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4 years ago

14

An avant-garde composer wants to use the Doppler effect in his new opera. As the soprano sings, he wants a large bat to fly towa

rd her from the back of the stage. The bat will be outfitted with a microphone to pick up the singer's voice and a loudspeaker to rebroadcast the sound toward the audience. The composer wants the sound the audience hears from the bat to be, in musical terms, one half-step higher in frequency than the note they are hearing from the singer. Two notes a half-step apart have a frequency ratio of 21/12 = 1.059. With what speed must the bat fly toward the singer?

1 answer:

Brilliant_brown [7]4 years ago

6 0

Answer:

18.94 m/s

Explanation:

The bat becomes the source of sound . Source is moving towards observer

apparent frequency = actual frequency x (V / V - V_s)

V is velocity of sound and V_s is velocity of source or bat.

apparent frequency / actual frequency = V / (V - V_s)

21 / 12 = 340 / (340 - V_s)

1.059(340 - V_s) = 340

340 x .059 = 1.059 V_s

V_s = 18.94 m/s

=

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A playground merry-go-round has a mass of 115 kg and a radius of 2.50 m and it is rotating with an angular velocity of 0.520 rev

tatuchka [14]

Answer:

$W_f = 2.319 rad/s$

Explanation:

For answer this we will use the law of the conservation of the angular momentum.

$L_i = L_f$

so:

$I_mW_m = I_sW_f$

where $I_m$ is the moment of inertia of the merry-go-round, $W_m$ is the initial angular velocity of the merry-go-round, $I_s$ is the moment of inertia of the merry-go-round and the child together and $W_f$ is the final angular velocity.

First, we will find the moment of inertia of the merry-go-round using:

I = $\frac{1}{2}M_mR^2$

I = $\frac{1}{2}(115 kg)(2.5m)^2$

I = 359.375 kg*m^2

Where $M_m$ is the mass and R is the radio of the merry-go-round

Second, we will change the initial angular velocity to rad/s as:

W = 0.520*2 $\pi$ rad/s

W = 3.2672 rad/s

Third, we will find the moment of inertia of both after the collision:

$I_s = \frac{1}{2}M_mR^2+mR^2$

$I_s = \frac{1}{2}(115kg)(2.5m)^2+(23.5kg)(2.5m)^2$

$I_s = 506.25kg*m^2$

Finally we replace all the data:

$(359.375)(3.2672) = (506.25)W_f$

Solving for $W_f$ :

$W_f = 2.319 rad/s$

7 0

3 years ago

Select all the correct answers.

sdas [7]

I believe there are two correct answers, and those answers are A and D

5 0

3 years ago

We can model a pine tree in the forest as having a compact canopy at the top of a relatively bare trunk. Wind blowing on the top

frez [133]

We need to consider for this exercise the concept Drag Force and Torque. The equation of Drag force is

$F_D = c_D A \frac{\rho V^2}{2}$

Where,

F_D = Drag Force

$c_D$ = Drag coefficient

A = Area

$\rho$ = Density

V = Velocity

Our values are given by,

$c_D = 0.5$ (That is proper of a cone-shape)

$A = 9m^2$

$\rho = 1.2Kg/m^3$

$V = 6.5m/s$

Part A ) Replacing our values,

$F_D = 0.5*9*\frac{1.2*6.5^2}{2}$

$F_D = 114.075N$

Part B ) To find the torque we apply the equation as follow,

$\tau = F*d$

$\tau = (114.075N)(7)$

$\tau = 798.525N.m$

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3 years ago

When two license plates are issued they need to be attached to the _________ of the vehicle they were issued for.

Sergio039 [100]

Answer:

Front and the rear bumpers.

Explanation:

When two license plates are issued they need to be attached to the Front and the rear bumpers of the vehicle they were issued for.

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3 years ago

An electric iron connected to a 110 V source draws 9 A of current. How much heat (in joules) does it generate in a minute?

meriva

Q = 110V x 9A x 60s = J

7 0

3 years ago