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1 year ago

C6H12O6 + 6O2 6CO2+6H2O + energy Which statement correctly compares the reactants and products of the equation?

Physics

1 answer:

ryzh [129]1 year ago

5 0

The true statement about the equation of respiration is:

The mass of the reactants is the same as the mass of the products; option C.

<h3>What is respiration?</h3>

Respiration is the process by which living organisms break down large molecules such as glucose into smaller molecules such a carbon dioxide and water to release energy in the form of ATP.

Respiration are of two types in living organisms:

Aerobic respiration which requires oxygen
anaerobic respiration that does not require oxygen

The equation of aerobic respiration is given below:

C₆H₁₂O₆ + 6 O₂ → 6 CO₂ + 6 H₂O + energy

In the given equation above, glucose reacts i the presence of oxygen is broken down into carbon dioxide and water and energy is released.

According to the law of conservation of mass, the mass of reactants is equal to the mass of the products.

Learn more about respiration at: brainly.com/question/22673336

#SPJ1

Note that the complete question is given below:

Respiration describes the process that living cells use to release energy by combining sugar and oxygen. The primary chemical changes that happen during respiration are shown in the equation below.

C₆H₁₂O₆ + 6 O₂ → 6 CO₂ + 6 H₂O + energy

Which statement correctly compares the reactants and products of the equation?

The mass of the reactants is less than the mass of the products.

The mass of the reactants is greater than the mass of the products.

The mass of the reactants is the same as the mass of the products.

The number of reactant molecules is greater than the number of products.

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3 years ago

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An automobile with an initial speed of 4.92 m/s accelerates uniformly at the rate of 3.2 m/s2 . Find the final speed of the car

Rudik [331]

Answer:19.32 m/s

Explanation:

Given

initial speed of car(u)=4.92 m/s

acceleration(a)= $3.2 m/s^2$

Speed of car after 4.5 s

using equation of motion

v=u+at

$v=4.92+3.2\times 4.5=4.92+14.4$

v=19.32 m/s

Displacement of the car after 4.5 s

$v^2-u^2=2as$

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3 years ago

A factory worker pushes a 30.0-kg crate a distance of 4.5 m along a level floor at constant velocity by pushing horizontally on

SIZIF [17.4K]

(a) 73.5 N

The velocity of the crate is constant: this means that the acceleration is zero (a=0), so according to Newton's second law

$\sum F = ma$

the resultant of the forces must be zero: $\sum F = 0$ (1)

The motion is along the horizontal direction, so we are only interested in the forces acting along this direction. There are two of them:

$F$ , the push applied by the worker

$F_f=-\mu mg$ , the force of friction, with $\mu=0.25$ being the coefficient of friction, $m=30.0 kg$ being the mass of the crate, and $g=9.8 m/s^2$ . The negative sign is due to the fact that the friction acts in the opposite direction to the motion. Eq.(1) then becomes

$F-\mu mg=0\\F=\mu mg=(0.25)(30.0 kg)(9.8 m/s^2)=73.5 N$

So, this is the force that the worker must apply.

(b) 330.8 J

The work done by the pushing force of the worker on the crate is given by:

$W=Fd cos \theta$

where

F = 73.5 N is the force

d = 4.5 m is the displacement

$\theta=0^{\circ}$ is the angle between the direction of the force and the displacement (0 degrees, since they are in same direction)

Substituting, we have

$W=(73.5 N)(4.5 m)(cos 0^{\circ})=330.8 J$

(c) -330.8 J

To calculate the work done by friction, we apply the same formula:

$W=F_f d cos \theta$

where

$F_f = \mu mg=(0.25)(30.0 kg)(9.8 m/s^2)=73.5 N$ is the magnitude of the force of friction

d = 4.5 m is the displacement

$\theta=180^{\circ}$ is the angle between the direction of the force of friction and the displacement (it is 180 degrees since the two are into opposite directions)

Substituting, we find

$W=(73.5 N)(4.5 m)(cos 180^{\circ})=-330.8 J$

So, the work done by friction is negative.

(d) 0 J

As before, the work done by any force on the crate is

$W=F_f d cos \theta$

We notice that both gravity and normal force are perpendicular to the displacement: therefore, $\theta=90^{circ}$ , and so

$cos \theta=0$

which means that the work done by both forces is zero.

(e) 0 J

The total work done on the crate is the sum of the work done by the four forces acting on it, so:

$W=W_{push} + W_{friction}+W_{gravity}+W_{normal}=330.8J-330.8J+0+0=0$

And this is in accordance with the work-energy theorem, which states that the variation of kinetic energy of the crate is equal to the work done on it: since the crate is moving at constant velocity, its variation of kinetic energy is zero, as well as the work done on it.

5 0

3 years ago