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Alisiya [41]
3 years ago
5

A person, with his ear to the ground, sees a huge stone strike the concrete pavement. A moment later two sounds are heard from t

he impact: one travels in the air and the other in the concrete, and they are 6.4 s apart. Part A How far away did the impact occur? (Use vair=343m/s , vconcrete=3000m/s )
Physics
1 answer:
marishachu [46]3 years ago
7 0

Answer:

The impact occured at a distance of 2478.585 meters from the person.

Explanation:

(After some research on web, we conclude that problem is not incomplete) The element "Part A" may lead to the false idea that question is incomplete. Correct form is presented below:

<em>A person, with his ear to the ground, sees a huge stone strike the concrete pavement. A moment later two sounds are heard from the impact: one travels in the air and the other in the concrete, and they are 6.4 seconds apart. How far away did the impact occur? (Sound speed in the air: 343 meters per second, sound speed in concrete: 3000 meters per second)</em>

Sound is a manifestation of mechanical waves, which needs a medium to propagate themselves. Depending on the material, sound will take more or less time to travel a given distance. From statement, we know this time difference between air and concrete (\Delta t), in seconds:

\Delta t = t_{A}-t_{C} (1)

Where:

t_{C} - Time spent by the sound in concrete, in seconds.

t_{A} - Time spent by the sound in the air, in seconds.

By suposing that sound travels the same distance and at constant speed in both materials, we have the following expression:

\Delta t = \frac{x}{v_{A}}-\frac{x}{v_{C}}

\Delta t = x\cdot \left(\frac{1}{v_{A}}-\frac{1}{v_{C}}  \right)

x = \frac{\Delta t}{\frac{1}{v_{A}}-\frac{1}{v_{C}}  } (2)

Where:

v_{C} - Speed of the sound in concrete, in meters per second.

v_{A} - Speed of the sound in the air, in meters per second.

x - Distance traveled by the sound, in meters.

If we know that \Delta t = 6.4\,s, v_{C} = 3000\,\frac{m}{s} and v_{A} = 343\,\frac{m}{s}, then the distance travelled by the sound is:

x = \frac{\Delta t}{\frac{1}{v_{A}}-\frac{1}{v_{C}}  }

x = 2478.585\,m

The impact occured at a distance of 2478.585 meters from the person.

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A 0.311 kg tennis racket moving 30.3 m/s east makes an elastic collision with a 0.0570 kg ball moving 19.2 m/s west. Find the ve
amm1812

<u>Answer</u>:

The velocity of the tennis racket after the collision 14.966 m/s.

<u>Step-by-step explanation:</u>

let the following:

m₁ = mass of tennis racket = 0.311 kg

m₂ = mass of the ball = 0.057 kg

u₁ = velocity of tennis racket before collision = 30.3 m/s

u₂ = velocity of the ball before collision = -19.2 m/s

v₁ = velocity of tennis racket after collision

v₂ = velocity of the ball after collision

Right (+) , Left (-)

An elastic collision is an encounter between two bodies in which the total kinetic energy of the two bodies remains the same.

So, the total kinetic energy before collision = the total kinetic energy after collision.

So, 0.5 m₁ u₁² + 0.5 m₂ u₂² = 0.5 m₁ v₁² + 0.5 m₂ v₂²  ⇒ (1)

Also, the total momentum before collision = the total momentum after collision.

So, m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂  ⇒ (2)

Solving (1) and (2):

∴ v₁ = [ u₁ * (m₁ - m₂) + u₂ * 2m₂ ]/ (m₁ + m₂)

      = ( 30.3 * (0.311 - 0.057) - 19.2 * 2 * 0.057 ) / ( 0.311 + 0.057)

      = 14.966 m/s.

So, the velocity of the tennis racket after the collision 14.966 m/s.

7 0
3 years ago
Studen
Vesna [10]

Complete Question

A football coach walks 18 meters westward, then 12 meters eastward, then 28 meters westward, and finally 14 meters eastward.

a

From this motion what is the distance covered

b

What is the magnitude and direction of the displacement

Answer:

a

  D =   72 \  m

b

Magnitude

             d =  - 20 \ m

 Direction  

West

Explanation:

From the question we are told that

The first distance covered westward is d_w_1  =  18 \  m /tex]     The  first distance covered eastward is [tex]d_e1 =  12 \  m /tex]      The second distance covered westward is [tex]d_w_2  =  28 \  m /tex]      The  second distance covered eastward is [tex]d_e2 =  14 \  m /tex]   Generally the distance covered is mathematically represented as       [tex]D =  d_w1 + d_w2 + d_e1 + d_e2

=> D =   18 + 28 + 12 +  14

=> D =   72 \  m

For the second question eastward is in the direction of the positive x-axis so it would be positive and westward is in the direction of the negative x-axis so it would be negative

The magnitude of the displacement is

d =  -d_w1 -d_w2 + d_e1 + d_e2

=>  d =  -18-28 + 12 + 14

=> d =  - 20 \ m

The direction is west

7 0
3 years ago
A wave has a frequency of 655 he And is traveling at 455m/sec. what is the wavelength
Vlada [557]

Answer:

v=wavelength ×f

wavelength=v/f=455/655=0.694m

3 0
3 years ago
¿Qué significa que la investigación debe ser replicable? Analiza.
Inga [223]

Significa que en su investigación, debe proporcionar suficiente información para que otras personas que lean su investigación puedan hacer la investigación nuevamente.

4 0
3 years ago
A motorist traveling with a constant speed of 15 m/s (about 34 mi/h) passes a school-crossing corner, where the speed limit is 0
nikklg [1K]

Answer:

(a) 10 s

(b) 30 m/s

(c) 150 m

Explanation:

The motorist's position at time t is:

x = 15t

The officer's position at time t is:

x = ½ (3) t² = 1.5 t²

(a) When they have the same position, the time is:

15t = 1.5 t²

t = 0 or 10 s

(b) The officer's speed is:

v = 3t

v = 30 m/s

(c) The position is:

x = 15t = 150 m

6 0
3 years ago
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