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dexar [7]
3 years ago
13

You should increase the distance between your car and the vehicle ahead when you:

Physics
1 answer:
ELEN [110]3 years ago
8 0
<span>You should increase the distance between your car and the vehicle ahead when you are being tailgated by another driver.
Tailgated is a condition when the vehicle behind you is very close to you that there is no distance between your vehicle and the vehicle behind you. In that case you should increase distance from the vehicle ahead to avoid collision in any case. These days people were mostly in hurry and everyone want to go ahead, so tailgating is very common.</span>
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Which of the following is an example of changing physical capital?
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a car moving with a speed of 10 metre per second is accelerator at the rate of 2 metre per second square find its velocity after
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a = ( v(2) - v(1) ) ÷ ( t(2) - t(1) )

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3 years ago
A piano wire with mass 2.95 g and length 79.0 cm is stretched with a tension of 29.0 N . A wave with frequency 105 Hz and amplit
Romashka-Z-Leto [24]

The concept needed to solve this problem is average power dissipated by a wave on a string. This expression ca be defined as

P = \frac{1}{2} \mu \omega^2 A^2 v

Here,

\mu = Linear mass density of the string

\omega =  Angular frequency of the wave on the string

A = Amplitude of the wave

v = Speed of the wave

At the same time each of this terms have its own definition, i.e,

v = \sqrt{\frac{T}{\mu}} \rightarrow Here T is the Period

For the linear mass density we have that

\mu = \frac{m}{l}

And the angular frequency can be written as

\omega = 2\pi f

Replacing this terms and the first equation we have that

P = \frac{1}{2} (\frac{m}{l})(2\pi f)^2 A^2(\sqrt{\frac{T}{\mu}})

P = \frac{1}{2} (\frac{m}{l})(2\pi f)^2 A^2 (\sqrt{\frac{T}{m/l}})

P = 2\pi^2 f^2A^2(\sqrt{T(m/l)})

PART A ) Replacing our values here we have that

P = 2\pi^2 (105)^2(1.8*10^{-3})^2(\sqrt{(29.0)(2.95*10^{-3}/0.79)})

P = 0.2320W

PART B) The new amplitude A' that is half ot the wavelength of the wave is

A' = \frac{1.8*10^{-3}}{2}

A' = 0.9*10^{-3}

Replacing at the equation of power we have that

P = 2\pi^2 (105)^2(0.9*10^{-3})^2(\sqrt{(29.0)(2.95*10^{-3}/0.79)})

P = 0.058W

8 0
3 years ago
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