Physics, 17.09.2019 00:40, jack104365

Can a particle moving with instantaneous speed 3.00 m/s on a path with radius of curvature 2.00 m have an acceleration of magnitude 6.00 m/s2? (b) can it have an acceleration of magnitude 4.00 m/s2? in each case, if the answer is yes, explain how it can happen; if the answer is no, explain why not.

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answered: gjfhjgg2848

Circular acceleration has 2 components, the centripetal and tangential acceleration.

The formula for calculating the centripetal acceleration of a particle is:

$a_{c}$ = v^2/r

Where v is the velocity or instantaneous speed and r is the radius.

$a_{c}$ = (3^2)/6 = 1.5 m/s^2

Assuming that the particle stays on the same circular path, another thing it can have is the tangential acceleration. While the centripetal acceleration is constant, tangential acceleration depends on the location of the particle in the circular path. So to answer A and B, yes we can have an acceleration of 6 and 4 respectively.

answered: mandilynn22

(a) So yes acceleration can have a magnitude $6 \rm m/s^{2}$

(b)acceleration can not have a value $4 \rm m/s^{2}$ as it is lesser than $4.5 \rm m/s^{2}$

Explanation:

Given information:

Instantaneous speed , $v=\rm 3.00\ m/s$

Radius of curvature of path $r=\rm 2\ m$

We know,

$\vec{a}_{total}= \vec{a}_c \ + \vec{ a}_t$

Where, a_c is centripetal acceleration and a_t is tangential acceleration

The centripetal acceleration of particle is

$a_c=\frac{v^{2} }{r} \rm m/s^{2}$ = $\frac{3^{2} }{2} \rm ms^{-2}$

$a_c=9/2=4.5\rm ms^{-2}$

Magnitude of total acceleration is

$\left | a \right |=\sqrt{a_c^{2} +a_t^{2} }$

$\sqrt{a_c^{2} +a_t^{2} } \geq a_x$ , which means total acceleration can be greater than or equal to $4.5 \rm m/s^{2}$

(a) So yes acceleration can have a magnitude $6 \rm m/s^{2}$

(b)acceleration can not have a value $4 \rm m/s^{2}$ as it is lesser than $4.5 \rm m/s^{2}$ .

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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.

answered: Guest

That would be the corect answer hope this

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Can a particle moving with instantaneous speed 3.00 m/s on a path with radius of curvature 2.00 m ha...