length of the hose closest to the end of the hose will cause a tear there. also the other factors that might lead to the failure of the garden hose includes :

hoop stress ( which acts along the circumference of the pipe):

αh = \frac{PD}{2T}     EQUATION 1

and Longitudinal stress ( acting along the length of the pipe )

αl = \frac{PD}{4T}       EQUATION 2

where p = water pressure inside the hose

          d = diameter of hose, T = thickness of hose

we can as well attribute the failure of the hose to the material used in making the hose .

assume for a thin cylindrical pipe material used to be

\frac{D}{T} ≥  20

insert this value into equation 1

αh = \frac{20 *30}{2}  = 60/2 = 30 psi

the allowable hoop stress was developed by the material which could have also led to the failure of the garden hose

8 0
2 years ago
Determine the angular acceleration of the uniform disk if (a) the rotational inertia of the disk is ignored and (b) the inertia
lukranit [14]

Answer:

α = 7.848 rad/s^2  ... Without disk inertia

α = 6.278 rad/s^2  .... With disk inertia

Explanation:

Given:-

- The mass of the disk, M = 5 kg

- The right hanging mass, mb = 4 kg

- The left hanging mass, ma = 6 kg

- The radius of the disk, r = 0.25 m

Find:-

Determine the angular acceleration of the uniform disk without and with considering the inertia of disk

Solution:-

- Assuming the inertia of the disk is negligible. The two masses ( A & B )  are hung over the disk in a pulley system. The disk is supported by a fixed support with hinge at the center of the disk.

- We will make a Free body diagram for each end of the rope/string ties to the masses A and B.

- The tension in the left and right string is considered to be ( T ).

- Apply newton's second law of motion for mass A and mass B.

                      ma*g - T = ma*a

                      T - mb*g = mb*a

Where,

* The tangential linear acceleration ( a ) with which the system of two masses assumed to be particles move with combined constant acceleration.

- g: The gravitational acceleration constant = 9.81 m/s^2

- Sum the two equations for both masses A and B:

                      g* ( ma - mb ) = ( ma + mb )*a

                      a =  g* ( ma - mb ) / ( ma + mb )

                      a = 9.81* ( 6 - 4 ) / ( 6 + 4 ) = 9.81 * ( 2 / 10 )

                      a = 1.962 m/s^2  

- The rope/string moves with linear acceleration of ( a ) which rotates the disk counter-clockwise in the direction of massive object A.

- The linear acceleration always acts tangent to the disk at a distance radius ( r ).

- For no slip conditions, the linear acceleration can be equated to tangential acceleration ( at ). The correlation between linear-rotational kinematics is given below :

                     a = at = 1.962 m/s^2

                     at = r*α      

Where,

           α: The angular acceleration of the object ( disk )

                    α = at / r

                    α = 1.962 / 0.25

                    α = 7.848 rad/s^2                                

- Take moments about the pivot O of the disk. Apply rotational dynamics conditions:

             

                Sum of moments ∑M = Iα

                 ( Ta - Tb )*r = Iα

- The moment about the pivots are due to masses A and B.

 

               Ta: The force in string due to mass A

               Tb: The force in string due to mass B

                I: The moment of inertia of disk = 0.5*M*r^2

                   ( ma*a - mb*a )*r = 0.5*M*r^2*α

                   α = ( ma*a - mb*a ) / ( 0.5*M*r )

                   α = ( 6*1.962 - 4*1.962 ) / ( 0.5*5*0.25 )

                   α = ( 3.924 ) / ( 0.625 )

                   α = 6.278 rad/s^2

6 0
2 years ago
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