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3 years ago

What are the products in the photosynthesis equation?

Physics

2 answers:

valkas [14]3 years ago

6 0

Answer:

Explanation:

Reil [10]3 years ago

4 0

Answer : C

Plants use a process called photosynthesis to make food. During photosynthesis, plants trap light energy with their leaves. Plants use the energy of the sun to change water and carbon dioxide( co2 ) into a sugar called glucose ( c6h12o6 ). Glucose is used by plants for energy and to make other substances like cellulose and starch.

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is the following sentence true or false? the faster the particles of a substance are moving, the more energy they have.

svetlana [45]

I think true. I'm pretty sure, but check w/ others too.

8 0

3 years ago

Read 2 more answers

Suppose you design a new thermometer called the "x" thermometer. on the x scale, the boiling point of water is 130.0 ox and the

Hoochie [10]

You've told us:

-- 130°x = 212°F

and

-- 10°x = 32°F

Thank you. Those are two points on a graph of °x vs °F . With those, we can figure out the equation of the graph, and easily convert ANY temperature on one scale to the equivalent temperature on the other scale.

-- If our graph is going to have °x on the horizontal axis and °F on the vertical axis, then the two points we know are (130, 212) and (10, 32) .

-- The slope of the line through these two points is

Slope = (32 - 212) / (10 - 130)

Slope = (-180) / (-120)

Slope = 1.5

So far, the equation of the graph is

F = 1.5 x + (F-intercept)

Plug one of the points into this equation. I'll use the second point (10, 32) just because the numbers are smaller:

32 = 1.5 (10) + F-intercept

32 = 15 + (F-intercept)

F-intercept = 17

So the equation of the conversion graph is

F = 1.5 x + 17

There you are ! Now you can plug ANY x temperature in there, and the F temperature jumps out at you.

The question is asking what temperature is the same on both scales. This seems tricky, but it's not too bad. Whatever that temperature is, since it's the same on both scales, you can take the conversion equation, and write the same variable in BOTH places.

We can write [ x = 1.5x + 17 ], solve it for x, and the solution will be the same temperature in F too.

We can write [ F = 1.5F + 17 ], solve it for F, and the solution will be the same temperature in x too.

F = 1.5F + 17

Subtract F from each side: 0.5F + 17 = 0

Subtract 17 from each side: 0.5F = -17

Multiply each side by 2 : F = -34

That should be the temperature that's the same number on both scales.

Let's check it out, using our handy-dandy conversion formula (the equation of our graph):

F = 1.5x + 17

Plug in -34 for x:

F = 1.5(-34) + 17

F = -51 + 17

It works ! -34 on either scale converts to -34 on the other one too. If the temperature ever gets down to -34, and you take both thermometers outside, they'll both read the same number.

6 0

3 years ago

Four particles are in a 2-d plane with masses, x- and y- positions, and x- and y- velocities as given in the table below: what i

Arte-miy333 [17]

I attached the picture of the missing table.
Center of mass is the point such that if you apply force to it, the system would move without rotating.
We can use following formula to calculate the center of mass:
$R=\frac{1}{M}\sum_{i=1}^{n=i}m_ir_i$
Where M is the sum of the masses of all particles.
Part 1
To calculate the x coordinate of the center of mass we will use this formula:
$R_x=\frac{1}{M}\sum_{i=1}^{n=i}m_ix_i$
I will do all the calculations in the google sheet that I will share with you.
For the x coordinate of the center of mass we get:
$R_x=0.96m$
Part 2
To calculate the y coordinate of the center of mass we will use this formula:
$R_y=\frac{1}{M}\sum_{i=1}^{n=i}m_iy_i$
I will do all the calculations in the google sheet that I will share with you.
For the x coordinate of the center of mass we get:
$R_y=-0.84m$
Part 3
We will calculate speed along x and y-axis separately and then will add them together.
$v_x=\frac{\sum_{i=1}^{n=i}m_iv_x_i}{M}$
$v_y=\frac{\sum_{i=1}^{n=i}m_iv_y_i}{M}$
Total velocity is:
$v=\sqrt{v_x^2+v_y^2}$
Once we calculate velocities we get:
$v_x=-1.08\frac{m}{s}\\ v_y=-0.03\frac{m}{s}\\ v=\sqrt{(-1.08)^2+(-0.03)^2}=1.08\frac{m}{s}$
Part 4
Because origin is left to our center of mass(please see the attached picture) placing fifth mass in the origin would move the center of mass to the left along the x-axis.
Part 5
If you place fifth mass in the center of the mass nothing would change. The center of mass would stay in the same place.
Here is the link to the spreadsheet:
https://docs.google.com/spreadsheets/d/1SkQHbI1BxiJnwpWbLmP0XWgcNPrGquH1K2MfN6cznVo/edit?usp=sharing

3 0

2 years ago

A plane electromagnetic wave travels northward. At one instant, its electric field has a magnitude of 9.6 V/m and points eastwar

lara31 [8.8K]

Answer:

The values is $B = 3.2 *10^{-8} \ T$

The direction is out of the plane

Explanation:

From the question we are told that

The magnitude of the electric field is $E = 9.6 \ V/m$

The magnitude of the magnetic field is mathematically represented as

$B = \frac{E}{c}$

where c is the speed of light with value

$B = \frac{ 9.6}{3.0 *10^{8}}$

$B = 3.2 *10^{-8} \ T$

Given that the direction off the electromagnetic wave( c ) is northward(y-plane ) and the electric field(E) is eastward(x-plane ) then the magnetic field will be acting in the out of the page (z-plane )

3 0

3 years ago

A small rock is thrown vertically upward with a speed of 21.0 m/sm/s from the edge of the roof of a 21.0-mm-tall building. The r

FromTheMoon [43]

Answer:

Explanation:

Given:

Vo = 21 m/s

Vf = 0 m/s

Using equation of Motion,

Vf^2 = Vo^2 - 2aS

S = (21^2)/2 × 9.8

= 22.5 m.

Given:

S = 22.5 + 21 mm

= 22.521 m

Vo = 0 m/s

Using the equation of motion,

S = Vo × t + 1/2 × a × t^2

22.521 = 0 + 1/2 × 9.8 × t^2

t^2 = (2 × 22.521)/9.8

= 4.6

t = 2.14 s

4 0

3 years ago