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____ [38]
1 year ago
14

Standing 38.6 m away from a rock wall, you yell. How much time in seconds will it take you to hear your echo to two significant

digits? Make sure to account for the travel from you to the wall and from the wall back to you.
Physics
1 answer:
weqwewe [10]1 year ago
8 0

It will take you 0.23 seconds to hear your echo

<h3>What is Echo ?</h3>

Echo can be simply defined as the reflection of sound wave.

Given that you are 38.6 m away from a rock wall, you yell. To know how much time in seconds it will take you to hear your echo to two significant digits, You must make sure to account for the travel from you to the wall and from the wall back to you.

  • The speed of sound V = 340 m/s
  • The distance D to and fro = 2 x 38.6 = 77.2 m
  • The time taken T = ?

Speed V is the distance per time

V = D / T

Substitute all the parameters into the formula

340 = 77.2 / T

Make T the subject of formula

T = 77.2 / 340

T = 0.2270

T = 0.23 s to two significant digits

Therefore, it will take you 0.23 seconds to hear your echo.

Learn more about Echo here: brainly.com/question/14090821

#SPJ1

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(a) a_s=0.62\frac{m}{s^2}

(b) a_s=0.13\frac{m}{s^2}

(c) x_f=2.6m

Explanation:

(a) According to Newton's second law, the acceleration of a body is directly proportional to the force exerted on it and inversely proportional to it's mass.

a_s=\frac{F}{m_s}\\a_s=\frac{5.2N}{8.4kg}\\a_s=0.62\frac{m}{s^2}

(b) According to Newton's third law, the force that the sled exerts on the girl is equal in magnitude but opposite in the direction of the force that the girl exerts on the sled:

a_g=\frac{F}{m_g}\\a_g=\frac{5.2N}{40kg}\\a_g=0.13\frac{m}{s^2}

(c) Using the kinematics equation:

x_f=x_0+v_0t \pm  \frac{at^2}{2}

For the girl, we have x_0=0 and v_0=0. So:

x_f_g=\frac{a_gt^2}{2}(1)

For the sled, we have v_0=0. So:

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When they meet, the final positions are the same. So, equaling (1) and (2) and solving for t:

x_0_s-\frac{a_st^2}{2}=\frac{a_st^2}{2}\\t^2(a_g+a_s)=2x_0_s\\t=\sqrt{\frac{2x_s_0}{a_g+a_s}}\\t=\sqrt{\frac{2(15m)}{0.13\frac{m}{s^2}+0.62\frac{m}{s^2}}}\\t=6.32s

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x_f_g=\frac{0.13\frac{m}{s^2}(6.32s)^2}{2}\\x_f_g=2.6m\\x_f=2.6m

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