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Sloan [31]
3 years ago
6

An "E" tuning fork with a frequency of 329 Hz is used to test the "E" note of a piano. Four beats are heard. This means _____.

Physics
2 answers:
inysia [295]3 years ago
8 0
The piano key is out of tune and has a frequency of either 325 or 333 Hz
o-na [289]3 years ago
7 0

The correct choice is

The piano key is out of tune and has a frequency of either 325 or 333 Hz

Beat frequency has been given as 4 Hz. hence the frequency must be above or below the given frequency of 329 Hz by 4 Hz. To get the upper level of frequency we add the beat frequency to the given frequency. To get the lower level, we subtract the beat frequency from the given frequency

f' = 329 + 4 = 333 Hz

and

f'' = 329 - 4 = 325 Hz

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Can cause dryness of the face of you excessively wash it it needs oils to keep it not dry
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3 years ago
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A Ferris wheel starts at rest and builds up to a final angular speed of 0.70 rad/s while rotating through an angular displacemen
PilotLPTM [1.2K]

Answer:

The average angular acceleration is 0.05 radians per square second.

Explanation:

Let suppose that Ferris wheel accelerates at constant rate, the angular acceleration as a function of change in angular position and the squared final and initial angular velocities can be clear from the following expression:

\omega^{2} = \omega_{o}^{2} + 2 \cdot \alpha\cdot (\theta-\theta_{o})

Where:

\omega_{o}, \omega - Initial and final angular velocities, measured in radians per second.

\alpha - Angular acceleration, measured in radians per square second.

\theta_{o}, \theta - Initial and final angular position, measured in radians.

Then,

\alpha = \frac{\omega^{2}-\omega_{o}^{2}}{2\cdot (\theta-\theta_{o})}

Given that \omega_{o} = 0\,\frac{rad}{s}, \omega = 0.70\,\frac{rad}{s} and \theta-\theta_{o} = 4.9\,rad, the angular acceleration is:

\alpha = \frac{\left(0.70\,\frac{rad}{s} \right)^{2}-\left(0\,\frac{rad}{s} \right)^{2}}{2\cdot \left(4.9\,rad\right)}

\alpha = 0.05\,\frac{rad}{s^{2}}

Now, the time needed to accelerate the Ferris wheel uniformly is described by this kinematic equation:

\omega = \omega_{o} + \alpha \cdot t

Where t is the time measured in seconds.

The time is cleared and obtain after replacing every value:

t = \frac{\omega-\omega_{o}}{\alpha}

If \omega_{o} = 0\,\frac{rad}{s},  \omega = 0.70\,\frac{rad}{s} and \alpha = 0.05\,\frac{rad}{s^{2}}, the required time is:

t = \frac{0.70\,\frac{rad}{s} - 0\,\frac{rad}{s} }{0.05\,\frac{rad}{s^{2}} }

t = 14\,s

Average angular acceleration is obtained by dividing the difference between final and initial angular velocities by the time found in the previous step. That is:

\bar \alpha = \frac{\omega-\omega_{o}}{t}

If \omega_{o} = 0\,\frac{rad}{s},  \omega = 0.70\,\frac{rad}{s} and t = 14\,s, the average angular acceleration is:

\bar \alpha = \frac{0.70\,\frac{rad}{s} - 0\,\frac{rad}{s} }{14\,s}

\bar \alpha = 0.05\,\frac{rad}{s^{2}}

The average angular acceleration is 0.05 radians per square second.

4 0
3 years ago
A box of mass m is held at rest on a frictionless surface with force F up the ramp. The ramp has an angle
frutty [35]

Answer:

Gv

Explanation:

8 0
3 years ago
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An uncharged series RC circuit is to be connected across a battery. For each of the following changes, determine whether the tim
slavikrds [6]

a) Increase

b) Unchanged

c) Increase

Explanation:

a)

The charge on a capacitor charging in a RC circuit connected to a battery follows the exponential equation:

Q(t)=Q_0 (1-e^{-\frac{t}{RC}})

where

Q_0 = CV is the final charge stored in the capacitor, where C is the capacitance and V is the voltage of the battery

t is the time

R is the resistance of the circuit

The capacitor reaches 90% of its final charge when

Q(t)=0.90Q_0

Substituting and re-arranging the equation, we find:

0.90Q_0 = Q_0(1-e^{-\frac{t}{RC}})\\0.90=1-e^{-\frac{t}{RC}}\\e^{-\frac{t}{RC}}=0.10\\-\frac{t}{RC}=ln(0.10)\\t=-RCln(0.10)=2.30RC

We see that if we double the RC constant, then (RC)'=2(RC)

So the time taken will double as well:

t'=2.30(RC)'=2.30(2RC)=2(2.30RC)=2t

So, the answer is "increase"

b)

In this second part, the battery voltage is doubled.

According to the equation written in part a),

Q_0 =CV

this means also that the final charge stored on the capacitor will also double.

However, the equation that gives us the time needed for the capacitor to reach 90% of its full charge is

t=2.30 RC

We see that this equation does not depend at all on the voltage of the battery.

Therefore, if the battery voltage is doubled, the final charge on the capacitor will double as well, but the time needed for the capacitor to reach 90% of its charge will not change.

So the correct answer is

"unchanged"

c)

In this case, a second resistor is added in series with the original resistor of the circuit.

We know that for two resistors in series, the total resistance of the circuit is given by the sum of the individual resistances:

R=R_1+R_2

Since each resistance is a positive value, this means that as we add new resistors, the total resistance of the circuit increases.

Therefore in this problem, if we add a resistor in series to the original circuit, this means that the total resistance of the circuit will increase.

The time taken for the capacitor to reach 90% of its final charge is still

t=2.30 RC

As we can see, this time is directly proportional to the resistance of the circuit, R: therefore, if we add a resistor in series, the resistance of the circuit will increase, and therefore this time will increase as well.

So the correct answer is

"increase"

8 0
3 years ago
Que la tarcer ley de newton
slava [35]

Explanation:

If you write it in English so I can help u if you need it

3 0
3 years ago
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