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alina1380 [7]
3 years ago
14

Which particle would have the slowest rate of deposition? A.round particle B.very large particle C.particle with sharp ends D.pa

rticle with a high density
Physics
2 answers:
Neko [114]3 years ago
8 0

Answer:

C.particle with sharp ends is the correct answer

Explanation:

particle with sharp ends would have the slowest rate of deposition.

The frictional force which is important that gives resistance to the movement will be higher for irregularly shaped particles and this is the reason sharp ends particles have the slowest rate of deposition.

Deposition happens when the particles settle down on a solid surface.

Density is directly proportional to deposition it means that the particles with high density settle faster and particle with sharp end has lower density so that the reason they have slowest deposition rate.

lisabon 2012 [21]3 years ago
3 0

The particle with sharp ends have the slowest rate of deposition  

Answer: Option C  

<u>Explanation:</u>

          As per aerosol physics, deposition is a process where aerosol particles accumulate or settle on solid surfaces. Thereby, it reduces the concentration of particles in the air. Deposition velocity (rate of deposition) defines from F = vc, where v is deposition rate, F denotes flux density and c refers concentration.  

          Deposition velocity is slowest for particles of intermediate-sized particles because the frictional force offers resistance to the flow. Density is directly proportional to the deposition rate so clearly shows that high-density particles settle faster. Due to friction, round and large-sized particles deposit faster than oval/flattened sediments.  

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vA 61.2-kg circus performer is fired from a cannon that is elevated at an angle of 57.8 ° above the horizontal. The cannon uses
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Answer:

The effective spring constant of the firing mechanism is 1808N/m.

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First, we can use kinematics to obtain the initial velocity of the performer. Since we know the angle at which he was launched, the horizontal distance and the time in which it's traveled, we can calculate the speed by:

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v_0=\frac{20.8m}{(2.60s)\cos57.8\°}\\\\v_0=15.0m/s

Now, we can use energy to obtain the spring constant of the firing mechanism. By the conservation of mechanical energy, considering the instant in which the elastic band is at its maximum stretch as t=0, and the instant in which the performer flies free of the bands as final time, we have:

E_0=E_f\\\\U_e=K\\\\\frac{1}{2}kx^2=\frac{1}{2}mv^2\\\\\implies k=\frac{mv^2}{x^2}

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