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3 years ago

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15 beats are heard in 5 seconds when a piano key and a tuning fork of slightly higher frequency (440 Hz) are struck at the same

time. What is the frequency of the sound emitted by the piano key?

1 answer:

chubhunter [2.5K]3 years ago

6 0

Answer:

437Hz

Explanation:

if two sound waves have a little difference in frequencies $f_{a}$ and $f_{a}$ interfere, the resulting sound wave produces a beat.

Given:

$f_{a}=440Hz$

$f_{b}=frequency\ of\ the\ sound\ emitted by\ the\ piano\ key$

$f_{beat}=frequency\ of\ beat$

where

$f_{beat}=\frac{number\ of\ beat}{time}$

$f_{beat}=\frac{15beats}{5}=3Hz$

$f_{beat}=f_{a}-f_{b}$

$f_{b}=f_{a}-f_{beat}$ = 440Hz - 3Hz = 437Hz

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Find the kenetic energy of a car of mass 700kg racing with a velocity of 10m/s

fiasKO [112]

Answer:

35000 KJ

Explanation:

The equation for the kinetic energy is given by the formula :

$E_{k} = \frac{1}{2} mv^{2}$

$E_{k} = \frac{1}{2} (700)(10)^{2}$

$E_{k} = \frac{1}{2} (700)(100)$

$E_{k} = (350)(100)$ OR $E_{k} = \frac{1}{2} (70000)$

$E_{k} = 35000$

Units will be kilojoules since the units of mass was kilograms .

Our final answer is 35000 KJ

Hope this helped and have a good day

5 0

2 years ago

Vector C has a magnitude of 24.6 m and points in the − y ‑ direction. Vectors A and B both have positive y ‑ components, and mak

frez [133]

Answer:

A= 61.35

B= -44.40

Explanation:

1. Using the components method we have:

$A_{x}= A cos \alpha\\B_{x}= B cos \beta\\C_{x}= 0\\\\A_{y}= A sin \alpha\\B_{y}= B sin \beta\\C_{y}= 24.6\\$

Considering that the vector sum $A+B+C=0$ , then:

$|V|=\sqrt{V_{x}^{2} +V_{y}^{2} }=0$

Then:

$V_{x} ^{2} =0; V_{x} =0\\V_{y} ^{2} =0; V_{y} =0$

It means the value of x and y component is 0.

2. Determinate the equations that describe each component:

$V_{x}= A cos \alpha -B cos \beta=0 (1)\\V_{y}= A sin \alpha +B sin \beta - C=0 (2)$

Form Eq. (1):

$A=B \frac{cos \beta}{cos \alpha} (3)$

Replacing A in Eq. (2):

$(B\frac{cos \beta}{cos \alpha})(sin \alpha)+ B sin\beta-C=0\\(B\frac{cos \beta}{cos \alpha})(sin \alpha)+ B sin\beta-=C\\\\B(\frac{cos \beta. sin \alpha}{cos \alpha}+ sin\beta)=C\\B=C(\frac{cos \beta. sin \alpha}{cos \alpha}+ sin\beta)^{-1} (4)$

Replacing values of C, α and β in (4):

$B= 24.6 (\frac{(cos 27.7)(sin 44.9)}{cos 44.9}+sin 27.7)^{-1} \\B= -44.4$

Replacing value of B in (3)

$A=-44.40\frac{cos 27.7}{sin 49.9} \\A= 61.35$

5 0

3 years ago

On a distance vs. time graph, how do you know when the object is moving away from its starting position?

Ainat [17]

U know by if they are in first place

8 0

3 years ago

Read 2 more answers

An internal combustion engine uses fuel, of energy content 44.4 MJ/kg, at a rate of 5 kg/h. If the efficiency is 28%, determine

Fudgin [204]

Efficiency = Power Output / Power Input

Power Input = Rate of Energy input = 44.4 MJ/kg * 5 kg/h

= 222 MJ/h

But 1 hour = 3600seconds

222 MJ/h = 222 MJ/3600s = 0.061667 MW J/s = Watts

Power input = 0.061667 MW = 61 667 W

From Efficiency = Power Output / Power Input

   28% = Power Output / 61667

   Power Output = 0.28 * 61667

   Power Output = 17266.76 W

Power Output = 17 267 W

Rate of heat Rejection = Power input - Power output

      = 61667 - 17267 = 44400 W

Rate of heat Rejection = 44 400 W.

C- Copyright.

5 0

3 years ago

A dolphin in an aquatic show jumps straight up out of the water at a velocity of 15.0 m/s. (a) List the knowns in this problem.

astra-53 [7]

Answer:

a)

Y0 = 0 m

Vy0 = 15 m/s

ay = -9.81 m/s^2

b) 7.71 m

c) 3.06 s

Explanation:

The knowns are that the initial vertical speed (at t = 0 s) is 15 m/s upwards. Also at that time the dolphin is coming out of the water, so its initial position is 0 m. And since we can safely assume this happens in Earth, the acceleration is the acceleration of gravity, which is 9.81 m/s^2 pointing downwards

Y(0) = 0 m

Vy(0) = 15 m/s

ay = -9.81 m/s^2 (negative because it points down)

Since acceleration is constant we can use the equation for uniformly accelerated movement:

Y(t) = Y0 + Vy0 * t + 1/2 * a * t^2

To find the highest point we do the first time derivative (this is the speed:

V(t) = Vy0 + a * t

We equate this to zero

0 = Vy0 + a * t

0 = 15 - 9.81 * t

15 = 9.81 * t

t = 0.654 s

At this time it will have a height of:

Y(0.654) = 0 + 15 * 0.654 - 1/2 * 9.81 * 0.654^2 = 7.71 m

The doplhin jumps and falls back into the water, when it falls again it position will be 0 again. So we can equate the position to zero to find how long it was in the air knowing that it started the jump at t = 0s.

0 = Y0 + Vy0 * t + 1/2 * a * t^2

0 = 0 + 15 * t - 1/2 * 9.81 t^2

0 = 15 * t - 4.9 * t^2

0 = t * (15 - 4.9 * t)

t1 = 0 This is the moment it jumped into the air

0 = 15 - 4.9 * t2

15 = 4.9 * t2

t2 = 3.06 s This is the moment when it falls again.

3.06 - 0 = 3.06 s

5 0

3 years ago