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Gekata [30.6K]
4 years ago
7

A 12kg turtle rests on the bed of a zookeeper's truck, which is traveling down a country road at 55mi/h. The zookeeper spots a d

eer in the road, and slows to a stop in 12s. Assuming constant acceleration, what is the minimum coefficient of static friction between the turtle and truck bed surface that is needed to prevent the turtle from sliding?
Physics
1 answer:
gulaghasi [49]4 years ago
3 0

Answer:0.21

Explanation:

Given

mass of turtle m=12 kg

velocity of Truck v=55 mi/h\approx 24.58 m/s

time taken to stop t=12 s

using v=u+at

where v=final velocity

u=initial velocity

a=acceleration

t=time

0=24.58+a\cdot 12

a=-\frac{24.58}{12}

a=-2.04 m/s^2

therefore 2.04 is the deceleration experienced by turtle which is countered by frictional force

a=\mu g  ,where \mu =coefficient of friction

2.04=\mu \times 9.8

\mu =0.209\approx 0.21

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(1.a) The surface area being vibrated by the time the sound reaches the listener is 5,026.55 m².

(1.b) The intensity of the sound wave as it reaches the person listening is 0.02 W/m².

(1.c) The relative intensity of the sound as heard by the listener is 103 dB.

(2.a) The speed of sound if the air temperature is 15⁰C is 340.3 m/s.

(2.b) The frequency of the sound heard by the suspect is 614.3 Hz.

<h3>Surface area being vibrated</h3>

The surface area being vibrated by the time the sound reaches the listener is calculated as follows;

A = 4πr²

A = 4π x (20)²

A = 5,026.55 m²

<h3>Intensity of the sound</h3>

The intensity of the sound is calculated as follows;

I = P/A

I = (100) / (5,026.55)

I = 0.02 W/m²

<h3>Relative intensity of the sound</h3>

B = 10log(\frac{I}{I_0} )\\\\B = 10 \times log(\frac{0.02}{10^{-12}} )\\\\B = 103 \ dB

<h3>Speed of sound at the given temperature</h3>

v= 331.3\sqrt{1 + \frac{T}{273} } \\\\v = 331.3\sqrt{1 + \frac{15}{273} } \\\\v = 340.3 \ m/s

<h3>Frequency of the sound</h3>

The frequency of the sound heard is determined by applying Doppler effect.

f_o = f_s(\frac{v \pm v_0}{v \pm v_s} )

where;

  • -v₀ is velocity of the observer moving away from the source
  • -vs is the velocity of the source moving towards the observer
  • fs is the source frequency
  • fo is the observed frequency
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f_0 = f_s(\frac{v-v_0}{v- v_s} )

f_0 = 512(\frac{340.3 - 10}{340.3 - 65} )\\\\f_0 = 614.3 \ Hz

Learn more about intensity of sound here: brainly.com/question/17062836

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An oil tanker has collided with a smaller vessel, resulting in an oil spill in a large, calm-water bay of the ocean. You are inv
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Complete Question

An oil tanker has collided with a smaller vessel, resulting in an oil spill in a large, calm-water bay of the ocean. You are investigating the environmental effects of the accident and need to know the area of the spill. The tanker captain informs you that 18000 liters of oil have escaped and that the oil has an index of refraction of n = 1.1. The index of refraction of the ocean water is 1.33. From the deck of your ship you note that in the sunlight the oil slick appears to be blue. A spectroscope confirms that the dominant wavelength from the surface of the spill is 485 nm. Assuming a uniform thickness, what is the largest total area oil slick

Answer:

The  largest total area of the oil slick  A = 8.257 *10^{9} \ m^2

Explanation:

From the question we are told that

     The volume of oil the escaped is  V  = 18000 \ L

    The refractive index of oil is n_o = 1.1

     The refractive index of water is n_w = 1.33

      The wavelength of the light  is \lambda = 485 \ nm =  485 * 10^{-9} \ m

         

Generally the thickness of the oil for condition of constructive interference between the oil and the water is mathematically represented as

          d = m *\frac{\lambda}{2n_w}

Where is the order of interference of the light and it value ranges from 1, 2, 3,...n

It is usually take as 1 unless stated otherwise by the question

substituting value

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The are can be mathematically evaluated as

        A = \frac{V}{d}

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