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zhuklara [117]
2 years ago
6

Forest protect the soil from erosion ? True or False

Physics
2 answers:
Deffense [45]2 years ago
6 0

Answer:

True

Explanation:

Forest protect the soil from erosion because the trees in the forest hold the soil together not allowing it to erode.

Ivenika [448]2 years ago
4 0

Answer:

<h2><em>True</em></h2>

Forests helps in binding the soil together. Thus, prevent soil from erosion.

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WHAT ARE THREE TECHNICAL ADVANCEMENTS
Snezhnost [94]

the iphone

the airplane

5G

6 0
2 years ago
In the Bohr model of the atom, atomic electrons approximatately 'orbit' the nucleus. The hydrogen atom consists of a proton of m
lara31 [8.8K]

Answer:

F=3.61\times 10^{-47}\ N

Explanation:

Mass of a proton, m_p=1.67\times 10^{-27}\ kg

Mass of an electron, m_e=9.11\times 10^{-31}\ kg

The distance between the electron and the proton is, r=5.3\times 10^{-11}\ m

We need to find the mutual attractive gravitational force between the electron and proton. The gravitational force is given by :

F=G\dfrac{m_em_p}{r^2}

Where G is the universal Gravitational constant

F=6.67\times 10^{-11}\times \dfrac{9.11\times 10^{-31}\times 1.67\times 10^{-27}}{(5.3\times 10^{-11})^2}\\\\F=3.61\times 10^{-47}\ N

So, the force between the electron and proton is 3.61\times 10^{-47}\ N.

7 0
3 years ago
Interactive Solution 6.39 presents a model for solving this problem. A slingshot fires a pebble from the top of a building at a
mariarad [96]

(a) 29.8 m/s

To solve this problem, we start by analyze the vertical motion first. This is a free fall motion, so we can use the following suvat equation:

v_y^2 - u_y^2 = 2as

where, taking upward as positive direction:

v_y is the final vertical velocity

u_y = 0 is the initial vertical velocity (zero because the pebble is launched horizontally)

a=g=-9.8 m/s^2 is the acceleration of gravity

s = -25.0 m is the displacement

Solving for vy,

v_y = \sqrt{u^2+2as}=\sqrt{0+2(-9.8)(-25)}=-22.1 m/s (downward, so we take the negative solution)

The pebble also have a horizontal component of the velocity, which remains constant during the whole motion, so it is

v_x = 20.0 m/s

So, the final speed of the pebble as it strikes the ground is

v=\sqrt{v_x^2+v_y^2}=\sqrt{20.0^2+(-22.1)^2}=29.8 m/s

(b) 29.8 m/s

In this case, the pebble is launched straight up, so its initial vertical velocity is

u_y = 20.0 m/s

So we can find the final vertical velocity using the same suvat equation as before:

v_y^2 - u_y^2 = 2as

v_y = \sqrt{u^2+2as}=\sqrt{(20.0)^2+2(-9.8)(-25)}=-29.8 m/s (downward, so we take the negative solution)

The horizontal speed instead is zero, since the pebble is initially launched vertically, so the final speed is just equal to the magnitude of the vertical velocity:

v = 29.8 m/s

(c) 29.8 m/s

This case is similarly to the previous one: the only difference here is that the pebble is launched straight down instead than up, therefore

u_y = -20.0 m/s

Using again the same suvat equation:

v_y^2 - u_y^2 = 2as

v_y = \sqrt{u^2+2as}=\sqrt{(-20.0)^2+2(-9.8)(-25)}=-29.8 m/s (downward, so we take the negative solution)

As before, the horizontal speed instead is zero, since the pebble is initially launched vertically, so the final speed is just equal to the magnitude of the vertical velocity:

v = 29.8 m/s

We notice that the final value of the speed is always the same in all the three parts, so it does not depend on the direction of launching. This is due to the law of conservation of energy: in fact, the initial mechanical energy of the pebble (kinetic+potential) is the same in all three cases (because the height h does not change, and the speed v does not change either), and the kinetic energy gained during the fall is also the same (since the pebble falls the same distance in all 3 cases), therefore the final speed must also be the same.

7 0
3 years ago
Find the volume of a rectangular prism that is 8m long, 4m wide, and 300cm high
konstantin123 [22]
  • L=8m
  • B=4m
  • H=300cm=3m

\\ \bull\tt\longmapsto Volume=LBH

\\ \bull\tt\longmapsto Volume=8(4)(3)

\\ \bull\tt\longmapsto Volume=96m^3

6 0
2 years ago
Read 2 more answers
In a lab, a student drags a shoe across the floor at constant speed. If the coefficient of static friction between the floor and
Svetllana [295]
<span>B) 0.6 N
   I suspect you have a minor error in your question. Claiming a coefficient of static friction of 0.30N is nonsensical. Putting the Newton there is incorrect. The figure of 0.25 for the coefficient of kinetic friction looks OK. So with that correction in mind, let's solve the problem. The coefficient of static friction is the multiplier to apply to the normal force in order to start the object moving. And the coefficient of kinetic friction (which is usually smaller than the coefficient of static friction) is the multiplied to the normal force in order to keep the object moving. You've been given a normal force of 2N, so you need to multiply the coefficient of static friction by that in order to get the amount of force it takes to start the shoe moving. So: 0.30 * 2N = 0.6N And if you look at your options, you'll see that option "B" matches exactly.</span>
7 0
2 years ago
Read 2 more answers
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