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Mice21 [21]
3 years ago
12

Two in-phase loudspeakers are 3.0 m apart. they emit sound with a frequency of 490 hz. a microphone is placed half-way between t

he speakers and then moved along the line joining the two speakers until the first point of constructive interference is found. at what distance from that midpoint is that first point? the speed of sound in air is 343 m/s.
Physics
1 answer:
stira [4]3 years ago
3 0

Answer:

0.35 m

Explanation:

Constructive interference occurs when the difference in the distance between the two speakers is  

0, λ, 2λ,...  

Here λ = v/f = 343/490 = 0.7m  

The first point of interference is the initial point which would mean the distance is the trivial solution of 0  .

A microphone is placed half-way between the speakers and then moved along the line joining the two speakers until the first point of constructive interference is found.

Let x be the distance from the midpoint to the next interference.  

This occurs when 1.5 + x -(1.5 - x) = 0.7

so 2x = 0.7 which means x = 0.7/2 = 0.35m

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Answer:

0.54m

Explanation:

Step one:

given data

length of seesaw= 3m

mass of man m1= 85kg

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W1= 85*10= 850N

mass of daughter m2= 35kg

W2= 35*10= 350N

distance from the center= (1.5-0.2)= 1.3m

Step two:

we know that the sum of clockwise moment equals the anticlockwise moment

let the distance the must sit to balance the system be x

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455=850x

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Hence the man must sit 0.54m from the right to balance the system

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There is a moon orbiting an Earth-like planet. The mass of the moon is 9.58 × 1022 kg, the center-to-center separation of the pl
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Answer:

= 4.38 × 10³⁴kgm²/s

Explanation:

Given that,

mass of moon m = 9.5 × 10²²kg

Orbital radius r = 4.28  × 10⁵km

Orbital period  T = 28.9days

T = 28.9  × 24 × 60 × 60

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Angular momentum of the moon about the planet

L = mvr

L = mr²w

L = mr^2\frac{2\pi }{T} \\\\L = \frac{9.5 \times 10^2^2 \times(4.28\times10^8)^2\times2\times3.14}{2496960} \\\\L = 4.389.5 \times 10^3^4kgm^2/s

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On a frictionless horizontal air table, puck A (with mass 0.254 kg ) is moving toward puck B (with mass 0.367 kg ), which is ini
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Answer:

v_a=0.8176 m/s

\Delta K=0.07969 J - 0.0849 J = -0.00521 J

Explanation:

According to the law of conservation of linear momentum, the total momentum of both pucks won't be changed regardless of their interaction if no external forces are acting on the system.

Being m_a and m_b the masses of pucks a and b respectively, the initial momentum of the system is

M_1=m_av_a+m_bv_b

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M_1=m_av_a

After the collision and being v'_a and v'_b the respective velocities, the total momentum is

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Both momentums are equal, thus

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Solving for v_a

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The change of kinetic energy is

\Delta K=0.07969 J - 0.0849 J = -0.00521 J

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