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ICE Princess25 [194]

2 years ago

6

(4%) Problem 10: The standard frequency of the musical pitch called "middle C" is 261.6 Hz. show answer No Attempt How fast, in

meters per second, would a source emitting a tone of middle C have to move toward you so that you would hear the pitch C# (C sharp) above middle C, with a frequency of 277.2 Hz? Assume that takes place in air at a high altitude, where the speed of sound is 298 m/s.

1 answer:

bearhunter [10]2 years ago

6 0

Answer:

16.77 m/s

Explanation:

Given that

Frequency of middle pitch, Fo = 261.6 Hz

Frequency of C sharp, f = 277.2 Hz

Velocity of sound in air, v = 298 m/s

Speed of sound from the source, Vs = ? m/s

Using the formula

f = Fo•(V + Vr)/(V + Vs)

← Doppler

Vr would be +ve if the receiver is moving toward source;

Vs would be -ve if source is moving toward the receiver

277.2 Hz = 261.6Hz * (298 + 0) / (298 - Vs)

277.2 = 77956.8 / (298 - Vs)

298 - Vs = 77956.8 / 277.2

298 - Vs = 281.23

Vs = 298 - 281.23

Vs = 16.77 m/s

Thus, the speed needed is 16.77 m/s

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2 years ago

A puck of mass m = 0.085 kg is moving in a circle on a horizontal frictionless surface. It is held in its path by a massless str

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Answer:

T = 11.93 N

Explanation:

Newton's second law to the puck in the circular path

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a : radial acceleration of the puck (m/s²)

Data:

m = 0.085 kg

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2 years ago

On a nice summer day,Kim takes her niece Madison for a walk in her stroller.If they start from rest and accelerate at a rate of

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2.5m/s

Explanation:

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2 years ago

A tank contains 100 gal of water and 50 oz of salt.water containing a salt concentration of 1 4 (1 1 2 sin t) oz/gal flows into

Alchen [17]

Answer:

Explanation:

Heres the possible full question and solution:

A tank contains 100 gal of water and 50 oz of salt. Water containing a salt concentration of ¼ (1 + ½ sin t) oz/gal flows ito the tank at a rate of 2 gal/min, and the mixture in the tank flows out at the same rate.

a. Find the amount of salt in the tank at any time.

b. Plot the solution for a time period long enough so that you see the ultimate behavior of the graph.

c. The long-time behavior of the solution is an oscillation about a certain constant level. What is this level? What is the amplitude of the oscillation?

solution

a)

Consider the tank contains 100gal of water and 50 oz of salt

Assume that the amount of salt in the tank at time t is Q(t).

Then, the rate of change of salt in the tank is given by $\frac{dQ}{dt}$ .

Here, $\frac{dQ}{dt}$ =rate of liquid flowing in the tank - rate of liquid flowing out.

Therefore,

$Rate_{in} =2gal/min \times \frac{1}{4} (1+ \frac{1}{2}sin t)oz/gal\\\\\\ \frac{1}{2} (1+ \frac{1}{2}sin t)oz/min\\\\\\Rate_{out}=2gal/min \times\frac{Q}{100}oz/gal\\\\\frac{Q}{50}oz/min$

Therefore,

$\frac{dQ}{dt}$ can be evaluated as shown below:

$\frac{dQ}{dt}=\frac{1}{2}(1+\frac{1}{2}\sin t)-\frac{Q}{50}\\\\\\\frac{dQ}{dt}+\frac{1}{50}Q=\frac{1}{2}+\frac{1}{4}\sin t$

The above differential equation is in standard form:

$\frac{dy}{dt}+Py=G$

Here, $P=\frac{1}{50},G=\frac{1}{2}+\frac{1}{4}\sin t$

The integrating factor is as follows:

$\mu(t)=e^{\int {P}dt}\\\mu(t)=e^{\int {\frac{1}{50}}dt}\\\mu(t)=e^{\frac{t}{50}}$

Thus, the integrating factor is $\mu(t)=e^{\frac{t}{50}}$

Therefore, the general solution is as follows:

$y\mu(t)=\int {\mu (t)G}dt\\\\Qe^{\frac{t}{50}}=\int {e^{\frac{t}{50}}(\frac{1}{2}+\frac{1}{4}\sin t) dt}\\\\Qe^{\frac{t}{50}}=\frac{1}{2}\int {e^{\frac{t}{50}}dt + \frac{1}{4}\int {\sin t {e^{\frac{t}{50}}} dt}\\\\\Qe^{\frac{t}{50}}=25 {e^{\frac{t}{50}} + \frac{1}{4}\int {\sin t {e^{\frac{t}{50}}} dt}+C...(1)$

Here, C is arbitrary constant of integration.

