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IrinaVladis [17]

3 years ago

13

Un diapasón vibra a 140 [Hz] y la onda emitida por él se propaga con una rapidez de 340 m/s.Considerando lo anterior, se puede a

firmar correctamente que la longitud de onda de la onda emitida por el diapasón es

1 answer:

Gnom [1K]3 years ago

5 0

The question is: A tuning fork vibrates at 140 [Hz] and the wave emitted by it propagates with a speed of 340 m / s. Considering the above, it can be correctly stated that the wavelength of the wave emitted by the tuning fork is

Answer: Wavelength of the wave emitted by the tuning fork is 2.43 m.

Explanation:

Given: Frequency = 140 Hz

Speed = 340 m/s

wavelength = ?

Formula used to calculate the wavelength is as follows.

$\lambda = \frac{v}{f}$

where,

$\lambda$ = wavelength

v = speed

f = frequency

Substitute the values into above formula.

$\lambda = \frac{v}{f}\\= \frac{340 m/s}{140 Hz} (1 Hz = 1 s^{-1})\\= 2.43 m$

Thus, we can conclude that wavelength of the wave emitted by the tuning fork is 2.43 m.

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Answer:

20 mangintiude beacuse

Explanation:

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3 years ago

(a) Consider the initial-value problem dA/dt = kA, A(0) = A0 as the model for the decay of a radioactive substance. Show that, i

murzikaleks [220]

Answer:

a) $t = -\frac{ln(2)}{k}$

b) See the proof below

$A(t) = A_o 2^{-\frac{t}{T}}$

c) $t = 3T \frac{ln(2)}{ln(2)}= 3T$

Explanation:

Part a

For this case we have the following differential equation:

$\frac{dA}{dt}= kA$

With the initial condition $A(0) = A_o$

We can rewrite the differential equation like this:

$\frac{dA}{A} =k dt$

And if we integrate both sides we got:

$ln |A|= kt + c_1$

Where $c_1$ is a constant. If we apply exponential for both sides we got:

$A = e^{kt} e^c = C e^{kt}$

Using the initial condition $A(0) = A_o$ we got:

$A_o = C$

So then our solution for the differential equation is given by:

$A(t) = A_o e^{kt}$

For the half life we know that we need to find the value of t for where we have $A(t) = \frac{1}{2} A_o$ if we use this condition we have:

$\frac{1}{2} A_o = A_o e^{kt}$

$\frac{1}{2} = e^{kt}$

Applying natural log we have this:

$ln (\frac{1}{2}) = kt$

And then the value of t would be:

$t = \frac{ln (1/2)}{k}$

And using the fact that $ln(1/2) = -ln(2)$ we have this:

$t = -\frac{ln(2)}{k}$

Part b

For this case we need to show that the solution on part a can be written as:

$A(t) = A_o 2^{-t/T}$

For this case we have the following model:

$A(t) = A_o e^{kt}$

If we replace the value of k obtained from part a we got:

$k = -\frac{ln(2)}{T}$

$A(t) = A_o e^{-\frac{ln(2)}{T} t}$

And we can rewrite this expression like this:

$A(t) = A_o e^{ln(2) (-\frac{t}{T})}$

And we can cancel the exponential with the natural log and we have this:

$A(t) = A_o 2^{-\frac{t}{T}}$

Part c

For this case we want to find the value of t when we have remaining $\frac{A_o}{8}$

So we can use the following equation:

$\frac{A_o}{8}= A_o 2^{-\frac{t}{T}}$

Simplifying we got:

$\frac{1}{8} = 2^{-\frac{t}{T}}$

We can apply natural log on both sides and we got:

$ln(\frac{1}{8}) = -\frac{t}{T} ln(2)$

And if we solve for t we got:

$t = T \frac{ln(8)}{ln(2)}$

We can rewrite this expression like this:

$t = T \frac{ln(2^3)}{ln(2)}$

Using properties of natural logs we got:

$t = 3T \frac{ln(2)}{ln(2)}= 3T$

8 0

3 years ago

Calculate the final speed of a 110-kg rugby player who is initially running at 8.00 m/s but collides head-on with a padded goalp

Contact [7]

<span>The answer is -0.8 m/s. We know acceleration is the average of final minus initial velocity over time (a = (vf-v0)/t). We also know that Force is equal to Mass times Acceleration (F = ma). Using our force equation, we know that the acceleration we get is negative 8.8 (-8.8). The force is acting in the opposite direction of the rugby player, hence the negative sign. From there, plug in that number for a in the velocity equation, and solve for vf, as v0 and t are known. We get 0.8 m/s in the opposite direction that the player was running.</span>

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3 years ago

You are given five transparent objects: a calcite crystal, a diamond, a piece of window glass, a sample of quartz, and a piece o

blsea [12.9K]

Answer:

Explanation:

The calcite Crystal can be identified by carrying out an acid test on it. This is done by bringing it in contact with a weak acid which cause crack in it structure. Hence a little of carbondioxide gas is released.

Since diamond is the hardest object,it can be identified on a Mohs scale. Likewise it can be tested by bringing it in contact with a newspaper, if the letters on the paper are not seen, that shows it is a pure diamond.

Window glass is identified by the code on it.

While Quartz Crystal is identified by scratch test. When it is used to scratch all other softer stones and metal, it leaves mark on them.

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3 years ago

what consistent physiological pattern are more common in women's teeth? A. Women's canines are generally sharper. B. Women have

irakobra [83]

Answer:

B

Explanation:

B. Women have straighter teeth.

5 0

3 years ago