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4 years ago

Which of these factors determines soil texture?

Physics

2 answers:

STatiana [176]4 years ago

8 0

It is B particle sizes

lys-0071 [83]4 years ago

5 0

Option B, particle sizes, is the right answer.

Soil texture is a distribution apparatus practiced both in the field and lab to prepare soil properties based on their natural texture. Soil texture can be circumscribed adopting qualitative techniques such as texture by touch, and quantitative approaches such as the hydrometer technique. Soil texture concentrates on the particles that are less than two millimeters in broadness which incorporate clay, silt and sand.

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According to the barometer in the attachment below, which statement is true?

mixer [17]

Your answer is A

Pls mark me brainiest and I sure hope this helps you

8 0

2 years ago

Read 2 more answers

A block of mass 1.5 hangs at the of end of a weight cord suspended from the ceiling.what is the tension in the cord, and with wh

Len [333]

The tension in the cord is 14.7 N and the force of pull of the cord is 14.7 N, assuming the block is stationary.

<h3>What is the tension in the cord?</h3>

The tension in the cord is calculated as follows;

T = ma + mg

where;

a is the acceleration of the block
g is acceleration due to gravity
m is mass of the block

T = m(a + g)

T = 1.5(a + 9.8)

T = 1.5a + 14.7

Thus, the tension in the cord is (1.5a + 14.7) N.

If the block is at rest, the tension is 14.7 N.

<h3>Force of the force</h3>

The force with which the cord pulls is equal to the tension in the cord

F = T = m(a + g)

F = (1.5a + 14.7) N

If the block is stationary, a = 0, the tension and force of pull of the cord = 14.7 N.

Thus, the tension in the cord is 14.7 N and the force of pull of the cord is 14.7 N, assuming the block is stationary.

Learn more about tension here: brainly.com/question/187404

#SPJ1

4 0

1 year ago

A distant galaxy is determined to be 150 million light years distant and moving away from us; using the Hubble law determine its

dlinn [17]

Your question kind of petered out there towards the end and you didn't specify
the terms, so I'll pick my own.

The "Hubble Constant" hasn't yet been pinned down precisely, so let's pick a
round number that's in the neighborhood of the last 20 years of measurements:

   <em>70 km per second per megaparsec</em>.

We'll also need to know that 1 parsec = about 3.262 light years.

So the speed of your receding galaxy is

   (Distance in LY) x (1 megaparsec / 3,262,000 LY) x (70 km/sec-mpsc) =

      (150 million) x (1 / 3,262,000) x (70 km/sec) =

   <em>3,219 km/sec </em>in the direction away from us (rounded)

4 0

3 years ago

A car accelerates at a rate of 13m/s^2[S]. If the car's initial velocity is 120km/h[N]. What will its final velocity be in m/s,

Delvig [45]

Answer:

the final velocity of the car is 59.33 m/s [N]

Explanation:

Given;

acceleration of the car, a = 13 m/s²

initial velocity of the car, u = 120 km/h = 33.33 m/s

duration of the car motion, t = 2 s

The final velocity of the car in the same direction is calculated as follows;

v = u + at

where;

v is the final velocity of the car

v = 33.33 + (13 x 2)

v = 59.33 m/s [N]

Therefore, the final velocity of the car is 59.33 m/s [N]

6 0

3 years ago

Each plate of a parallel‑plate capacitor is a square of side 3.63 cm, and the plates are separated by 0.473 mm. The capacitor is

NARA [144]

Answer:

E= 55.53 x 10³ V/m

Explanation:

Given that

a= 3.63 cm

Area ,A= a²

distance ,d= 0.473 mm

Stored energy ,U = 8.49 nJ

Value of capacitor given as

$C=\dfrac{\varepsilon _oA}{d}$

By putting the values

$C=\dfrac{8.85\times 10^{-12}\times 3.63^2\times 10^{-4}}{0.473\times 10^{-3}}$

C=2.46 x 10⁻¹¹ F

$U=\dfrac{1}{2}CV^2$

V=Voltage difference

$V=\sqrt{\dfrac{2U}{C}}$

$V=\sqrt{\dfrac{2\times 8.49\times 10^{-9}}{2.46\times 10^{-11}}}$

V=26.27 V

V= E d

E=Electric filed

26.27 = E x 0.473 x 10⁻³

E= 55.53 x 10³ V/m

7 0

3 years ago