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Sladkaya [172]
3 years ago
8

A runner of mass 51.8 kg starts from rest and accelerates with a constant acceleration of 1.31 m/s^2 until she reaches a velocit

y of 5.47 m/s. She then continues running at this constant velocity. How long does the runner take to travel 165 m?
Physics
1 answer:
Stells [14]3 years ago
6 0

Answer:32.24 s

Explanation:

Given

mass of runner (m)=51.8 kg

Constant acceleration(a)=1.31 m/s^2

Final velocity (v)=5.47 m/s

Time taken taken to reach 5.47 m/s

v=u+at

5.47=0+1.31\times t

t=\frac{5.47}{1.31}=4.17 s

Distance traveled during this time is

s=ut+\frac{1}{2}at^2

s=\frac{1}{2}\times 1.31\times 4.17^2=11.42 m

So remaining distance left to travel with constant velocity=153.57 m

thus time =\frac{distance}{speed}

t_2=\frac{153.57}{5.47}=28.07 s

Total time=28.07+4.17=32.24 s

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1 \times 10^{-10.1} \mathrm{Wm}^{-2} is the intensity of the sound.

Answer: Option B

<u>Explanation:</u>

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This is the goal of the decibel scale (dB).  Because logarithm has the property of recording a large number and returning a small number, the dB scale is based on a logarithmic scale. The scale is defined so that the hearing threshold has intensity level of sound as 0.

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Where,

I = Intensity of the sound produced

I_{0} = Standard Intensity of sound of 60 decibels = 1 \times 10^{-12} \mathrm{Wm}^{-2}

So for 19 decibels, determine I as follows,

                   19 d B=(10 d B) \times \log _{10}\left(\frac{I}{1 \times 10^{-12} W m^{-2}}\right)

                  \log _{10}\left(\frac{1}{1 \times 10^{-12} \mathrm{Wm}^{-2}}\right)=\frac{19}{10}

                  \log _{10}\left(\frac{1}{1 \times 10^{-12} \mathrm{Wm}^{-2}}\right)=1.9

When log goes to other side, express in 10 to the power of that side value,

                  \left(\frac{I}{1 \times 10^{-12} W m^{-2}}\right)=10^{1.9}

                  I=1 \times 10^{-12} \mathrm{Wm}^{-2} \times 10^{1.9}=1 \times 10^{-12-1.9}=1 \times 10^{-10.1} \mathrm{Wm}^{-2}

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The whistling sound from the hearing aids represents that your hearing aids is working perfectly ad is known as the "feedback". So, the given statement is true.

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<u>Explanation:</u>

It's often sounds irritating when a hearing aids of your grandpa or Grandma whistles. especially, when they put them out of their ears. Actually, this feedback sound from hearing aids occur when the sounds from the outer side  bounces back to the microphone of the hearing aids.

The sound bounces back when it doesn't gets inside of your ear canal so that one can hear the sound through the hearing aid. When the sounds bounces back in the hearing aid, it get re-amplified and thus we hear the whistle sound which is known as the feedback of the device.

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