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Bad White [126]
4 years ago
14

A light source simultaneously emits light of two wavelengths, 480 nm and 560 nm, respectively. The source is used in a double-sl

it interference experiment where the slit spacing is a 0.040 mm, and the distance between double slits and the screen is 1.2 m. What is the separation between the second-order bright fringes of the two wavelengths as they appear on the screen
Physics
1 answer:
Assoli18 [71]4 years ago
6 0

Answer:

0.48 cm

Explanation:

given data

wavelength = 480 nm

wavelength = 560 nm

slit spacing = 0.040 mm

distance between double slits and the screen  = 1.2 m

solution

we know that  (1 nm= 10^{-9} m)

we wil take here equation of equations of interference that is

ym = R × (m λ)/d    ..........................1

here m = 2 R  i.e distance of screen and slit

 so put here value and we get

separation between the second-order bright fringes = 0.48 cm

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A sign is attached to the side of a building as seen below. The sign is attached to a steel rod. A wire connects to the end of t
Kruka [31]

Answer:

500N

Explanation:

50kgx10N/kg

= 500N

3 0
3 years ago
If fuel consumption is 80 pounds per hour and groundspeed is 180 knots, how much fuel is required for an airplane to travel 477
lisov135 [29]

Answer:

212 pounds

Explanation:

477 nm / 180 nm/hr     *   80 #/hr = 212 #

5 0
2 years ago
A car moves at a constant velocity of 30 m/s and has 3.6 × 105 J of kinetic energy. The driver applies the brakes and the car st
LekaFEV [45]

The force needed to the stop the car is -3.79 N.

Explanation:

The force required to stop the car should have equal magnitude as the force required to move the car but in opposite direction. This is in accordance with the Newton's third law of motion. Since, in the present problem, we know the kinetic energy and velocity of the moving car, we can determine the mass of the car from these two parameters.

So, here v = 30 m/s and k.E. = 3.6 × 10⁵ J, then mass will be

K.E = \frac{1}{2} * m*v^{2}  \\\\m = \frac{2*KE}{v^{2} } = \frac{2*3.6*10^{5} }{30*30}=800 kg

Now, we know that the work done by the brake to stop the car will be equal to the product of force to stop the car with the distance travelled by the car on applying the brake.Here it is said that the car travels 95 m after the brake has been applied. So with the help of work energy theorem,

Work done = Final kinetic energy - Initial kinetic energy

Work done = Force × Displacement

So, Force × Displacement = Final kinetic energy - Initial Kinetic energy.

Force * 95 = 0-3.6*10^{5}\\ \\Force =\frac{-3.6*10^{5} }{95}=-3.79 N

Thus, the force needed to the stop the car is -3.79 N.

5 0
3 years ago
I don’t understand the 4th one please help someone.
Gnoma [55]

Answer:

From point A to point D is 20

The final displacement is 32

Explanation:

AB=8

BC=8

CD=4

DE=8

EF=4

DE= 8 because the object moves 4 meters in each direction.

AB+BC+CD=20(A to D)

AB+BC+CD+DE+EF=32(Final displacement)

7 0
3 years ago
If the rods with diameters and lengths listed below are made of the same material, which will undergo the largest percentage len
seraphim [82]

Answer:

The highest percentage of change corresponds to the thinnest rod, the correct answer is a

Explanation:

For this exercise we are asked to change the length of the bar by the action of a force applied along its length, in this case we focus on the expression of longitudinal elasticity

               F / A = Y ΔL/L

where F / A is the force per unit length, ΔL / L is the fraction of the change in length, and Y is Young's modulus.

In this case the bars are made of the same material by which Young's modulus is the same for all

              ΔL / L = (F / A) / Y

the area of ​​the bar is the area of ​​a circle

               A = π r² = π d² / 4

               A = π / 4 d²

we substitute

              ΔL / L = (F / Y) 4 /πd²

changing length

               ΔL = (F / Y 4 /π) L / d²

The amount between paracentesis are all constant in this exercise, let's look for the longitudinal change

a) values ​​given d and 3L

               ΔL = cte 3L / d²

               ΔL = cte L /d²  3

To find the percentage, we must divide the change in magnitude by its value and multiply by 100.

                ΔL/L % = [(F /Y  4/π 1/d²) 3L ] / 3L 100

                ΔL/L  % = cte 100%

 

b) 3d and L value, we repeat the same process as in part a

               ΔL = cte L / 9d²

               ΔL = cte L / d² 1/9

               ΔL / L% = cte 100/9

               ΔL / L% = cte 11%

   

c) 2d and 2L value

               ΔL = (cte L / d ½ )/ 2L

               ΔL/L% = cte 100/4

               ΔL/L% = cte 25%

d) value 4d and L

               ΔL = cte L / d² 1/16

                ΔL/L % = cte 100/16

                ΔL/L % = cte 6.25%

   

The highest percentage of change corresponds to the thinnest rod, the correct answer is a

5 0
3 years ago
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