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Crank
2 years ago
8

Identify the stage in cellular respiration that produces carbon dioxide as a waste product

Physics
1 answer:
ollegr [7]2 years ago
6 0
During cellular respiration, organisms use oxygen to turn glucose into carbon dioxide, water, and energy in the form of ATP. The process has three stages: glycolysis , the Krebs cycle, and the electron transport chain. Glycolysis in the cytoplasm ), breaks down 1 glucose into 2 pyruvate and 2 ATP. The Krebs cycle (in the mitochondrion's matrix), provides the hydrogen and electrons needed for the electron transport chain. Another 2 are formed here. The electron transport chain (on the inner mitochondrial membrane) forms 32 ATP through oxidative phosphorylation .
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1. The 10-min unithydrograph for a 2.25 km2 urban catchment is given by:(a) estimate the run off hydrograph for a 10-min rainfal
vivado [14]

Answer:

A ) Find the direct runoff depth; d =V A 22,698 _ 2.25x106 =0.01m =1cm The above result shows that the given hydrograph qualifies as a unit hydrograph. .

Explanation:

Brainlyist

5 0
3 years ago
Physics help, thank you guys so much!
katrin2010 [14]

Answer:

 Δt = 5.85 s

Explanation:

For this exercise let's use Faraday's Law

           emf = -  \frac{d \phi}{dt} -  d fi / dt

           \phi = B. A

           \phi = B A cos θ

The bold are vectors. It indicates that the area of the body is A = 0.046 m², the magnetic field B = 1.4 T, also iindicate that the normal to the area is parallel to the field, therefore the angle θ = 0 and cos 0 =1.

suppose a linear change of the magnetic field

            emf = - A \frac{B_f - B_o}{ \Delta t}

             Dt = - A  \frac{B_f - B_o}{emf}

the final field before a fault is zero

       

let's calculate

            Δt = - 0.046 (0- 1.4) / 0.011

            Δt = 5.85 s

4 0
2 years ago
A wire is formed into a circle having a diameter of 10.0cm and is placed in a uniform magnetic field of 3.00mT . The wire carrie
Paul [167]

The range of potential energies of the wire-field system for different orientations of the circle are -

θ                  U

0°             375 π x 10^{-7}

90°              0

180°        - 375 π x 10^{-7}

We have current carrying wire in a form of a circle placed in a uniform magnetic field.

We have to the range of potential energies of the wire-field system for different orientations of the circle.

<h3>What is the formula to calculate the Magnetic Potential Energy?</h3>

The formula to calculate the magnetic potential energy is -

U = M.B = MB cos $\theta

where -

M is the Dipole Moment.

B is the Magnetic Field Intensity.

According to the question, we have -

U = M.B = MB cos $\theta

We can write M = IA (I is current and A is cross sectional Area)

U = IAB cos $\theta

U = Iπr^{2}B cos $\theta

For $\theta = 0° →

U(Max) = MB cos(0) = MB =  Iπr^{2}B = 5 × π × ( 0.05 ) ^{2} × 3 × 10^{-3} =

375 π x 10^{-7}.

For $\theta = 90° →

U = MB cos (90) = 0

For $\theta = 180° →

U(Min) = MB cos(0) = - MB =  - Iπr^{2}B = - 5 × π × ( 0.05 ) ^{2} × 3 × 10^{-3} =

- 375 π x 10^{-7}.

Hence, the range of potential energies of the wire-field system for different orientations of the circle are -

θ                  U

0°             375 π x 10^{-7}

90°              0

180°        - 375 π x 10^{-7}

To solve more questions on Magnetic potential energy, visit the link below-

brainly.com/question/13708277

#SPJ4

3 0
1 year ago
The fluid in a grdulated cylinder should be read at the _____ of the meniscus.
galina1969 [7]
<span>The fluid in a graduated cylinder should be read at the BOTTOM of the meniscus.</span>
5 0
3 years ago
Read 2 more answers
An object is placed in front of a convex lens of a length 10cm. What is the nature of the image formed if the object distance is
Lady_Fox [76]

{\mathfrak{\underline{\purple{\:\:\: Given:-\:\:\:}}}} \\ \\

\:\:\:\:\bullet\:\:\:\sf{Focal\:length=10\:cm}

\:\:\:\:\bullet\:\:\:\sf{Object \ distance = -15\:cm}

\\

{\mathfrak{\underline{\purple{\:\:\:To \:Find:-\:\:\:}}}} \\ \\

\:\:\:\:\bullet\:\:\:\sf{Nature \: of \:the\:image}

\\

{\mathfrak{\underline{\purple{\:\:\: Solution:-\:\:\:}}}} \\ \\

<h3>☯ <u>By using formula of Lens</u> </h3>

\\

\dashrightarrow\:\: {\boxed{\sf{\dfrac{1}{u} + \dfrac{1}{v} = \dfrac{1}{f}}}}

\\

\dashrightarrow\:\: \sf{\dfrac{1}{v}-\dfrac{1}{-15}=\dfrac{1}{10}}

\\

\dashrightarrow\:\: \sf{\dfrac{1}{v}+\dfrac{1}{15}=\dfrac{1}{10}}

\\

\dashrightarrow\:\: \sf{\dfrac{1}{v} = \dfrac{1}{10} - \dfrac{1}{15}}

\\

\dashrightarrow\:\: \sf{\dfrac{1}{v} = \dfrac{1}{30}}

\\

\dashrightarrow\:\: \sf{ v = 30 \ cm}

\\

<h3>☯ <u>Now, Finding the magnification </u></h3>

\\

\dashrightarrow\:\: \sf{ m = \dfrac{-30}{-15}}

\\

\dashrightarrow\:\: \sf{m = -2}

\\

<h3>☯ <u>Hence</u>,\\</h3>

\:\:\:\:\star\:\:\:\sf{Image \ distance = 30 \ cm}

\:\:\:\:\star\:\:\:\sf{Nature = Real \ \& \ inverted}

3 0
2 years ago
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