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Lady_Fox [76]
3 years ago
14

A soil sample is 25 percent sand, 55 percent clay, and 20 percent silt. Use the following soil texture triangle to determine the

type of soil present in this sample.
Select one:
a. Clay
b. Silty loam
c. Sandy clay
d. Sandy clay loam

Physics
1 answer:
vova2212 [387]3 years ago
5 0
I believe the answer is A. clay
The amount of clay alone determine which type of soil it would be according to the Triangle.
At that triangle, we could see that the maximum amount of clay to be considered as 'sandy clay' is 50.
Since the amount of sand surpassed that, it automatically fall into clay type even without having to look to other composition.
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A train travels 55 kilometers in 2 hours, and then 52 kilometers in 5 hours. What is its average speed?
bulgar [2K]

The formula for working out speed is distance ÷ time.

55 km ÷ 2 hours = 27.5 km/h (average speed for first part of journey)

52km ÷ 5 hours = 10.4 km/h (average speed for second part of journey)

(27.5 + 10.4) ÷ 2 = 18.95 km/h (average speed throughout the journey)

4 0
3 years ago
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A roller coaster car starts from rest at the top of a hill 15 m high and rolls down to ground level. From there it starts into a
Softa [21]

Answer:

955.5N

Explanation:

The normal force is given by the difference between the centripetal force and gravity at the top of the loop:

F_N = F_C - F_G = m\frac{v^{2} }{r} - mg

mass m = 65kg

radius of the loop r = 4m

velocity v = ?

g = 9.8 m/s²

To find the centripetal force, you need to find the velocity of the car at the top of the loop.

Use energy conservation:

E_{tot}=mgh + \frac{1}{2} mv^{2}

At the top of the hill:

E_{tot}= mgh_{hill}

At the top of the loop:

E_{tot}=mgh_{loo}_p +\frac{1}{2} m v^{2}

Setting both energies equal and canceling the mass m gives:

gh_{hill} = gh_{loo}_p + \frac{1}{2} v^{2}

Solving for v:

v^{2} = 2g(h_{hill}-h_{loo}_p)

Using v in the first equation:

F_N = \frac{2mg(h_{hill}-h_{loo}_p)}{r} - mg

F_N = 955.5N

7 0
3 years ago
particle of mass 59 g and charge 51 µC is released from rest when it is 32 cm from a second particle of charge −14 µC. Determine
AleksAgata [21]

Answer:

The initial acceleration of the 59g particle is 1062.7\frac{m}{s^{2}}

Explanation:

Newton's second laws relates acceleration (a), net force(F) and mass (m) in the next way:

F=ma (1)

We already know the mass of the particle so we should find the electric force on it to use on (1), the magnitude of the electric force between two charged objects by Columb's law is:

F=k\frac{\mid q_{1}q_{2}\mid}{r^{2}}

with q1 and q2 the charge of the particles, r the distance between them and k the constant k=9.0\times10^{9}\,\frac{Nm^{2}}{C^{2}}. So:

F=(9.0\times10^{9})\frac{\mid (51\times10^{-6})(-14\times10^{-6})\mid}{0.32^{2}}

F=62.7 N

Using that value on (1) and solving for a

a=\frac{F}{m}=\frac{62.7}{0.059}=1062.7\frac{m}{s^{2}}

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3 years ago
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Keith_Richards [23]

vertical! or in other words, up!


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3 years ago
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3 years ago
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