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3 years ago

What is the difference between interplanting and intercropping?

Physics

1 answer:

Marysya12 [62]3 years ago

5 0

Answer:

Interplanting is the practice of planting a fast-growing crop between a slower-growing one to make the most of your garden space. ... Intercropping enables you to boost the health of all plants because it can enhance soil fertility and cooperation among different plants.

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The nonreflective coating on a camera lens with an index of refraction of 1.21 is designed to minimize the reflection of 570-nm

lord [1]

Answer: 117.8 nm

Explanation:

Given,

Nonreflective coating refractive index : n = 1.21

Index of refraction: $n_0$ = 1.52

Wave length of light = λ = 570 nm = $570\times10^{-9}\ m$

$\text{ Thickness}=\dfrac{\lambda}{4n}$

$=\dfrac{570\times10^{-9}\ m}{4\times1.21}\\\\\approx\dfrac{117.8\times 10^{-9}\ m}{1}\\\\=117.8\text{ nm}$

Hence, the minimum thickness of the coating that will accomplish= 117.8 nm

5 0

4 years ago

At which angle must a laser beam enter the water for no refraction to occur?

abruzzese [7]

Light that enters the new medium <em>perpendicular to the surface</em> keeps sailing straight through the new medium unrefracted (in the same direction).

Perpendicular to the surface is the "normal" to the surface. So the angle of incidence (angle between the laser and the normal) is zero, and the law of refraction (just like the law of reflection) predicts an angle of zero between the normal and the refracted (or the reflected) beam.

Moral of the story: If you want your laser to keep going in the same direction after it enters the water, or to bounce back in the same direction it came from when it hits the mirror, then shoot it <em>straight on</em> to the surface, perpendicular to it.

5 0

3 years ago

A 2000-kg truck traveling at a speed of 6.0 m/s slows down to 4.0 m/s along a straight road. What

antiseptic1488 [7]

Answer: -4000 kg • m/s

3 0

3 years ago

The Hubble Space Telescope in orbit above the Earth has a 2.4 m circular aperture. The telescope has equipment for detecting ult

strojnjashka [21]

Answer:

Option d

The minimum angular separation between two objects that the Hubble Space Telescope can resolve is $4.8x10^{-8}rad$ .

Explanation:

The resulting image in a telescope that will be gotten from an object is a diffraction pattern instead of a perfect point (point spread function (PSF)).

That diffraction pattern is gotten because the light encounters different obstacles on its path inside the telescope (interacts with the walls and edges of the instrument).

The diffraction pattern is composed by a central disk, called Airy disk, and diffraction rings.

The angular resolution is defined as the minimal separation at which two sources can be resolved one for another, or in other words, when the distance between the two diffraction pattern maxima is greater than the radius of the Airy disk.

The angular resolution can be determined in analytical way by means of the Rayleigh criterion.

$\theta = 1.22\frac{\lambda}{D}$ (1)

Where $\lambda$ is the wavelength and D is the diameter of the telescope.

Notice that it is necessary to express the wavelength in the same units than the diameter.

$\lambda = 95nm \cdot \frac{1x10^{-9}m}{1nm}$ ⇒ $9.5x10^{-8}m$

Finally, equation 1 can be used.

$\theta = 1.22(\frac{9.5x10^{-8}m}{2.4m})$

$\theta = 4.8x10^{-8}rad$

Hence, the minimum angular separation between two objects that the Hubble Space Telescope can resolve is $4.8x10^{-8}rad$ .

5 0

4 years ago

Suppose you lived in a pre-industrial society and needed to lift a heavy (20 kg) block a height of 5 m and had two choices for h

igomit [66]

Let's break the question into two parts:

1) The force needed in Ramp scenario.
2) The effort force needed in the lever scenario.

1. Ramp Scenario:
In an incline, the only component of cart's weight(mg) that is in the direction of motion is $mgsin \alpha$ . Therefore the effort force in this case must be equal or greater than $mgsin \alpha$ .

Now we need to find $\alpha$ . $\alpha$ is the angle between the incline of the ramp and the ground.

Since the height is 5m and the length of the ramp is 8m, $sin \alpha$ would be 5/8 or 0.625. Now that you have $sin \alpha$ , mutiple it with mg.

=> m*g* $sin \alpha$ = 20 * 10 * 5 / 8. (Taking g = 10 m/s² for simplicity) = 125N
Therefore, the minimum Effort force you would require in this case is 125N.

2. Lever Scenario:
Just apply "moment action" in this case, which is:
$F_{e} d_{e} = F_{r} d_{r}$

$F_{e}$ = ?

$F_{r}$ = mg = 20 * 10 = 200N
$d_{e}$ = 10m
$d_{r}$ = 1m

Plug-in the values in the above equation:
$F_{e}$ = 200/10= 20N

As 20N << 125N, the best choice is to use lever.

4 0

3 years ago