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ZanzabumX [31]
2 years ago
15

What are some characteristics of soil? ​

Physics
2 answers:
Anarel [89]2 years ago
6 0

Soils are composed of organic matter (stuff that used to be alive, like plants and animals) and small inorganic matter. There are three basic soil types: sand, silt, and clay. Sand is comprised of tiny rock fragments and is the roughest in texture. Clay becomes sticky or greasy when wet, and very hard when dry.

jeka942 years ago
4 0

Explanation:

1. Soils provide Support to plants

2. Helpful in Economic Activity (Farming) and many more...

3. Soil are home for decomposers.

4. They're helpful for increasing Greenery..

hope it helps

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8. A 2 kg flower pot weighing 20 N falls from a window ledge.
Alina [70]

The force of the air resistance is 4 N.

The given parameters;

  • mass of the flower pot, m = 2 kg
  • weight of the flower pot, W = 20 N

Let the air resistance = F

Apply Newton's second law of motion to determine the force of the air resistance acting upward to oppose the motion of the pot falling downwards.

\Sigma F = ma\\\\W - F = ma\\\\a = \frac{W - F}{m} \\\\8 = \frac{20 - F}{2} \\\\20 - F = 16\\\\F = 20 - 16\\\\F = 4 \ N

Thus, the force of the air resistance is 4 N.

Learn more here: brainly.com/question/19887955

8 0
3 years ago
How many electrons does eaach shell hold?
postnew [5]
Each electron holds 2(n2)
3 0
3 years ago
While at a construction site, you see a crane lift a 1000kg steel beam up to a height of 10m in a time period of 5.0 seconds. A
storchak [24]

Answer:

Car has more power output than crane      

Explanation:

We have given that mass of the crane m = 1000 kg

Height through which crane lift the steel beam h = 10 m

Acceleration due to gravity g=9.8m/sec^2

So work done by crane W=mgh=1000\times 9.8\times 10=98000j

Time period is given as t = 5 sec

We know that power P=\frac{W}{t}=\frac{98000}{5}=19600watt

Now mass of the car = 1000 kg

Initial velocity u = 0 m /sec

Final velocity v = 10 m/sec

We know that work done is equal to the change in kinetic energy

So work done =\frac{1}{2}mv^2-\frac{1}{2}mu^2

=\frac{1}{2}\times 1000\times 10^2-\frac{1}{2}\times 1000\times 0^2=50000j

Time ids given as t = 2 sec

So power P=\frac{W}{t}=\frac{50000}{2}=25000watt

So car has more power output than crane

8 0
3 years ago
List down atleast five activities where friction gave you an undesirable experience
Step2247 [10]

Answer:

Rug burn, Indian burn done to you by a friend, friction from the road causes your car to accelerate at a slower rate, The cylinder heads in an engine, When trying to move a heavy object across a rough surface

Explanation:

7 0
3 years ago
Read 2 more answers
A projectile is launched horizontally from the top a 35.2m high cliff and lands a distance of 107.6m from the base of the cliff.
tankabanditka [31]

Answer:

v_o=40.14\ m/s

Explanation:

<u>Horizontal Launch </u>

It happens when an object is launched with an angle of zero respect to the horizontal reference. It's characteristics are:

  • The horizontal speed is constant and equal to the initial speed v_o
  • The vertical speed is zero at launch time, but increases as the object starts to fall
  • The height of the object gradually decreases until it hits the ground
  • The horizontal distance where the object lands is called the range

We have the following formulas

\displaystyle v_x=v_o

\displaystyle x=v_o.t

\displaystyle v_y=g.t

\displaystyle y=\frac{gt^2}{2}

Where v_o is the initial horizontal speed, v_y is the vertical speed, t is the time, g is the acceleration of gravity, x is the horizontal distance, and y is the height.

If we know the initial height of the object, we can compute the time it takes to hit the ground by using

\displaystyle y=\frac{gt^2}{2}

Rearranging and solving for t

\displaystyle 2y=gt^2

\displaystyle t^2=\frac{2\ y}{g}

\displaystyle t=\sqrt{\frac{2\ y}{g}}

We then replace this value in

\displaystyle x=v_o.t

To get

\displaystyle v_o=\frac{x}{t}

\displaystyle v_o=\frac{x}{\sqrt{\frac{2y}{g}}}

\displaystyle v_o=\sqrt{\frac{g}{2y}}.x

The initial speed depends on the initial height y=32.5 m, the range x=107.6 m and g=9.8 m/s^2. Computing v_o

\displaystyle v_o=\sqrt{\frac{9.8}{2(35.2)}}\ 107.6

The launch velocity is  

\boxed{v_o=40.14\ m/s}

7 0
3 years ago
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