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diamong [38]
3 years ago
6

Rafael lives near a road at the bottom of the hill. His parents are concerned that soil will wash off the hill and rocks will fa

ll onto the road. Rafeal converts an investigation to find out if grass growing on a hillside will help stop soil erosion. He collects two samples of the same size and type of soil. One sample of soil has grass growing on it and the other does not. He places each sample of soil in a small tray. For his investigation, Rafael pours the same amount of water onto both samples of soil. He uses a large tray to collect water that may flow through the soil. Which practice shows the best way for Rafael to set up this investigation?

Physics
1 answer:
timofeeve [1]3 years ago
7 0

Answer:

The first picture/option

Explanation:

<em>The first picture/option shows the best way for Rafael to set up the investigation.</em>

The experimental set-up must be in a sloppy position in order to simulate the condition that is characteristic of the bottom of the hill where he stays. Hence, the two soil samples must be tilted in order to determine if the grass can help to stop the washing down of the soil from the top of the hill.

The two options following the first option do not fulfill the condition of the same treatment. Both soil samples must be subjected to the same treatment condition in order for the outcome to be valid. The last option could have also been admissible but did not simulate the hill/slope conditions of where Rafael stays.

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1. Bone has a Young’s modulus of about
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#1

As we know that

Y = \frac{stress}{strain}

now plug in all data into this

1.8\times 10^{10} = \frac{1.68 \times 10^8}{strain}

strain = 9.33 \times 10^{-3}

now from the formula of strain

strain = \frac{\Delta L}{L}

9.33 \times 10^{-3} = \frac{\Delta L}{0.54}

\Delta L = 5.04 \times 10^{-3} m

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#2

As we know that

pressure * area = Force

here we know that

Area = 3.53 \times 11.6 = 40.95 m^2

P = 0.2 atm = 0.2 \times 1.01 \times 10^5 = 2.02\times 10^4 Pa

now force is given as

F = 40.95 \times (2.02\times 10^4) = 8.27 \times 10^5 N

#3

As we know that density of water will vary with the height as given below

\rho = \frac{\rho_0}{1 - \frac{\Delta P}{B}}

here we know that

\Delta P = 2600 atm = 2600 \times 1.01 \times 10^5 = 2.63\times 10^8 Pa

B = 2.3 \times 10^9 N/m^2

now density is given as

\rho = \frac{1050}{1 - \frac{2.63\times 10^8}{2.3 \times 10^9}}

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#4

as we know that pressure changes with depth as per following equation

P = P_o + \rho g h

here we know that

P = 3 P_0

now we will have

3P_0 = P_0 + \rho g h

2P_0 = \rho g h

2(1.01 \times 10^5) = 1025 (9.81)(h)

here we will have

h = 20.1 m

so it is 20.1 m below the surface

#5

Here net buoyancy force due to water and oil will balance the weight of the block

so here we will have

mg = \rho_1V_1g + \rho_2V_2g

A(0.0476)979 = 922(A)(0.0476 - x) + 1000(A)(x)

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x = 3.48 cm

so it is 3.48 cm below the interface

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