Solve $\int {\sin te^{\frac{t}{50}}} dt}$

Here $u = e^{\frac{t}{50}}$ and $v =\sin t$ .

Substitute u , v in the below formula:

$\int{u,v}dt=u\int{v}dt-\int\frac{du}{dt}\int{v}dt\dot dt\\\\\int {e^{\frac{t}{50}}\sin t}dt=-e^{\frac{t}{50}}\cos t + \frac{1}{50}\int{e^{\frac{t}{50}}\cos t}dt...(2)$

Now, take $u = e^{\frac{t}{50}}$ , $v =\sin t$

Therefore, $\int{e^{\frac{t}{50}}\cos t} dt=\int {e^{\frac{t}{50}}\sin t}dt - \frac{1}{50}\int{e^{\frac{t}{50}}\sin t}dt...(3)$

Use (3) in equation(2)

$\int {e^{\frac{t}{50}}\sin t}dt=-e^{\frac{t}{50}}\cos t + \frac{e^{\frac{t}{50}}}{50}\sin t - \frac{1}{2500}\int{e^{\frac{t}{50}}\sin t}dt\\\\\frac{2501}{2500}\int{e^{\frac{t}{50}}\sin t}dt={e^{\frac{t}{50}}\cos t}+\frac{e^{\frac{t}{50}}}{50}\sin t\\\\\int{e^{\frac{t}{50}}\sin t}dt=\frac{2500}{2501}{e^{\frac{t}{50}}\cos t}+\frac{50}{2501}e^{\frac{t}{50}}\sin t...(4)$

Use (4) in equation(l) .

$Qe^{\frac{t}{50}}=25 e^{\frac{t}{50}} - \frac{625}{2501}e^{\frac{t}{50}}\cos t +\frac{25}{5002}e^{\frac{t}{50}}\sin t+C$

Apply the initial conditions $t =0, Q = 50$ .

$50=25-\frac{625}{2501}+c\\\\c=\frac{63150}{2501}$

So, $Qe^{\frac{t}{50}}=25 e^{\frac{t}{50}} - \frac{625}{2501}e^{\frac{t}{50}}\cos t +\frac{25}{5002}e^{\frac{t}{50}}\sin t+\frac{63150}{2501}$

Therefore, the amount of salt in the tank at any time is as follows:

$Qe^{\frac{t}{50}}=25 e^{\frac{t}{50}} - \frac{625}{2501}e^{\frac{t}{50}}\cos t +\frac{25}{5002}e^{\frac{t}{50}}\sin t+\frac{63150}{2501}e^{\frac{-t}{50}}$

b)

sketch the solution curve as shown in attachment as graph 1:

CHECK COMMENT FOR C

3 0

2 years ago

A large solar panel on a spacecraft in Earth orbit produces 1.0 kW of power when the panel is turned toward the sun. What power

Mandarinka [93]

Answer:

e*P_s = 11 W

Explanation:

Given:

- e*P = 1.0 KW

- r_s = 9.5*r_e

- e is the efficiency of the panels

Find:

What power would the solar cell produce if the spacecraft were in orbit around Saturn

Solution:

- We use the relation between the intensity I and distance of light:

I_1 / I_2 = ( r_2 / r_1 ) ^2

- The intensity of sun light at Saturn's orbit can be expressed as:

I_s = I_e * ( r_e / r_s ) ^2

I_s = ( 1.0 KW / e*a) * ( 1 / 9.5 )^2

I_s = 11 W / e*a

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P_s = 11 W / e

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2 years ago