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anastassius
2 months ago
12

Narrow, bright fringes are observed on a screen behind a diffraction grating. The entire experiment is then immersed in water. D

o the fringes on the screen get closer together, get farther apart, remain the same, or disappear? Explain.
Physics
1 answer:
Ostrovityanka [3.2K]2 months ago
8 0

Response:

n (a sin θ) = m λ₀

Since n > 1, this indicates that the fringes separate further apart

Clarification:

In a diffraction experiment, the equation for constructive interference fringes is provided by

a sin θ = m λ₀

It is presumed that the air has been evacuated from the experiment, setting n = 1

When this experiment is conducted in water, the wavelength alters

λₙ = λ₀ / n

for achieving constructive interference

a sin θ = m λₙ

we replace

a sin θ = m λ / n

n (a sin θ) = m λ₀

Given that water's refractive index is n = 1.33, the distance between the fringes increases due to n > 1, causing the fringes to move apart

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A 450g mass on a spring is oscillating at 1.2Hz. The totalenergy of the oscillation is 0.51J. What is the amplitude.
Sav [3153]

Response:

A=0.199

Clarification:

We know that  

Mass of spring=m=450 g==\frac{450}{1000}=0.45 kg

Where 1 kg=1000 g

Frequency of oscillation=

\nu=1.2Hz

Energy for oscillation is 0.51 J

To determine the amplitude of oscillations.

Energy for oscillator=E=\frac{1}{2}m\omega^2A^2

Where \omega=2\pi\nu=Angular frequency

A=Amplitude

\pi=\frac{22}{7}

Using the formula

0.51=\frac{1}{2}\times 0.45(2\times \frac{22}{7}\times 1.2)^2A^2

A^2=\frac{2\times 0.51}{0.45\times (2\times \frac{22}{7}\times 1.2)^2}=0.0398

A=\sqrt{0.0398}=0.199

Therefore, the amplitude of oscillation=A=0.199

4 0
2 months ago
During a snowball fight two balls with masses of 0.4 and 0.6 kg, respectively, are thrown in such a manner that they meet head-o
Yuliya22 [3333]

Answer:

The snowball's speed after the impact is 3 m/s

Explanation:

Given the following:

mass of each ball

m₁ = 0.4 Kg

m₂ = 0.6 Kg

initial speed of both balls = v₁ = 15 m/s

Speed of 1 Kg mass post-collision =?

Applying conservation of momentum

m₁ v₁ - m₂ v₁ = (m₁+m₂) V

A negative velocity indicates that the second ball moves in the opposite direction.

0.4 x 15 - 0.6 x 15 = (1) V

Therefore,

V = - 3 m/s

Consequently,

The snowball's speed following the collision is 3 m/s

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2 months ago
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A 5-kg concrete block is lowered with a downward acceleration of 2.8 m/s2 by means of a rope. The force of the block on the Eart
Maru [3345]

The moment the body impacts the ground, two types of Forces are produced: the gravitational pull and the Normal Force. This aligns with Newton's third law, indicating that every action has an equal and opposite reaction. If the downward force of gravity is directed toward the earth, the reactionary force from the block acts upwards, equivalent to its weight:

F = mg

Where,

m = mass

g = gravitational acceleration

F = 5*9.8

F = 49N

Consequently, the answer is E.

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