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Jlenok [28]
3 years ago
13

. When two speakers play different sounds with a constant frequency, beats are heard with a frequency of 5 Hz. If the frequency

of sound played by one of the speakers is 1050 Hz, what is the frequency of sound played by the other speaker
Physics
1 answer:
morpeh [17]3 years ago
3 0

The beat frequency is the concept required to develop this process. The phenomenon is generated when you have two waves but their frequencies do not differ greatly. So the beat frequency will be the difference between those two waves. For this case we have the difference and one of the waves, therefore,

F_b = |F_1-F_2|

Our values are given as,

F_b = 5Hz

F_1 = 1050Hz

Using the formula of the frequency I will have,

F_b = |F_1-F_2|

Replacing we have,

5 = |1050-F_2|

F_2 = 1050 \pm 5

F_2 = 1045,1055 Hz

The possible values are two: 1045Hz and 1055Hz

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n explosion breaks an object initially at rest into two pieces, one of which has 2.0 times the mass of the other. If 8400 J of k
ANTONII [103]

Answer:

The heavier piece acquired 2800 J  kinetic energy

Explanation:

From the principle of conservation of linear momentum:

0 = M₁v₁ - M₂v₂

M₁v₁ = M₂v₂

let the second piece be the heavier mass, then

M₁v₁ = (2M₁)v₂

v₁  = 2v₂ and v₂ = ¹/₂ v₁

From the principle of conservation of kinetic energy:

¹/₂ K.E₁ + ¹/₂ K.E₂ = 8400 J

¹/₂ M₁(v₁)² + ¹/₂ (2M₁)(¹/₂v₁)² = 8400

¹/₂ M₁(v₁)² + ¹/₄M₁(v₁)² = 8400

K.E₁ + ¹/₂K.E₁ = 8400

Now, we determine K.E₁ and note that K.E₂ = ¹/₂K.E₁

1.5 K.E₁ = 8400

K.E₁ = 8400/1.5

K.E₁ = 5600 J

K.E₂ = ¹/₂K.E₁ = 0.5*5600 J = 2800 J

Therefore, the heavier piece acquired 2800 J  kinetic energy

3 0
3 years ago
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Find the equivalent resistance Req between terminals a and b if terminals c and d are open and again if terminals c and d are sh
adoni [48]

Answer:

351 ohm

720 ohm

Explanation:

When c and d are open:

Terminals c and d are open.. If you  redraw the circuit as below, you can see that the two resistors in the first  column are in parallel as, they are connected together at both pairs of terminals  (due to the short).

Hence, we have a pair of parallel resistors:

Req1 = (R1*R2)/ (R1 + R2) = 360*540/(360+540) = 216 ohms

Req2 = (R3*R4)/ (R3 + R4) = 180*540/(180+540) = 135 ohms

Now these two sets are  in series with another Hence,

Req = Req1 + Req2 = 216 + 135 = 351 ohms

Answer: 351 ohms

When c and d are shorted:

The current will flow through the least resistant path naturally from resistors R3 and R1 or R4.

Both of these resistor lie in a single path placing the resistors in series to one another, hence

Req = R3 + R1 = 180 + 540 = 720 ohms

Answer:720 ohms

7 0
3 years ago
Two particles execute simple harmonic motion of the same amplitude and frequency along close parallel lines. They pass each othe
deff fn [24]

Answer:

\theta_2 - \theta_1 = 156.93 degree

Explanation:

As we know that the displacement of the particle from the mean position is 1/5 times of its amplitude

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so now we have

\frac{A}{5} = A sin\omega t

now we have

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so the phase other particle in opposite direction is given as

\theta_2 = 180 - 11.53 = 168.46 degree

so we have phase difference given as

\theta_2 - \theta_1 = 168.46 - 11.53

\theta_2 - \theta_1 = 156.93 degree

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