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Brums [2.3K]
3 years ago
9

Sound travels about 750 miles per hour. if you stand in a canyon and sound a horn, you will hear an echo. suppose it takes about

3.5 seconds to hear the echo. how far away is the canyon wall, in feet
Physics
2 answers:
diamong [38]3 years ago
5 0
The answer to this question would be:3850ft

To answer this question, you need to convert the speed velocity from miles/hour into feet/second. The equation would be: 750 miles/hour x 5280 foot/mile x 1 hour/3600second = 1100 ft/s
Then multiply the time with the velocity= 3.5 second x 1100 ft/s= 3850ft
umka21 [38]3 years ago
3 0

The canyon is 3.8 ×10³ feet if the echo is heard after 3.5 seconds.

Further Explanation

Speed is how fast an object moves or how far an object travels per unit time. If the distance traveled and the total time of travel are known, the speed can be calculated using the formula:

speed = \frac{distance}{time}

In the problem, we are given:

  • speed = 750 mi/h
  • time = 3.5 seconds

We are looking for the distance traveled by sound which can be represented by the variable <em>d.</em>

We can solve for d by manipulating the speed formula to get the equation:

distance \ (d) \ = speed \times time

But before we can plug in the given into the formula, the speed must be converted in terms of seconds since the given time is also in seconds:

speed \ = \frac{750 \ mi}{1 \ h} \times \frac{1 \ h}{60 min} \times \frac{1 \ min}{60 sec}\\\\\boxed {speed = 0.2083 \ \frac{mi}{sec}}

Now that the speed in miles per second is known, the distance can now be calculated using the formula above:

distance \ = speed \times \ time\\distance \ = 0.2083 \frac{mi}{sec} \times 3.5 sec\\\boxed {distance = 0.729 \ mi/tex]Now, since the problem requires the distance in feet, the conversion factor is used: 1 mi = 5280 ft[tex]0.729 mi \times \frac{5280 \ ft}{1 \ mi} = \boxed {3,849.12 \ ft}\\

Since the given has two significant digits, the answer must also have two. Therefore, the final answer must be 3.8 × 10³ feet.

<h3>LEARN MORE</h3>
  • Learn more about Velocity brainly.com/question/862972
  • Learn more about Acceleration brainly.com/question/4134594
  • Learn more about Distance - Time Graphs brainly.com/question/1378025

Keywords: speed, distsance, time

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A capacitor with initial charge q0 is discharged through a resistor. a) In terms of the time constant τ, how long is required fo
-BARSIC- [3]

Answer:

It would take \tau(\ln 9 - \ln 8) time for the capacitor to discharge from q_0 to \displaystyle \frac{8}{9} \, q_0.

It would take \tau(\ln 9 - \ln 7) time for the capacitor to discharge from q_0 to \displaystyle \frac{7}{9}\, q_0.

Note that \ln 9 = 2\,\ln 3, and that\ln 8 = 3\, \ln 2.

Explanation:

In an RC circuit, a capacitor is connected directly to a resistor. Let the time constant of this circuit is \tau, and the initial charge of the capacitor be q_0. Then at time t, the charge stored in the capacitor would be:

\displaystyle q(t) = q_0 \, e^{-t / \tau}.

<h3>a)</h3>

\displaystyle q(t) = \left(1 - \frac{1}{9}\right) \, q_0 = \frac{8}{9}\, q_0.

Apply the equation \displaystyle q(t) = q_0 \, e^{-t / \tau}:

\displaystyle \frac{8}{9}\, q_0 = q_0 \, e^{-t/\tau}.

The goal is to solve for t in terms of \tau. Rearrange the equation:

\displaystyle e^{-t/\tau} = \frac{8}{9}.

Take the natural logarithm of both sides:

\displaystyle \ln\, e^{-t/\tau} = \ln \frac{8}{9}.

\displaystyle -\frac{t}{\tau} = \ln 8 - \ln 9.

t = - \tau \, \left(\ln 8 - \ln 9\right) = \tau(\ln 9 - \ln 8).

<h3>b)</h3>

\displaystyle q(t) = \left(1 - \frac{1}{9}\right) \, q_0 = \frac{7}{9}\, q_0.

Apply the equation \displaystyle q(t) = q_0 \, e^{-t / \tau}:

\displaystyle \frac{7}{9}\, q_0 = q_0 \, e^{-t/\tau}.

The goal is to solve for t in terms of \tau. Rearrange the equation:

\displaystyle e^{-t/\tau} = \frac{7}{9}.

Take the natural logarithm of both sides:

\displaystyle \ln\, e^{-t/\tau} = \ln \frac{7}{9}.

\displaystyle -\frac{t}{\tau} = \ln 7 - \ln 9.

t = - \tau \, \left(\ln 7 - \ln 9\right) = \tau(\ln 9 - \ln 7).

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Answer:

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Explanation:

La expresión tiene un error por omisión, su forma correcta queda descrita a continuación:

<em>"Una prenda de 320 gramos de ropa gira en el interior de una lavadora si dicha lavadora tiene un radio de 40 centímetros y gira con una frecuencia de 4 hertz. Halle </em><em>a)</em><em> el período, </em><em>b) </em><em>la velocidad angular, </em><em>c) </em><em>la fuerza con la que gira la prenda y </em><em>d) </em><em>la velocidad lineal de la lavadora."</em>

El tambor gira a velocidad angular constante (\omega), en radianes por segundo, lo cual significa que la prenda experimenta una aceleración centrífuga (a), en metros por segundo al cuadrado. En primer lugar, calculamos el período de rotación del tambor (T), en segundos:

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(f = 4\,hz)

T = \frac{1}{4\,hz}

T = 0.25\,s

Ahora determinamos la fuerza aplicada sobre la prenda (F), en newtons:

F = m\cdot a (2)

F = \frac{4\pi^{2}\cdot m \cdot r}{T^{2}} (2b)

Donde:

m - Masa de la prenda, en kilogramos.

r - Radio interior del tambor, en metros.

(m = 0.32\,kg, r = 0.4\,m, T = 0.25\,s)

F = \frac{4\pi^{2}\cdot (0.32\,kg)\cdot (0.4\,m)}{(0.25\,s)^{2}}

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(r = 0.4\,m, T = 0.25\,s)

v = \frac{2\pi\cdot (0.4\,m)}{0.25\,s}

v \approx 10.053\,\frac{m}{s}

Y la velocidad angular del tambor de la lavadora:

\omega = \frac{2\pi}{T}

(T = 0.25\,s)

\omega = \frac{2\pi}{0.25\,s}

\omega \approx 25.133\,\frac{rad}{s}

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