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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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7 0
3 years ago
Derive an expression for the energy needed to launch an object from the surface of Earth to a height h above the surface.Ignorin
Sergio [31]

Answer:

Explanation:

Potential energy on the surface of the earth

= - GMm/ R

Potential at height h

=  - GMm/ (R+h)

Potential difference

= GMm/ R -  GMm/ (R+h)

= GMm ( 1/R - 1/ R+h )

= GMmh / R (R +h)

This will be the energy needed  to launch an object from the surface of Earth to a height h above the surface.

Extra  energy is needed to get the same object into orbit at height h

= Kinetic energy of the orbiting object at height h

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= 1/2 x GMm / ( R + h)

8 0
4 years ago
In a certain region of space, the electric field is zero. from this fact, what can you conclude about the electric potential in
eduard
The answer is that it is constant. The relation between electric field and electric potential is given as, E=  -gradient (V).  The E, the partial rate of change of Electric potential, in the equation implies that the V, the partial differential of the potential of the three-dimensional space (assuming it is considered) is constant. 
5 0
3 years ago
A 11.0 kg satellite has a circular orbit with a period of 1.80 h and a radius of 7.50 × 106 m around a planet of unknown mass. I
Anuta_ua [19.1K]

Answer:

Explanation:

Expression for times period of a satellite can be given as follows

Time period T = 1.8 x 60 x 60

= 6480

T² = \frac{4\times \pi^2\times r^3}{GM} where T is time period , r is radius of orbit , G is gravitational constant and M is mass of the satellite.

6480² = 4 x 3.14² x 7.5³ x 10¹⁸ / GM

GM = 4 x 3.14² x 7.5³ x 10¹⁸ / 6480²

= 3.96 X 10¹⁴

Expression for acceleration due to gravity

g = GM / R² where R is radius of satellite

20 = 3.96 X 10¹⁴ / R²

R² = 3.96 X 10¹⁴ / 20

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8 0
3 years ago
a 28.7 KG sled is pulled forward with a 63 net force across the ground with MK equals 0.169 what is the acceleration of the sled
abruzzese [7]

Answer:

0.54\ \text{m/s}^2

Explanation:

F = Force on the sled = 63 N

m = Mass of sled = 28.7 kg

\mu_k = Coefficient of kinetic friction = 0.169

g = Acceleration due to gravity = 9.81\ \text{m/s}^2

The force balance of the system is given by

F-\mu_k mg=ma\\\Rightarrow a=\dfrac{F-\mu_k mg}{m}\\\Rightarrow a=\dfrac{63-0.169\times 28.7\times 9.81}{28.7}\\\Rightarrow a=0.54\ \text{m/s}^2

The acceleration of the sled is 0.54\ \text{m/s}^2.

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3 years ago
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