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andrey2020 [161]
2 years ago
12

Two identical speakers are set some distance apart in a large open field. Both are producing sound, in unison, with a wavelength

of 4 m. An observer wandering around the field notices that the sound is surprisingly loud at certain locations and unusually quiet at others. The observer concludes that this is the effect of interference between the two sources of sound waves. For each location, indicate which sort of interference would occur at that point.
Physics
1 answer:
Marrrta [24]2 years ago
6 0

Answer:

Constructive interference occurs at the surprisingly loud locations and destructive interference occurs at the unusually quiet locations.

Explanation:

This is because, constructive interference tends to combine the effects of the wave when they are in phase (that is, moving in the same direction), which thus amplifies the effect and produces surprisingly loud sounds at those locations, while destructive interference occurs when the waves are out of phase with each other(that is, move in opposite directions) and thus, their effects tend to cancel out thus producing locations of unusually quiet sounds.

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The correct answer to the question is : A) The velocity of the cart after it hits the wall.

EXPLANATION:

Before answering this question, first we have to understand impulse.

Impulse of a body is defined as the change in momentum or the product of force with time.

Mathematically impulse = m ( v- u ).

Here, v is the final momentum and u is the initial momentum.

Hence, we need the velocity of the cart after it hits the wall in order to calculate the impulse of the lab cart.

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Answer:

As we need to use a nested loop in our function,hence push $ra

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What provides the force necessary to start a building or bridge oscillating?
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3 years ago
A toy car having mass m = 1.10 kg collides inelastically with a toy train of mass M = 3.55 kg. Before the collision, the toy tra
kkurt [141]

Answer:

V_{ft}= 317 cm/s

ΔK = 2.45 J

Explanation:

a) Using the law of the conservation of the linear momentum:

P_i = P_f

Where:

P_i=M_cV_{ic} + M_tV_{it}

P_f = M_cV_{fc} + M_tV_{ft}

Now:

M_cV_{ic} + M_tV_{it} = M_cV_{fc} + M_tV_{ft}

Where M_c is the mass of the car, V_{ic} is the initial velocity of the car, M_t is the mass of train, V_{fc} is the final velocity of the car and V_{ft} is the final velocity of the train.

Replacing data:

(1.1 kg)(4.95 m/s) + (3.55 kg)(2.2 m/s) = (1.1 kg)(1.8 m/s) + (3.55 kg)V_{ft}

Solving for V_{ft}:

V_{ft}= 3.17 m/s

Changed to cm/s, we get:

V_{ft}= 3.17*100 = 317 cm/s

b) The kinetic energy K is calculated as:

K = \frac{1}{2}MV^2

where M is the mass and V is the velocity.

So, the initial K is:

K_i = \frac{1}{2}M_cV_{ic}^2+\frac{1}{2}M_tV_{it}^2

K_i = \frac{1}{2}(1.1)(4.95)^2+\frac{1}{2}(3.55)(2.2)^2

K_i = 22.06 J

And the final K is:

K_f = \frac{1}{2}M_cV_{fc}^2+\frac{1}{2}M_tV_{ft}^2

K_f = \frac{1}{2}(1.1)(1.8)^2+\frac{1}{2}(3.55)(3.17)^2

K_f = \frac{1}{2}(1.1)(1.8)^2+\frac{1}{2}(3.55)(3.17)^2

K_f = 19.61 J

Finally, the change in the total kinetic energy is:

ΔK = Kf - Ki = 22.06 - 19.61 = 2.45 J

4 0
3 years ago
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