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3 years ago

14

Wind erosion can be reduced by _____.

1 answer:

yanalaym [24]3 years ago

4 0

Using land according to its capability. protect the soil surface with some form of cover. control runoff before it develops into an erosive force

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The drawing shows a large cube (mass = 28.6 kg) being accelerated across a horizontal frictionless surface by a horizontal force

MrRissso [65]

Answer:

P= 454.11 N

Explanation:

Since P is the only horizontal force acting on the system, it can be defined as the product of the acceleration by the total mass of the system (both cubes).

$P= (M+m)*a\\a = \frac{P}{28.6 +4.3}\\a = \frac{P}{32.9}$

The friction force between both cubes (F) is defined as the normal force acting on the smaller cube multiplied by the coefficient of static friction. Since both cubes are subject to the same acceleration:

$F = m * a*\mu \\F= 4.3*0.710*\frac{P}{32.9}\\F=3.053*\frac{P}{32.9}$

In order for the small cube to not slide down, the friction force must equal the weight of the small cube:

$3.053*\frac{P}{32.9} = 4.3 * g\\\\P = \frac{4.3*9.8*32.9}{3.053} \\P= 454.11 N$

The smallest magnitude that P can have in order to keep the small cube from sliding downward is 454.11 N

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3 years ago

Can anyone explain how tides work

Mila [183]

The position of the sun and the moon affect how high the tide is

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3 years ago

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if you knew the number of valence electrons in a nonmetal atom, how would you determine the valence of the element?

Vanyuwa [196]

It would be funny because . I will not be good

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3 years ago

A car left point a at 7:30 am and arrived at point b, 162 miles away at 10:30 am. what was its average speed in miles per hour?5

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3 years ago

There is a moon orbiting an Earth-like planet. The mass of the moon is 9.58 × 1022 kg, the center-to-center separation of the pl

kaheart [24]

Answer:

= 4.38 × 10³⁴kgm²/s

Explanation:

Given that,

mass of moon m = 9.5 × 10²²kg

Orbital radius r = 4.28 × 10⁵km

Orbital period T = 28.9days

T = 28.9 × 24 × 60 × 60

= 2,496,960s

Angular momentum of the moon about the planet

L = mvr

L = mr²w

$L = mr^2\frac{2\pi }{T} \\\\L = \frac{9.5 \times 10^2^2 \times(4.28\times10^8)^2\times2\times3.14}{2496960} \\\\L = 4.389.5 \times 10^3^4kgm^2/s$

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3 years ago

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