Only the smallest particles of soil can be displaced by suspension because

2 answers:

mote1985 [20]3 years ago

7 0

The choices can be found elsewhere and as follows:

a. they are so small that they stay close to the ground due to the attractive properties of charged soil particles.

b. they are easily carried by the wind.

c. they easily dissolve in liquid droplets.

d. it is easier for then to roll along the small crevices in the ground.

I think the correct answer from the choices listed above is option B. Only the smallest particles of soil can be displaced by suspension because they are so small that they are easily carried by the wind. Hope this answers the question. Have a nice day. Feel free to ask more questions.

densk [106]3 years ago

3 0

the answer is b-they are easily carried by wind.

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Julianne went to a restaurant to have a taste of her favorite fried chicken and spaghetti. She drove 2 km, east and then 8.5 km,

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Julianne’s displacement from her origin is equal to 10.015 kilometers.

Given the following data:

Distance A = 2 km, East.

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To calculate Julianne’s displacement from her origin:

<h3>How to calculate displacement.</h3>

We would denote the two (2) unit vectors along the East and Northeast directions by i and j respectively.

Note: Northeast is at angle of 45° with the East.

In terms of vectors, the distances becomes:

Distance A = 2i

$Distance\;B=8.5 [(cos 45i + sin 45j)]\\\\Distance\;B=(\frac{8.5}{\sqrt{2} } i \;+\;\frac{8.5}{\sqrt{2} } j)$

For the resultant displacement:

$D^2 = [(2+\frac{8.5}{\sqrt{2} } )^2+ (\frac{8.5}{\sqrt{2} } )^2\\\\D =\sqrt{[(2+\frac{8.5}{\sqrt{2} } )^2+ (\frac{8.5}{\sqrt{2} } )^2} \\\\D=2+\frac{8.5}{\sqrt{2} } + \frac{8.5}{\sqrt{2} }$