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2 years ago

What is the oscillation period of your eardrum when you are listening to the A4 note on a piano (frequency 440 Hz)?

Physics

2 answers:

Ulleksa [173]2 years ago

5 0

Answer:

<em>T </em>= 0.002 <em>s</em>.

Explanation:

Given data:

Frequency, $f = 440 \ \rm Hz$

The time period of oscillation of the eardrum can be expressed as,

$T = \frac{1}{f}$

$T = \frac{1}{440 \ \rm Hz}$

$T = 0.002 \ s.$

Contact [7]2 years ago

4 0

The period is the inverse of the frequence:

$T=\dfrac{1}{f}=\dfrac{1}{440}=0.0022\overline{72}$

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Answer:

4. The direct sunlight received by creosote bush in the desert area (in kWh/m2) during a 12 month period

Explanation:

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In order to test the hypothesis the student must record the direct sunlight received by creosote bush in the desert area (in kWh/m2) during a 12 month period. If the plants receive sunlight less than the above amount the plants should start dying. If not then the hypothesis is false.

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2 years ago

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2 years ago

An amount of work W is done on an object of mass m initially at rest, and as result it winds up moving at speed v. Suppose inste

sesenic [268]

Answer:

Explanation:

Given

W amount of work is done on the system such that it acquires v velocity after operation(initial velocity)

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$W=\frac{1}{2}mv_f^2-\frac{1}{2}mv^2-----2$

From 1 and 2 we get

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2 years ago

A signal generator has an output voltage of 2.0 V with no load. When a 600 Ω load is connected to it, the output drops to 1.0 V.

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Answer:

600 Ω

Explanation:

when there is no load attached to the generator the circuit is open and zero current flows through the circuit, hence voltage drop across the internal resister is zero.

when 600Ω load is connected current starts flowing through the circuit and some voltage will drop across the internal resister.

voltage across the load resister is 1 V, so the current through it will be:

I=V/R

I=1/600 A= 1.67mA

voltage drop in the internal resister is:

V= input voltage - output voltage

V=2 V-1 V

⇒V=1 V

Now by using ohm's law

R=V/ I

R= $\frac{1}{1.67*10^{-3} }$ (I= 1.67mA, as the resistors are connected in series)