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2 years ago

Which describes the role of oxygen in photosynthesis? *

Physics

2 answers:

Inessa [10]2 years ago

6 0

Answer:

It is a product, so it is released by the plant.

Explanation:

$6CO2 + 6H2O$ → $C6H12O6 + 6O2$

The plant uses water and carbon dioxide to create glucose and oxygen. Glucose is plant food, and Oxygen is given off as a byproduct.

igomit [66]2 years ago

5 0

Answer:

It is a product, so it is released by the plant.

Explanation:

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If a man walks 17m east and then 17m south , the magnitude of the man's displacement is

In-s [12.5K]

Hey here is your answer u r looking for
24.041m

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2 years ago

Read 2 more answers

Using dimensional analysis, construct a constant, with units of length only, out of all three of the following fundamental const

ludmilkaskok [199]

The quantity with units of length only is $\sqrt{\frac{hG}{c^3}}$

Explanation:

We have to combine the following constants:

- $h$ , Planck constant, with units $[m^2][kg][s^{-1}]$

- G, the Newton's gravitational constant, with units $[m^3][kg^{-1}][s^{-2}]$

- $c$ , the speed of light, with units $[m][s^{-1}]$

The combination of these constant should have units of length only, so with meters (m).

First, we notice that $h$ has [kg] in its units, while G has $[kg^{-1}]$ in its units, so in order to make the [kg] disappear, we have to multiply them and they should have same power, so:

$hG = [m^{2+3}][kg^{1-1}][s^{-1-2}]=[m^5][s^{-3}]$

Now we have to make the seconds, [s], disappear. We do that by dividing the new quantity by $c^3$ , so that the new units are:

$\frac{hG}{c^3}=\frac{[m^5][s^{-3}]}{([m][s^{-1}])^3}=\frac{[m^5][s^{-3}]}{[m^3][s^{-3}]}=[m^2]$

We are almost done: now the quantity has units of an area, squared meters. Therefore, in order to make it have it units of length, we just take its square root:

$\sqrt{\frac{hG}{c^3}}=\sqrt{[m^2]}=[m]$

Learn more about gravitational constant:

brainly.com/question/1724648

brainly.com/question/12785992

#LearnwithBrainly

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2 years ago

A runner makes one lap around a 200m track of 25s. What were the runners average speed and average velocity?

kobusy [5.1K]

Here is the answer. In order for us to get the speed, we are just going to use the formula speed is equal to distance over time. Given that the distance s 200m and the time is 25s, this would give us the answer of 8m/sec. Hope this answers your question.

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2 years ago

A force of 10 N acts on an object with a mass of 10 kg. What is its acceleration? A. 1 m/s2 B. 10 m/s2 C. 2 m/s2 D. 5 m/s2

Akimi4 [234]

The acceleration exerted by the object of mass 10 kg is $\mathbf{1} m / \boldsymbol{s}^{2}$

Answer: Option A

<u>Explanation:</u>

According to Newton’s second law of motion, any external force acting on a body will be directly proportional to the mass of the body as well as acceleration exerted by the body. So, the net external force acting on any object will be equal to the product of mass of the object with acceleration exerted by the object. Thus,

$Force = Mass \times Acceleration$

So,

$Acceleration=\frac{\text {Force}}{\text {Mass}}$

As the force acting on the object is stated as 10 N and the mass of the object is given as 10 kg, then the acceleration will be

$Acceleration =\frac{10 \mathrm{N}}{10 \mathrm{kg}}=1 \mathrm{m} / \mathrm{s}^{2}$

So, the acceleration exerted by the object of mass 10 kg is $\mathbf{1} m / \boldsymbol{s}^{2}$

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2 years ago

In midair an M = 145 kg bomb explodes into two pieces of m1 = 115 kg and another, respectively. Before the explosion, the bomb w

Daniel [21]

Answer:

$v_2=-133.17m/s$ , the minus meaning west.

Explanation:

We know that linear momentum must be conserved, so it will be the same before ( $p_i$ ) and after ( $p_f$ ) the explosion. We will take the east direction as positive.

Before the explosion we have $p_i=m_iv_i=Mv_i$ .

After the explosion we have pieces 1 and 2, so $p_f=m_1v_1+m_2v_2$ .

These equations must be vectorial but since we look at the instants before and after the explosions and the bomb fragments in only 2 pieces the problem can be simplified in one dimension with direction east-west.

Since we know momentum must be conserved we have:

$Mv_i=m_1v_1+m_2v_2$

Which means (since we want $v_2$ and $M=m_1+m_2$ ):

$v_2=\frac{Mv_i-m_1v_1}{m_2}=\frac{Mv_i-m_1v_1}{M-m_1}$

So for our values we have:

$v_2=\frac{(145kg)(24m/s)-(115kg)(65m/s)}{(145kg-115kg)}=-133.17m/s$

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2 years ago