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2 years ago

The difference in particle size of the soil constituents determine its texture.

1 answer:

3 0

Answer:Soil texture depends on the size of individual soil particles and is determined by the relative proportions of particle sizes that make up the soil. The texture affects the amount of water that can be absorbed for use by plants and animals.

Explanation:

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If the displacement of a horizontal mass-spring system was doubled, the elastic potential energy in the system would change by a

bixtya [17]

Answer:

K'=4K

Explanation:

The electric potential energy is given by :

$E=\dfrac{1}{2}kx^2$

Where

k is spring constant

x is compression or extension in the spring

If the displacement of a horizontal mass-spring system is doubled, x'= 2x

New elastic potential energy :

$E'=\dfrac{1}{2}kx'^2\\\\E'=\dfrac{1}{2}k(2x)^2\\\\=\dfrac{1}{2}k\times 4x^2\\\\K'=4\times \dfrac{1}{2}kx^2\\\\K'=4K$

So, new elastic potential energy 4 times the initial elastic potential energy.

3 0

3 years ago

A student is investigating potential and kinetic energy by stretching a spring across a table when the student lets go the sprin

nikitadnepr [17]

Answer: could be B.

Explanation:

7 0

3 years ago

A 40.0 kg wheel, essentially a thin hoop with radius 0.810 m, is rotating at 438 rev/min. It must be brought to a stop in 21.0 s

garik1379 [7]

Answer:

(a) Workdone = -27601.9J

(b) Average required power = 1314.4W

Explanation:

Mass of hoop,m =40kg

Radius of hoop, r=0.810m

Initial angular velocity Winitial=438rev/min

Wfinal=0

t= 21.0s

Rotation inertia of the hoop around its central axis I= mr²

I= 40 ×0.810²

I=26.24kg.m²

The change in kinetic energy =K. E final - K. E initail

Change in K. E =1/2I(Wfinal² -Winitial²)

Change in K. E = 1/2 ×26.24[0-(438×2π/60)²]

Change in K. E= -27601.9J

(a) Change in Kinetic energy = Workdone

W= 27601.9J( since work is done on hook)

(b) average required power = W/t

=27601.9/21 =1314.4W

4 0

3 years ago

Read 2 more answers

HELPPP PLEASE !!!!!!!

vlabodo [156]

Explanation:

a) d = ½.a.t²

200 = ½(4)t²

200 = 2t²

t² = 200/2

t² = 100

t =√100 = 10 s

b) Vt = a. t

= 4(10)

= 40 m/s

c) V av. = d/t = 200/10 = 20m/s

6 0

2 years ago

Two manned satellites approaching one another at a relative speed of 0.550 m/s intend to dock. The first has a mass of 2.50 ✕ 10

NNADVOKAT [17]

Answer: Their final relative velocity is -0.412 m/s.

Explanation:

According to the law of conservation,

$m_{1}v_{1} + m_{2}v_{2} = (m_{1} + m_{2})v$

Putting the given values into the above formula as follows.

$m_{1}v_{1} + m_{2}v_{2} = (m_{1} + m_{2})v$

$2.50 \times 10^{3} kg \times 0 m/s + 7.50 \times 10^{3} kg \times -0.550 m/s = (2.50 \times 10^{3} kg + 7.50 \times 10^{3} kg)v$

$-4.12 \times 10^{3} kg m/s = (10^{4} kg) v$

v = $\frac{-4.12 \times 10^{3} kg m/s}{10^{4} kg}$

= -0.412 m/s

Thus, we can conclude that their final relative velocity is -0.412 m/s.

8 0

3 years ago

The difference in particle size of the soil constituents determine its texture.​

The difference in particle size of the soil constituents determine its texture.