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Anvisha [2.4K]
3 years ago
12

Two beams of coherent light travel different paths, arriving at point P. If the maximum destructive interference is to occur at

point P, what should be the path difference between the two waves?Two beams of coherent light travel different paths, arriving at point P. If the maximum destructive interference is to occur at point P, what should be the path difference between the two waves?The path difference between the two waves should be one and one-quarter of a wavelengths.The path difference between the two waves should be four wavelengths.The path difference between the two waves should be one-half of a wavelength.The path difference between the two waves should be one-quarter of a wavelength.The path difference between the two waves should be two wavelengths.The path difference between the two waves should be one wavelength.
Physics
1 answer:
Nimfa-mama [501]3 years ago
5 0

Answer:

The path difference between the two waves should be one-half of a wavelength

Explanation:

When two beams of coherent light travel different paths, arriving at point P. If the maximum destructive interference is to occur at point P , then the condition for it is that the path difference of two beams must be odd multiple of half wavelength. Symbolically

path difference = ( 2n+1 ) λ / 2

So path difference may be λ/2 , 3λ/ 2,  5λ/ 2 etc .

Hence right option is

The path difference between the two waves should be one-half of a wavelength.

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In an elastic collision, a 400-kg bumper car collides directly from behind with a second, identical bumper car that is traveling
kirza4 [7]

Answer:

v₁ = 3.5 m/s

v₂ = 6.4 m/s

Explanation:

We have the following data:

m₁ = mass of trailing car = 400 kg

m₂ = mass of leading car = 400 kg

u₁ = initial speed of trailing car = 6.4 m/s

u₂ = initial speed of leading car = 3.5 m/s

v₁ = final speed of trailing car = ?

v₂ = final speed of leading car = ?

The final speed of the leading car is given by the following formula:

v_2=\frac{2m_1}{m_1+m_2}u_1-\frac{m_1-m_2}{m_1+m_2}u_2\\\\v_2=\frac{(2)(400\ kg)}{400\ kg+400\ kg}(6.4\ m/s)-\frac{400\ kg-400\ kg}{400\ kg + 400\ kg}(3.5\ m/s)

<u>v₂ = 6.4 m/s</u>

The final speed of the leading car is given by the following formula:

v_1=\frac{m_1-m_2}{m_1+m_2}u_1+\frac{2m_2}{m_1+m_2}u_2\\\\v_1=\frac{400\ kg-400\ kg}{400\ kg + 400\ kg}(6.4\ m/s)+\frac{(2)(400\ kg)}{400\ kg+400\ kg}(3.5\ m/s)

<u>v₁ = 3.5 m/s</u>

4 0
3 years ago
A field measuring 12 meters by 16 meters is to have a brick paver walkway installed all around it, increasing the total area to
kolezko [41]

Answer:

1.5 m

Explanation:

Length. L = 12 m

Width, W = 16 m

Area, A = 12 x 16 = 192 m^2

Let the width of pavement be d.

The new length, L' = 12 + 2d

the new width, W' = 16 + 2d  

New Area, A' = L' x W' = (12 + 2d)(16 + 2d) = 192 + 56 d + 4d^2

Difference in area = A' - A

285 =  192 + 56 d + 4d^2 - 192

93 =  56 d + 4d^2

4d^2 + 56 d - 93 = 0

d = \frac{-56\pm \sqrt{56^{2}+4\times 4\times 93}}{8}

d=\frac{-56\pm 87.72}{8}\

d = 1.5 m

Thus, the width of the pavement is 1.5 m.

6 0
3 years ago
Convert 700kg into grams<br>​
Verdich [7]

Answer:

700,000g

Explanation:

1kg = 1000 Grams (or any kg x 1000)

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Answer:

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esto se debe a que transponemos la fórmula acelerada y obtenemos Distancia = velocidad × tiempo

también recuerda transponer los segundos a horas viendo que la velocidad es por hora

También tenga en cuenta que no hablo español, así que esto fue extremadamente difícil

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