Choose the CORRECT statements. The superposition of two waves.

I. refers to the effects of waves at great distances.

Il. refers to how displacements of the two waves add together.

Ill. results into constructive interference and destructive interference

IV. results into minimum amplitude when crest meets trough.

V. results into destructive interference and the waves stop propagating.

A. I and II
B. II and III
C. I, II and III
D. II, III and IV
E. III, IV and V
F. II, III, IV and V​

Answers

Answer 1

Answer:

A

Explanation:

I guess not that much confidential!


Related Questions

A boy pushes his little brother on a sled. The sled accelerates from rest to (4 m/s). If the combined mass of his brother and the sled is (40.0 kg) and (20 W) of power is developéd, how long time does boy push the sled?


16s

300s

15s

23s​

Answers

300 because the mass and weight

The boy pushed the sled for 16 seconds.

We have a boy who pushes his little brother on a sled.

We have to determine for how long time does boy push the sled.

State Work - Energy Theorem.

The Work - Energy theorem states that the work done by the sum of all forces acting on a particle equals the change in the kinetic energy of the particle.

According to the question -

The sled is initially at rest → initial velocity (u) = 0.

Final velocity (v) = 4 m/s

Mass of boy and sled (M) = 40 kg

Power developed (P) = 20 W = 20 Joules/sec

According to work - energy theorem -

Work done (W) = Δ E(K) = E(f) - E(i)

Therefore -

W = ([tex]\frac{1}{2} \times 40 \times 4 \times 4 - \frac{1}{2}[/tex] x 40 x 0) = 320 Joule

Now, Power is defined as the rate of doing work -

P = [tex]\frac{dW}{dt}[/tex] = [tex]\frac{W}{t}[/tex]

20 = [tex]\frac{320}{t}[/tex]

t = 16 seconds

Hence, the boy pushed the sled for 16 seconds.

To solve more questions on Work, Energy and Power, visit the link below -

https://brainly.com/question/208670

#SPJ2

For an object with a given mass on Earth, calculate the weight of the object with the mass equal in magnitude to the number representing the day given in part 3 in kilograms using the formula F=W=mg. On the surface of the Earth g=9.8m/s^2

Answers

Answer: The weight of the object is 29.4 N

Explanation:

To calculate the weight of the object, we use the equation:

[tex]W=m\times g[/tex]

where,

m = mass of the object = 3 kg

g = acceleration due to gravity = [tex]9.8m/s^2[/tex]

Putting values in above equation, we get:

[tex]W=3kg\times 9.8m/s^2\\\\W=29.4N[/tex]

Hence, the weight of the object is 29.4 N

How fast much an 816kg Volkswagen travel to have the same momentum as (a) a 2650kg Cadillac going 16.0 km/h? (b) a 9080-kg truck also going 16.0 km/hr?

Answers

Answer:

Explanation:

From the given information:

the car's momentum = momentum of the truck

(a) 816 kg × v = 2650 kg × 16.0 km/h

v = (2650 kg × 16.0 km/h) /  816 kg

v = 51.96 km/hr

(b) 816 kg × v = 9080 kg × 16.0 km/h

v = (9080 kg × 16.0 km/h) /  816 kg

v = 178.04 km/hr

A man pulls his dog (m=20kg) on a sled with a force of 100N at a 60° angle from the horizontal. What is the horizontal component of the force?

A) 100N

B) 196N

C) 50N

D) 86N

show your work please

Answers

Answer:

the horizontal component of the force is 50 N

Explanation:

Given;

force applied by the man, F = 100 N

angle of inclination of the force, θ = 60⁰

mass of the dog, m = 20 kg

The horizontal component of the force is calculated as;

[tex]F_x = F\times cos(\theta)\\\\F_x = 100 \ N \times cos(60^0)\\\\F_x = 100\ N \times 0.5\\\\F_x = 50 \ N[/tex]

Therefore, the horizontal component of the force is 50 N

The wavelength of visible light range of 400 to 750mm .what is the corresponding range of photon energies for visible light

Answers

Answer:

The range of the photon energies is between:

2.652 x 10⁻²⁵ J    to    4.973 x 10⁻²⁵ J

Explanation:

The energy of a photon is calculated using the following equation;

E = hf

where;

h is Planck's constant = 6.63 x 10⁻³⁴ Js

f is frequency of the photon

[tex]E = h \frac{c}{\lambda} \\\\where;\\\\\lambda \ is \ the \ wavelength\\\\c \ is \ the \ speed \ of \ light \ = 3\times 10^8 \ m/s\\\\When \ \lambda = 400 \ mm = 400 \ \times 10^{-3} \ m\\\\E = \frac{(6.63 \times 10^{-34})(3\times 10^8)}{400 \times 10^{-3}} \\\\E = 4.973 \times 10^{-25} \ J[/tex]

[tex]When \ \lambda = 750 \ mm = 750 \ \times 10^{-3} \ m\\\\E = \frac{(6.63 \times 10^{-34})(3\times 10^8)}{750 \times 10^{-3}} \\\\E = 2.652 \times 10^{-25} \ J[/tex]

The range of the photon energies is between:

2.652 x 10⁻²⁵ J    to    4.973 x 10⁻²⁵ J

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