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Licemer1 [7]
4 years ago
9

A 1056-hertz tuning fork is struck at the same time as a note on the piano and you hear 2 beats/second. You tighten the piano st

ring very slightly and then hear 3 beats/second. What is the frequency of the piano string?
Physics
1 answer:
elena-14-01-66 [18.8K]4 years ago
6 0

Answer:

The frequency of the piano string is <em>1059 Hz</em>.

Explanation:

The frequency beat (fb), 2 beats/second, is the absolute difference between the frequency of the tuning fork (1056 Hz) and the frequency of the piano string.

As the piano string gets tightened, the frequency beat becomes 3 beats/second.

Therefore,

fb = fb = fpiano - ftuning fork\\ 3 Hz=fpiano-1056Hz\\ fpiano=1056Hz+3Hz\\ fpiano=1059Hz

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Simple machines are widely used in our daily life. why?<br>​
HACTEHA [7]

Answer:

Simple machines are useful because they reduce effort or extend the ability of people to perform tasks beyond their normal capabilities. Simple machines that are widely used include the wheel and axle, pulley, inclined plane, screw, wedge and lever.

7 0
3 years ago
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There is evidence that elephants communicate via infrasound, generating rumbling vocalizations as low as 14 Hz that can travel u
amid [387]

Answer:

L = 96.2 dB

Explanation:

As we know that the level of intensity is given as

\beta = 10Log(\frac{I}{I_o})

here we know

\beta = 104 dB

104 = 10Log(\frac{I}{10^{-12}})

I = 0.025 W/m^2

now the sound is travelling in all possible directions so we have

\frac{I_1}{I_2} = \frac{r_2^2}{r_1^2}

\frac{0.025}{I_2} = \frac{10^2}{4.1^2}

I_2 = 4.2 \times 10^{-3} W/m^2

now for level of sound we have

L = 10Log(\frac{4.2 \times 10^{-3}}{10^{-12}})

L = 96.2 dB

5 0
3 years ago
The stopping distance d of a car after the brakes are applied varies directly as the square of the speed r. If a car travelling
Crank
270/70^2  = x/80^2

Cross multiply

270 (6400) = 4900x   

x = 270(6400)/4900

352 and 32/49 feet

Hope this helps


4 0
3 years ago
(b) Released from rest at the same height, a can of frozen juice rolls to the bottom of the same hill. What is the translational
Oksanka [162]

Answer:

Explanation:

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mgh = 0.5mv^2 + 0.5Iω^2

I(moment of inertia) (basket ball) = (2/3)mr^2

mgh = 0.5mv^2 + 0.5( 2/3mr^2) ( v^2/r^2)

gh = 1/2v^2 + 1/3v^2

gh = v^2( 5/6)

v =  \sqrt{\frac{6gh}{5} }

putting the values we get

6.6 ^{2} = \frac{6\times9.8h}{5}

solving for h( height)

h = 3.704 m apprx

b) velocity of solid cylinder

mgh = 0.5mv^2 + 0.5( mr^2/2)( v^2/r^2) where ( I ofcylinder = mr^2/2)

g*h = 1/2v^2 + 1/4v^2

g*h = 3/4v^2

putting the value of h and g we get

v= = 6.957 m/s apprx

5 0
4 years ago
The rate constant of a reaction is 7.8 × 10−3 s−1 at 25°C, and the activation energy is 33.6 kJ/mol. What is k at 75°C? Enter yo
Alexus [3.1K]

Answer:

k_2 = 7.815 * 10^-3 s^-1

Explanation:

Given:

- rate constant of reaction k_1 = 7.8 * 10^-3 s^-1    @  T_1 = 25 C

- rate constant of reaction k_2 = ?    @  T_2 = 75 C

- The activation energy E_a = 33.6 KJ/mol

- Gas constant R = 8.314472 KJ / mol . K

Find:

- rate of reaction k_2 @ T_2 = 75 C

Solution:

- we will use a combined form of Arrhenius equations that relates rate constants k as function of E_a and temperatures as follows:

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- Evaluate             k_2 = 7.8 * 10^-3* e^[(33.6 / 8.314472)*(1/298 -1/348)

- Hence,                k_2 = 7.815 * 10^-3 s^-1

4 0
4 years ago
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