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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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irius, the brightest star in the sky, is 2.6 parsecs (8.6 light-years) from Earth, giving it a parallax of 0.379 arcseconds. Ano
Norma-Jean [14]

The actual distance of Regulus from Earth is 23.81 parsecs.

Given:

Parallax of Regulus, p = 0.042 arc seconds

Calculation:

When an observer changes their position, an apparent change in the object's position takes place. This change can be calculated using the angle ( or semi-angle) made by the observer and object i.e. the angle made between the two lines of observation from the object to the observer.

Thus from the relation of parallax of a celestial body we get:

S = 1/ tan p ≈ 1 / p

where S is the actual distance between the object and the observer

            p is the parallax angle observed

Here for Regulus, we get:

S = 1 / p

  = 1 / (0.042)                                     [ 1 parsecs = 1 arcseconds ]

  = 23.81 parsecs

We know that,

1 parsecs = 3.26 light-years = 206,000 AU

Converting the actual distance into light years we get:

23.81 parsecs = 23.81 × (3.26 light yrs) = 77.658 light-years

Therefore, the actual distance of Regulus from Earth is 23.81 parsecs which is 77.658 in light years.

Learn more about astronomical units here:

<u>brainly.com/question/16471213</u>

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6 0
2 years ago
Plz help me to find my science matching... <br>​
s2008m [1.1K]

Answer:

fibrous =potato

taproot =radish

stilt =maize and sugar cane

5 0
3 years ago
A circular loop of wire lies flat on a level table top in a region where the magnetic field vector points straight upward. The m
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The magnetic field direction and direction of induced current in a wire are related by the right hand grip rule. Since the magnetic field was upwards, the thumb points upwards and the fingers curl around it. When viewed from above, it is seen as a current flowing in the counter clockwise direction.
4 0
3 years ago
As a quality test of ball bearings, you drop bearings (small metal balls), with zero initial velocity, from a height of 1.94 m i
11Alexandr11 [23.1K]

Answer:

The average acceleration of the bearings is 0.77\times10^{3}\ m/s^2

Explanation:

Given that,

Height = 1.94  m

Bounced height = 1.48 m

Time interval t=14.86\times10^{-3}\ s

Velocity of the ball bearing just before hitting the steel plate

We need to calculate the velocity

Using conservation of energy

mgh=\dfrac{1}{2}mv^2

Put the value into the formula

9.8\times1.94=\dfrac{1}{2}\times v^2

v=\sqrt{2\times9.8\times1.94}

v=6.166\ m/s

Negative as it is directed downwards

After bounce back,

We need to calculate the velocity

Using conservation of energy

mgh=\dfrac{1}{2}mv^2'

Put the value into the formula

9.8\times1.48=\dfrac{1}{2}\times v^2'

v'=\sqrt{2\times9.8\times1.48}

v'=5.38\ m/s

We need to calculate the average acceleration of the bearings while they are in contact with the plate

Using formula of acceleration

a=\dfrac{v-v'}{t}

Put the value into the formula

a=\dfrac{5.38-(-6.166)}{14.86\times10^{-3}}

a=776.98\ m/s^2

a=0.77\times10^{3}\ m/s^2

Hence,The average acceleration of the bearings is 0.77\times10^{3}\ m/s^2

6 0
3 years ago
What does the length of a vector arrow represent?
nalin [4]
The length of a vector arrow represents an magnitude
6 0
2 years ago
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