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3 years ago

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Physics

1 answer:

cestrela7 [59]3 years ago

8 0

Answer:

1. Photosynthesis releases oxygen gas as a product.

2. Subtracting cellular respiration from gross primary productivity

3. True

Explanation:

1. Photosynthesis is the process by which autotrophic organisms synthesize their food in form of sugars in the presence of sunlight. Photosynthetic process combines carbon dioxide (CO2) and water (H2O) to form glucose (C6H12O6) and oxygen gas as products. The equation is as follows:

6CO2 + 6H2O → C6H12O6 + 6O2

2. Net primary productivity, denoted by NPP, refers to the amount of biomass present in an ecosystem. It can be calculated by subtracting the amount of CO2 lost via cellular respiration from the amount produced called Gross primary productivity (GPP). That is;

NPP = GPP - cellular respiration

3. The process by which organic matter is synthesized from inorganic ones by organisms called AUTOTROPHS or PRIMARY PRODUCERS is called PRIMARY PRODUCTIVITY. Examples of autotrophic organisms are green plants, algae etc. Hence, in an ecosystem that have a high primary productivity, there would be a rapid rate of conversion of solar energy to biomass via PHOTOSYNTHESIS.

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The particle moves with constant speed in a circular path, so its acceleration vector always points toward the circle's center.

At time $t_1$ , the acceleration vector has direction $\theta_1$ such that

$\tan\theta_1=\dfrac{4.00}{6.00}\implies\theta_1=33.7^\circ$

which indicates the particle is situated at a point on the lower left half of the circle, while at time $t_2$ the acceleration has direction $\theta_2$ such that

$\tan\theta_2=\dfrac{-6.00}{4.00}\implies\theta_2=-56.3^\circ$

which indicates the particle lies on the upper left half of the circle.

Notice that $\theta_1-\theta_2=90^\circ$ . That is, the measure of the major arc between the particle's positions at $t_1$ and $t_2$ is 270 degrees, which means that $t_2-t_1$ is the time it takes for the particle to traverse 3/4 of the circular path, or 3/4 its period.

Recall that

$\|\vec a_{\rm rad}\|=\dfrac{4\pi^2R}{T^2}$

where $R$ is the radius of the circle and $T$ is the period. We have

$t_2-t_1=(5.00-2.00)\,\mathrm s=3.00\,\mathrm s\implies T=\dfrac{3.00\,\rm s}{\frac34}=4.00\,\mathrm s$

and the magnitude of the particle's acceleration toward the center of the circle is

$\|\vec a_{\rm rad}\|=\sqrt{\left(6.00\dfrac{\rm m}{\mathrm s^2}\right)^2+\left(4.00\dfrac{\rm m}{\mathrm s^2}\right)^2}=7.21\dfrac{\rm m}{\mathrm s^2}$

So we find that the path has a radius $R$ of

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3 years ago

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Marrrta [24]

Answer:

Explanation:

<h2><u><em>Drought</em></u></h2>

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3 years ago

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fredd [130]

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3 years ago

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A girl (mass M) standing on the edge of a frictionless merry-go-round (radius R, rotational inertia I) that is not moving. She t

vladimir1956 [14]

a) $\omega=\frac{-mvR}{I+MR^2}$

b) $v=\frac{-mvR^2}{I+MR^2}$

Explanation:

Since there are no external torques acting on the system, the total angular momentum must remain constant.

At the beginning, the merry-go-round and the girl are at rest, so the initial angular momentum is zero:

$L_1=0$

Later, after the girl throws the rock, the angular momentum will be:

$L_2=(I_M+I_g)\omega +L_r$

where:

$I$ is the moment of inertia of the merry-go-round

$I_g=MR^2$ is the moment of inertia of the girl, where

M is the mass of the girl

R is the distance of the girl from the axis of rotation

$\omega$ is the angular speed of the merry-go-round and the girl

$L_r=mvR$ is the angular momentum of the rock, where

m is the mass of the rock

v is its velocity

Since the total angular momentum is conserved,

$L_1=L_2$

So we find:

$0=(I+I_g)\omega +mvR\\\omega=\frac{-mvR}{I+MR^2}$

And the negative sign indicates that the disk rotates in the direction opposite to the motion of the rock.

The linear speed of a body in rotational motion is given by

$v=\omega r$

where

$\omega$ is the angular speed

r is the distance of the body from the axis of rotation

In this problem, for the girl, we have:

$\omega=\frac{-mvR}{I+MR^2}$ is the angular speed

$r=R$ is the distance of the girl from the axis of rotation

Therefore, her linear speed is:

$v=\omega R=\frac{-mvR^2}{I+MR^2}$

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3 years ago

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Tamiku [17]

The problem should only have one part to it, but this one has two.
Before I can do the mass/energy conversion, I have to go and
look up the proton mass for myself ... go out and collect the straw
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makes it worth it. They make us do everything around here.

OK. In my Physics book⁽¹⁾, the proton rest mass is

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The formula that relates mass to the equivalent energy is

E = m c² .

The method of applying the formula is known as "plug in what you know",
as follows:

   E = (mass) x (speed of light)²

= (1.67 x 10⁻²⁷ kg) x (3 x 10⁸ m/s)²

   = (1.67 X 10⁻²⁷ Kg) x (9 x 10¹⁶ m²/s²)

=    (1.5 x 10⁻¹⁰) (kg-m²/s²)

=    1.5 x 10⁻¹⁰ joule .
____________________________________

⁽¹⁾ Halliday, David and Resnick, Robert, Physics , John Wiley & Sons,
Inc., 1960, inside front cover, "SELECTED PHYSICAL CONSTANTS".

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