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Morgarella [4.7K]
2 years ago
12

What could people do to DECREASE erosion of the Earth’s surface? A) Cut down more trees. B) Use more land to plant crops. C) All

ow natural vegetation to grow. D) Allow cattle to graze more often.
Physics
2 answers:
Free_Kalibri [48]2 years ago
6 0

What could people do to DECREASE erosion of the Earth’s surface?

  Well, People can plant crops or they can grow plants and if your looking at the options only B and C has that, but lets make sure. So A) "Cut down more trees" does not help DECREASE erosion, if you want to decrease erosion then you have to have roots to make earth's surface stay. A is eliminated, going to B, B says "Use more land to plant crops." that's better because crops have roots so it can keep the surface/land sucked in and will not get erode away. Anyways moving on to C, C "Allow natural vegetation to grow." that kinda helps decrease erosion; so imma keep that in mind, and finally D "Allow cattle to graze more often." does not help at all. So D and A is eliminated, So theres only C and B left, C will kinda help but B will help more; So it will leave us to the final answer which is....

      Answer: B) Use more land to plant crops.

I hope this helps you and I hope you understand what i'm talking about...


From The Special, Idk, my name is weired I just picked this name. I wish I could change it.

Bad White [126]2 years ago
3 0
B) Use more land to plant crops.

By planting crops, the roots can hold ground, thus preventing erosion.
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A 110-g bullet is fired from a rifle having a barrel 0.636 m long. Choose the origin to be at the location where the bullet begi
astraxan [27]
<span>Data:
mass =  110-g bullet
d = 0.636 m
Force = 13500 + 11000x - 25750x^2, newtons.

a) Work, W

W = ∫( F* )(dx) =∫[13500+ 11000x - 25750x^2] (dx) =

W =  13500x + 5500x^2 - 8583.33 x^3 ] from 0 to 0.636  =

W = 8602.6 joule

b) x= 1.02 m

</span><span><span>W =  13500x + 5500x^2 - 8583.33 x^3 ] from</span> 0 to 1.02

W = 10383.5

c) %

[W in b / W in a] = 10383.5 / 8602.6 = 1.21 => W in b is 21% more than work in a.


</span>
7 0
3 years ago
A rifle that shoots bullets at 477 m/s is to be aimed at a target 45.5 m away. If the center of the target is level with the rif
Free_Kalibri [48]

Answer:

The rifle barrel must be pointed at a height of 4.45cm above the target so that the bullet hits dead center.

Explanation:

First, we need to sketch the situation so we can have a better idea of what the problem looks like (Refer to uploaded picture).

So as you may see in the drawing, when pointing the rifle to the target, we can see it as a triangle, but in reality, the bullet will have a parabolic trajectory. Both points of view will help us determine what the height must be. In order to find it, we need to first determine at what angle the bullet should be shot. In order to do so we can use the range formula, which looks like this:

R=\frac{v^{2}sin(2\theta)}{g}

Where R is the range of the bullet (this is how far it goes before it has the

same height it was shot from), v is the original speed of the bullet, θ is the angle at which the bullet is shot and g is the acceleration of gravity.

We can solve this equation for theta, so we get:

gR=v^{2}sin(2\theta)

\frac{gR}{v^{2}}=sin(2\theta)

sin^{-1}(\frac{gR}{v^{2}})=2\theta

\theta=\frac{sin^{-1}(\frac{gR}{v^{2}})}{2}

so now we can substitute the given data:

\theta=\frac{sin^{-1}(\frac{(9.8m/s^{2})(45.5m)}{(477m/s)^{2}})}{2}

so we get:

θ=0.05614°

once we get the angle, we can look at the triangle diagram. From the drawing we can see that we can use the tan function to find the height:

tan \theta = \frac{h}{45.5m}

so we can solve this for h, so we get:

h=45.5m*tan(0.05614^{o})

which yields:

h=0.0445m

or

h=4.45cm

5 0
2 years ago
What are different ways to describe speed and velocity
ioda
Speed is a constant direction of where you go, constant motion, and velocity is the displacement of a objects direction, plus the direction the object is traveling to.
8 0
2 years ago
In a thundercloud there may be electric charges of 45.0 C near the top of the cloud and -45.0 C near the bottom of the cloud. Th
Gala2k [10]

Answer:

The electric force on the top charge is F=3.44\times 10^6\ N.

Explanation:

Given that,

Electric charges in a thundercloud, q_1=q_2=45\ C

The distance between charges, d = 2.3 km = 2300 m

Let F is the electric force on the top charge. The electric force is given by the formula as :

F=\dfrac{kq^2}{d^2}

F=\dfrac{9\times 10^9\times (45)^2}{(2300)^2}

F=3445179.58\ N

or

F=3.44\times 10^6\ N

So, the electric force on the top charge is F=3.44\times 10^6\ N. Hence, this is the required solution.                                                

7 0
3 years ago
A waiter is carrying a tray above his head and walking at a constant velocity. If he applies a force of 5.0 newtons on the tray
Darya [45]
In physical terms, a force does work if it moves an object in the direction the force is pointing towards.

In your example, the force is pushing a tray up. But the force doesn't move the tray any further up. It remains at the same height. 

The is no force done by this force.
3 0
3 years ago
Read 2 more answers
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