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Leto [7]
3 years ago
8

Which is a product of photosynthesis?

Physics
1 answer:
evablogger [386]3 years ago
8 0

<u>Answer</u>:-

The main product is glucose despite as the matter of fact it produces several products such as:

•Oxygen

&

•Water too.

Explanation:

The reaction of photosynthesis is given below:

6CO2 + 12H2O + (in the presence of sunlight) → C6H12O6 + 6O2 + 6H2O

Carbon dioxide + Water + Sunlight → Glucose + Oxygen + Water.

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At the moment t = 0, a 20.0 V battery is connected to a 5.00 mH coil and a 6.00 Ω resistor. (a) Immediately thereafter, how does
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(a) On the coil: 20 V, on the resistor: 0 V

The sum of the potential difference across the coil and the potential difference across the resistor is equal to the voltage provided by the battery, V = 20 V:

V = V_R + V_L

The potential difference across the inductance is given by

V_L(t) = V e^{-\frac{t}{\tau}} (1)

where

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At time t=0,

V_L(0) = V e^0 = V = 20 V

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(b) On the coil: 0 V, on the resistor: 20 V

Here we are analyzing the situation several seconds later, which means that we are analyzing the situation for

t >> \tau

Since \tau is at the order of less than milliseconds.

Using eq.(1), we see that for t >> \tau, the exponential becomes zero, and therefore the potential difference across the coil is zero:

V_L = 0

Therefore, the potential difference across the resistor will be

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(c) Yes

The two voltages will be equal when:

V_L = V_R (2)

Reminding also that the sum of the two voltages must be equal to the voltage of the battery:

V=V_L +V_R

And rewriting this equation,

V_R = V-V_L

Substituting into (2) we find

V_L = V-V_L\\2V_L = V\\V_L=\frac{V}{2}=10 V

So, the two voltages will be equal when they are both equal to 10 V.

(d) at t=5.77\cdot 10^{-4}s

We said that the two voltages will be equal when

V_L=\frac{V}{2}

Using eq.(1), and this last equation, this means

V e^{-\frac{t}{\tau}} = \frac{V}{2}

And solving the equation for t, we find the time t at which the two voltages are equal:

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(e-a) -19.2 V on the coil, 19.2 V on the resistor

Here we have that the current in the circuit is

I_0 = 3.20 A

The problem says this current is stable: this means that we are in a situation in which t>>\tau, so the coil has no longer influence on the circuit, which is operating as it is a normal circuit with only one resistor. Therefore, we can find the potential difference across the resistor using Ohm's law

V=I_0 R = (3.20 A)(6.0 \Omega)=19.2 V

Then the battery is removed from the circuit: this means that the coil will discharge through the resistor.

The voltage on the coil is given by

V_L(t) = -V e^{-\frac{t}{\tau}} (1)

which means that it is maximum at the moment when the battery is disconnected, when t=0:

V_L(0)=.V

And V this time is the voltage across the resistor, 19.2 V (because the coil is now connected to the resistor, not to the battery). So, the voltage across the coil will be -19.2 V, and the voltage across the resistor will be the same in magnitude, 19.2 V (since the coil and the resistor are connected to the same points in the circuit): however, the signs of the potential difference will be opposite.

(e-b) 0 V on both

After several seconds,

t>>\tau

If we use this approximation into the formula

V_L(t) = -V e^{-\frac{t}{\tau}} (1)

We find that

V_L = 0

And since now the resistor is directly connected to the coil, the voltage in the resistor will be the same as the coil, so 0 V. This means that the coil has completely discharged, and current is no longer flowing through the circuit.

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