Physics, 24.06.2019 19:00, xxcecccc

Human centrifuges are used to train military pilots and astronauts in preparation for high-g maneuvers. a trained, fit person wearing a g-suit can withstand accelerations up to about 9g (88.2 m/s2) without losing consciousness. hint (a) if a human centrifuge has a radius of 5.15 m, what angular speed (in rad/s) results in a centripetal acceleration of 9g? rad/s (b) what linear speed (in m/s) would a person in the centrifuge have at this acceleration? m/s

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answered: alexabdercmur

(a) 4.14 rad/s^2

The relationship beween centripetal acceleration and angular speed is

$a=\omega^2 r$

where

$\omega$ is the angular speed

r is the radius of the circular path

Here we gave

a = 9g = 88.2 m/s^2 is the centripetal acceleration

r = 5.15 m is the radius

Solving for $\omega$ , we find:

$\omega = \sqrt{\frac{a}{r}}=\sqrt{\frac{88.2 m/s^2}{5.15 m}}=4.14 rad/s^2$

(b) 21.3 m/s

The relationship between the linear speed and the angular speed is

$v=\omega r$

where

v is the linear speed

$\omega$ is the angular speed

r is the radius of the circular path

In this problem we have

$\omega=4.14 rad/s$

r = 5.15 m

Solving the equation for v, we find

v=(4.14 rad/s)(5.15 m)=21.3 m/s

answered: hilllz4793

Explanation:

(a) Centripetal acceleration, $a=9g=88.2\ m/s^2$

Radius, r = 5.15 m

Let $\omega$ is the angular speed. The relation between the angular speed and angular acceleration is given by :

$a=\omega^2 r$

$\omega=\sqrt{\dfrac{a}{r}}$

$\omega=\sqrt{\dfrac{88.2}{5.15}}$

$\omega=4.13\ rad/s$

(b) Let v is the linear speed of the person in the centrifuge have at this acceleration. It is given by :

$v=r\times \omega$

$v=5.15\times 4.13$

v = 21.26 m/s

Hence, this is the required solution.

answered: Guest

f=ma newton

f=100x2.5=250n

answered: Guest

scattering of light by tiny particles in air to produce the colors we see on sky, is dependent on the following factors:

(i) wavelength of light

(ii) size of the scattering particle

(iii) distance traveled by light through the atmosphere.

according to rayleigh's law of scattering, the amount of scattering is inversely proportional to the fourth power of its wavelength. shorter the wavelength, greater is its scattering. hence violet/blue light is scattered more than red.

sun's light is a composite light ( visible light) with wavelengths in the range 400 nm (violet) to 700 nm (red). as sunlight passes through our atmosphere, the molecules of air scatter light according to their wavelengths. during sunrise and sunset, the sun's position is below the horizon. hence light from the sun travels a greater distance in the atmosphere when compared to its position a noon.

as sunlight enters the earth's atmosphere, blue light is scattered away and at the end of its journey in the sky, only red and orange wavelengths remain in the beam.

hence at sunrise and sunset, the sky appears reddish orange.

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Human centrifuges are used to train military pilots and astronauts in preparation for high-g maneuve...