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3 years ago

Is soil a product of photosynthesis?

Physics

1 answer:

zmey [24]3 years ago

8 0

Answer:

yes

Explanation:

The plants use water (H2O) from the soil and carbon dioxide (CO2) from the air and recombine them to form carbohydrates (CH2O) and oxygen (O2).

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Three collinear forces,F1=45N west,F2=63N east and an unknown force F3 are applied to an object.The resultant force of the three

Greeley [361]

Take east to be the positive direction. Then the resultant force from adding F₁ and F₂ is

F₁ + F₂ = (-45 N) + 63 N = 18 N

which is positive, so it's directed east.

To this we add a third force F₃ such that the resultant is 12 N pointing west, making it negative, so that

18 N + F₃ = -12 N

F₃ = -30 N

So F₃ has a magnitude of 30 N and points west.

6 0

3 years ago

Which of these results in kinetic energy of an object? (1 point)

Aleksandr [31]

Motion I’m pretty sure

7 0

3 years ago

Read 2 more answers

3. When two things are the same temperature:

JulijaS [17]

The answer is definitely C.) their molecules move at the same average speed

3 0

3 years ago

What is the wave length if the distance from the central bright region to the sixth dark fringe is 1.9 cm . Answer in units of n

Yuri [45]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The wavelength is $\lambda = 622 nm$

Explanation:

From the question we are told that

The distance of the slit to the screen is $D = 5 \ m$

The order of the fringe is m = 6

The distance between the slit is $d = 0.9 \ mm = 0.9 *10^{-3} \ m$

The fringe distance is $Y = 1.9 \ cm = 0.019 \ m$

Generally the for a dark fringe the fringe distance is mathematically represented as

$Y = \frac{[2m - 1 ] * \lambda * D }{2d}$

=> $\lambda = \frac{Y * 2 * d }{[2*m - 1] * D}$

substituting values

=> $\lambda = \frac{0.019 * 2 * 0.9*10^{-3} }{[2*6 - 1] * 5}$

=> $\lambda = 6.22 *10^{-7} \ m$

$\lambda = 622 nm$

8 0

3 years ago

you stand on a straight desert road at night and observe a vehicle approaching. this vehicle is equipped with two small headligh

inysia [295]

Answer:

The distance that you marginally able to discern that there are two headlights rather than a single light source is 6.084 km

Explanation:

Given:

d = distance = 0.679 m

λ = wavelength of the light = 537 nm = 537x10⁻⁹m

dp = pupil diameter = 4.81 mm = 0.00481 m

Question: What distance, in kilometers, are you marginally able to discern that there are two headlights rather than a single light source, dx = ?

For the separation of the peak from the central maximum it is:

$sin\theta =\frac{\lambda }{d_{p} } =\frac{537x10^{-9} }{0.00481} =1.116x10^{-4}$

In this case, the two small sources of the headlights have the same angle as the images that form inside the eye

$d_{x} =\frac{d}{sin\theta } =\frac{0.679}{1.116x10^{-4} } =6.084x10^{3} m=6.084km$

7 0

4 years ago