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3 years ago

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At time t = 0 s, a wheel has an angular displacement of zero radians and an angular velocity of +26 rad/s. The wheel has a const

ant acceleration of -0.43 rad/s2. In this situation, the time t (after t = 0 s), at which the kinetic energy of the wheel is twice the initial value, is closest to...
150s How do you solve for this?? I keep getting a negative time value

1 answer:

Alex777 [14]3 years ago

6 0

Answer:

t= 25.04 s

Explanation:

Given that

Initial angular speed ,ωi= 26 rad/s

Angular acceleration ,α=0.43 rad/s²

We know that kinetic energy of the wheel given as

$KE=\dfrac{1}{2}I\omega^2$

I=Mass monent of inertia

ω=Angular speed

If kinetic energy is two time then the final angular speed

$\omega_f=\sqrt{2}\ \omega_i$

We know that

$\omega_f = \omega_i + \alpha\ t$

$\sqrt{2}\ \omega_i= \omega_i +0.43\times t$

t= 25.04 s

After 25.04 s then kinetic energy will become two times of the initial kinetic energy.

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A sound is recorded at 19 decibels. What is the intensity of the sound?

sp2606 [1]

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Answer: Option B

<u>Explanation:</u>

The range of sound intensity that people can recognize is so large (including 13 magnitude levels). The intensity of the weakest audible noise is called the hearing threshold. (intensity about $1 \times 10^{-12} \mathrm{Wm}^{-2}$ ). Because it is difficult to imagine numbers in such a large range, it is advisable to use a scale from 0 to 100.

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

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

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

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

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

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