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2 years ago

Why do farmers need to use strategies like crop rotation?

Physics

2 answers:

Mrrafil [7]2 years ago

6 0

Answer:

Crops drain the soil of nutrients

Explanation:

i took the test:) yw

alexdok [17]2 years ago

4 0

Answer:

Crops drain the soil of nutrients

Explanation:

Trust me fam

You might be interested in

Please Help!! When do ionic bonds occur?

iren2701 [21]

A is the answer. I think

8 0

2 years ago

A 100-m-long wire carrying a current of 4.0 A will be accompanied by a magnetic field of what strength at a distance of 0.050 m

solong [7]

Answer:

1.6 x 10^-5 T

Explanation:

i = 4 A

r = 0.05 m

The magnetic field due to long wire at a distance r is given by

$B = \frac{\mu _{0}}{4\pi }\times \frac{2i}{r}$

B = 10^-7 x 2 x 4 / 0.05

B = 1.6 x 10^-5 T

3 0

3 years ago

If the net force acting on a moving object CAUSES NO CHANGE IN ITS VELOCITY, what happens to the object's momentum?

SOVA2 [1]

If the net force acting on a moving object causes no change in its velocity, the object's momentum will stay the same.

<h3>What is momentum?</h3>

Momentum of a body in motion refers to the tendency of a body to maintain its inertial motion.

The momentum is the product of its mass and velocity.

This suggests that if the net force acting on a moving object causes no change in its velocity, the momentum of the object will remain the same.

Therefore, if the net force acting on a moving object causes no change in its velocity, the object's momentum will stay the same.

Learn more about momentum at: brainly.com/question/13554527

#SPJ1

5 0

1 year ago

This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive an

nalin [4]

Answer:

The maximum volume is 1417.87 $inch^3$

Explanation:

<u>Optimization Using Derivatives</u>

We have a 24x30 inch piece of metal and we need to make a rectangular box by cutting a square from each corner of the piece and bending up the sides. The width of the piece is 24 inches and its length is 30 inches

When we cut a square of each corner of side x, the base of the box (after bending up the sides) will be (24-2x) and (30-2x), width and length respectively. The volume of the box is

$V=(24-2x)(30-2x)x$

Operating

$V=4x^3-108x^2+720x$

To find the maximum value of V, we compute the first derivative and equate it to zero

$V'=12x^2-216x+720=0$

Simplifying by 12

$x^2-18x+60=0$

Completing squares

$x^2-18x+81-81+60=0$

$(x-9)^2=21$

We have two values for x

$x=9+\sqrt{21}=13.58\ inch$

$x=9-\sqrt{21}=4.42\ inch$

The first value is not feasible because it will produce a negative width (24-2(13.58))=-6.16

We'll keep only the solution

$x=4.42\ inch$

The width is

$w=(24-2(4.42))=15.16\ inch$

The length is

$l=(30-2(4.42))=21.16\ inch$

And the height

$x=4.42\ inch$

The maximum volume is

$V=(15.16)(21.16)(4.42)=1417.87\ inch^3$

4 0

3 years ago

An insect 5.25 mm tall is placed 25.0 cm to the left of a thin planoconvex lens. The left surface of this lens is flat, the righ

Zigmanuir [339]

Answer:

(A) therefore the image is

63 cm to the right of the lens
the image size is -13.22 cm
it is real
it is inverted

(B) therefore the image is

63 cm to the right of the lens
the image size is -13.22 cm
it is real
it is inverted

Explanation:

height of the insect (h) = 5.25 mm = 0.525 cm

distance of the insect (s) = 25 cm

radius of curvature of the flat left surface (R1) = ∞

radius of curvature of the right surface (R2) = -12.5 cm (because it is a planoconvex lens with the radius in the direction of the incident rays)

index of refraction (n) = 1.7

(A) we can find the location of the image by applying the formula below

$\frac{1}{f} =\frac{1}{s'} +\frac{1}{s}$ where

s' = distance of the image
f = focal length

but we first need to find the focal length before we can apply this formula

$\frac{1}{f} =(n-1)(\frac{1}{R1} -\frac{1}{R2} )$

$\frac{1}{f} =(1.7-1)(\frac{1}{∞} -\frac{1}{-12.5} )$

$\frac{1}{f} =(0.7)(0 + \frac{1}{12.5} )$

$\frac{1}{f} =\frac{0.7}{12.5}$

f = $\frac{12.5}{0.7}$

f = 17.9 cm

now that we have the focal length we can apply $\frac{1}{f} =\frac{1}{s'} +\frac{1}{s}$

$\frac{1}{f} - \frac{1}{s}$ =\frac{1}{s'}

$\frac{1}{17.9} - \frac{1}{25}$ =\frac{1}{s'}

$\frac{25 - 17.9}{17.9 x 25} =\frac{1}{s'}$

$\frac{7.1}{447.5} =\frac{1}{s'}$

s' = \frac{447.5}{7.1}[/tex] = 63 cm to the right of the lens

magnification = $\frac{-s'}{s} =\frac{y'}{y}$ where y' is the height of the image, therefore

$\frac{-s'}{s} =\frac{y'}{y}$

$\frac{-63}{25} =\frac{y'}{52.5}$

y' = $\frac{-63}{25}$ x 0.525 = -13.22 cm

therefore the image is

63 cm to the right of the lens
the image size is -13.22 cm
it is real
it is inverted

(B) if the lens is reversed, the radius of curvatures would be interchanged

radius of curvature of the flat left surface (R1) = ∞

radius of curvature of the right surface (R2) = 12.5 cm

we can find the location of the image by applying the formula below

$\frac{1}{f} =\frac{1}{s'} +\frac{1}{s}$ where

s' = distance of the image
f = focal length

but we first need to find the focal length before we can apply this formula

$\frac{1}{f} =(n-1)(\frac{1}{R1} -\frac{1}{R2} )$

$\frac{1}{f} =(1.7-1)(\frac{1}{12.5} -\frac{1}{∞} )$

$\frac{1}{f} =(0.7)( \frac{1}{12.5} - 0)$

$\frac{1}{f} =\frac{0.7}{12.5}$

f = $\frac{12.5}{0.7}$

f = 17.9 cm

now that we have the focal length we can apply $\frac{1}{f} =\frac{1}{s'} +\frac{1}{s}$

$\frac{1}{f} - \frac{1}{s}$ =\frac{1}{s'}

$\frac{1}{17.9} - \frac{1}{25}$ =\frac{1}{s'}

$\frac{25 - 17.9}{17.9 x 25} =\frac{1}{s'}$

$\frac{7.1}{447.5} =\frac{1}{s'}$

s' = \frac{447.5}{7.1}[/tex] = 63 cm to the right of the lens

magnification = $\frac{-s'}{s} =\frac{y'}{y}$ where y' is the height of the image, therefore

$\frac{-s'}{s} =\frac{y'}{y}$

$\frac{-63}{25} =\frac{y'}{52.5}$

y' = $\frac{-63}{25}$ x 0.525 = -13.22 cm

therefore the image is

63 cm to the right of the lens
the image size is -13.22 cm
it is real
it is inverted

7 0

3 years ago