All categories

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.

You might be interested in

explain the fleming left-hand rule with the diagram and what will the direction of the induced current in the figure if the magn

iVinArrow [24]

Hi sorry

I don’t know

but maybe next time

4 0

2 years ago

A ball is batted straight up into the air and reaches a maxium height 65.6 m (a) How long did it take to reach this height? (b)

kondaur [170]

Answer:

a) 3.65 seconds

b) 35.87 m/s

Explanation:

s = Displacement = 65.6 m

u = Initial velocity

v = Final velocity

t = Time taken

a = Acceleration due to gravity = 9.81 m/s² (downward direction is taken as positive and upward is taken as negative)

b) Equation of motion

$v^2-u^2=2as\\\Rightarrow 0^2-u^2=2\times -9.81\times 65.6\\\Rightarrow u=\sqrt{2\times 9.81\times 65.6}\\\Rightarrow u=35.87\ m/s$

Initial pop up velocity is 35.87 m/s

$v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-35.87}{-9.81}\\\Rightarrow t=3.65\ s$

It took 3.65 seconds to reach this height

6 0

3 years ago

A baseball is hit with a bat. The direction of the ball is completely reversed and its speed is doubled. If the actual contact w

alexdok [17]

Before the impact, let the velocity of the baseball was v m/s.

After being hit by the bat its velocity is -2v
So, change in velocity, Deltav=v-(-2v)=3v
Acceleration is defined as the rate of change in velocity, i.e. actual change in velocity divided by the time taken to change it. Time taken to change velocity is the time of actual contact of the bat and ball, i.e. 0.31 s.

a=(Deltav)/(Deltat)
=(3v)/0.37
Therefore, a/v=3/0.31=9.7 s^-1
So, the ratio of acceleration of the baseball to its original velocity is 9.7.

8 0

3 years ago

Read 2 more answers

A 70.0 kg sprinter starts a race with an acceleration of 1.60 m/s^2, What is the net external force (in N) on him? (Enter the ma

Ierofanga [76]

Answer:

External force on him will be 112 N

Explanation:

We have given the mass of the sprinter m =70 kg

Acceleration of the sprinter $a=1.6m/sec^2$

We have to find the net external force

According to second law of motion force = mass ×acceleration

Force is dependent on the mass and acceleration

So $F=70\times 1.6=112 N$

So external force will be 112 N

6 0

3 years ago

A coil of 40 turns is wrapped around a long solenoid of cross-sectional area 7.5×10−3m2. The solenoid is 0.50 m long and has 500

defon

To solve this problem it is necessary to apply the concepts related to mutual inductance in a solenoid.

This definition is described in the following equation as,

$M = \frac{\mu_0 N_1 N_2A_1}{l_1}$

Where,

$\mu =$ permeability of free space

$N_1 =$ Number of turns in solenoid 1

$N_2 =$ Number of turns in solenoid 2

$A_1=$ Cross sectional area of solenoid

l = Length of the solenoid

Part A )

Our values are given as,

$\mu_0 = 4\pi *10^{-7}H/m$

$N_1 = 500$

$N_2 = 40$

$A = 7.5*10^{-4}m^2$

$l = 0.5m$

Substituting,

$M = \frac{\mu_0 N_1 N_2A_2}{l_1}$

$M = \frac{(4\pi *10^{-7})(500)(40)(7.5*10^{-4})}{0.5}$

$M = 3.77*10^{-4}H$

PART B) Considering that many of the variables remain unchanged in the second solenoid, such as the increase in the radius or magnetic field, we can conclude that mutual inducantia will appear the same.

8 0

3 years ago