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2 years ago

One of the important waste products of photosynthesis is oxygen gas. Where does this oxygen gas come from?

Physics

2 answers:

Pani-rosa [81]2 years ago

8 0

Answer:

Explanation:

Equation of photosynthesis:

6CO2(g) + 6H2O(l) --> C6H12O6(s) + 6O2(g)

From the above equation, on the RHS, The oxygen during photosynthesis comes from split water molecules. During photosynthesis, the plant absorbs water and carbon dioxide. After the absorption, the water molecules are disassembled or broken down to extract electrons and protons in the presence of ultraviolet light and converted into simple sugar and oxygen.

Equation of the water breakdown:

2H2O --> O2 + 4H+

chubhunter [2.5K]2 years ago

5 0

Answer:

End product of photosynthesis.

Explanation:

Photosynthesis is a process that plants undergo in the manufacture of their food. This is done in the presence of sunlight which is trapped by their coloring pigment called chlorophyll and reactants such as Carbon dioxide and Water.

6CO2+6H2O= C6H12O6+ 6O2

The oxygen gas which is a waste product of photosynthesis is released into the atmosphere and used by animals in respiration.

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According to the law of refraction, light passing from air into a piece of glass at an angle of 30 degrees will cause the light

pantera1 [17]

Answer:

bend toward the normal line

Explanation:

When light passes from a less dense to a more dense substance, (for example passing from air into water), the light is refracted (or bent) towards the normal. In your question the light is moving from rarer to denser medium

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3 years ago

Which statement about the sun's energy is correct?

Vera_Pavlovna [14]

I believe it’s A. I know for sure it isn’t D.

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2 years ago

A bowling ball of mass 5 kg rolls down a slick ramp 20 meters long at a 30 degree angle to the horizontal. What is the work done

Elena-2011 [213]

Answer:

The work done by gravity during the roll is 490.6 J

Explanation:

The work (W) is:

$W = F*d$

<em>Where</em>:

F: is the force

d: is the displacement = 20 m

The force is equal to the weight (W) in the x component:

$F = W_{x} = mgsin(\theta)$

<em>Where:</em>

m: is the mass of the bowling ball = 5 kg

g: is the gravity = 9.81 m/s²

θ: is the degree angle to the horizontal = 30°

$F = mgsin(\theta) = 5 kg*9.81 m/s^{2}*sin(30) = 24.53 N$

Now, we can find the work:

$W = F*d = 24.53 N*20 m = 490.6 J$

Therefore, the work done by gravity during the roll is 490.6 J.

I hope it helps you!

6 0

2 years ago

What kind of light results when the total spectrum of refracted light is recombined? violet green white red?

ElenaW [278]

Hello
The final light will be white. In fact, each color of the visible spectrum is an electromagnetic wave with its own specific frequency and wavelength. White, instead, does not have a specific frequency: it is the sum of all the different wavelengths of the visible spectrum. Therefore, when recombining the spectrum of the refracted light all the different frequencies recombine together, and their sum gives white light.

(edited)

4 0

3 years ago

Steam flows steadily through an adiabatic turbine. The inlet conditions of the steam are 4 MPa, 500◦C, and 80 m/s, and the exit

Cerrena [4.2K]

Answer:

a) ΔEC=-23.4kW

b)W=12106.2kW

c) $A=0.01297m^2$

Explanation:

The kinetic energy is defined as:

$\frac{m*vel^2}{2}$ (vel is the velocity, to differentiate with v, specific volume).

The kinetic energy change will be: Δ ( $\frac{mvel^2}{2}$ )= $\frac{m*vel_2^2}{2}-\frac{m*vel_1^2}{2}$

Δ ( $\frac{mvel^2}{2}$ )= $\frac{m}{2}*(vel_2^2-vel_1^2)$

Where 1 and 2 subscripts mean initial and final state respectively.

Δ( $\frac{mvel^2}{2}$ )= $\frac{12\frac{kg}{s}}{2}*(50^2-80^2)\frac{m^2}{s^2}=-23400W=-23.4kW$

This amount is negative because the steam is losing that energy.

Consider the energy balance, with a neglective height difference: The energy that enters to the turbine (which is in the steam) is the same that goes out (which is in the steam and in the work done).

$H_1+\frac{m*vel_1^2}{2}=H_2+\frac{m*vel_2^2}{2}+W\\W=m*(h_1-h_2)+\frac{m}{2} *(vel_1^2-vel_2^2)$

We already know the last quantity: $\frac{m}{2} *(vel_1^2-vel_2^2)$ = $-$ Δ ( $\frac{mvel^2}{2}$ )=23400W

For the steam enthalpies, review the steam tables (I attach the ones that I used); according to that, $h_1=h(T=500C,P=4MPa)=3445.3\frac{kJ}{kg}$

The exit state is a liquid-vapor mixture, so its enthalpy is:

$h_2=h_f+xh_{fg}=289.23+0.92*2366.1=2483.4\frac{kJ}{kg}$

Finally, the work can be obtained:

$W=12\frac{kg}{s}*(3445.3-2438.4)\frac{kJ}{kg} +23.400kW)=12106.2kW$

C) For the area, consider the equation of mass flow:

$m=p*vel*A$ where $p$ is the density, and A the area. The density is the inverse of the specific volume, so $m=\frac{vel*A}{v}$

The specific volume of the inlet steam can be read also from the steam tables, and its value is: $0.08643\frac{m^3}{kg}$ , so:

$A=\frac{m*v}{vel}=\frac{12\frac{kg}{s}*0.08643\frac{m^3}{kg}}{80\frac{m}{s}}=0.01297m^2$

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7 0

3 years ago