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Anika [276]
1 year ago
9

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?

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

A.

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

B.

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

C.

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

D.

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

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Answer:

A

Explanation:

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A 980 kg roller coaster cart is traveling along a track at 17 m/s before it rolls down a 30 m tall hill (Point A). What will be
MrRissso [65]

The kinetic energy halfway the hill is 2.86\cdot 10^5 J

Explanation:

If there are no friction forces acting on the cart, we can apply the law of conservation of energy: the mechanical energy of the cart (which is the sum of potential energy + kinetic energy) must be conserved. So we can write:

U_A +K_A = U_B + K_B

where

U_A=mgh_A is the initial potential energy, at point A, with

m = 980 kg (mass of the cart)

g=9.8 m/s^2 (acceleration of gravity)

h_A = 30 m (height at point A)

K_A=\frac{1}{2}mv_A^2 is the initial kinetic energy, at point A , with

v_A=17 m/s (velocity at point A)

U_B=mgh_B is the final potential energy, at point B, where

h_B = 15 m (height at point B)

K_B=\frac{1}{2}mv_B^2 is the final kinetic energy, at point B, where

v_B is the velocity at point B

Here we are interested in finding K_B, so by re-arranging the equation and substituting we find:

K_B = U_A+K_B-U_B = mg(h_A-h_B)+\frac{1}{2}mv_A^2=(980)(9.8)(30-15)+\frac{1}{2}(980)(17)^2=2.86\cdot 10^5 J

Learn more about kinetic energy:

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8 0
3 years ago
As a physics instructor hurries to the bus stop, her bus passes her, stops ahead, and begins loading passengers. She runs at 6.0
Alika [10]

velocity of the physics instructor with respect to bus

v = 6 m/s

acceleration of the bus is given as

a = 2 m/s^2

acceleration of instructor with respect to bus is given as

a = -2 m/s^2

now the maximum distance that instructor will move with respect to bus is given as

v_f^2 - v_i^2 = 2 a d

0 - 6^2 = 2(-2)(d)

-36 = - 4 d

d = 9 m

so the position of the instructor with respect to door is exceed by

\delta x = 9 - 6 = 3 m

so it will be moved maximum by 3 m distance

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3 years ago
A bicyclist in the Tour de France crests a mountain pass as he moves at 18 km/h. At the bottom, 4.0 km farther, his speed is at
Allisa [31]

We are given:

v0 = initial velocity = 18 km/h

d = distance = 4 km

v = final velocity = 75 km/h

a =?

<span>
We can solve this problem by using the formula:</span>

v^2 = v0^2 + 2 a d

 

75^2 = 18^2 + 2 (a) * 4

5625 = 324 + 8a

<span>a = 662.625 km/h^2</span>

6 0
3 years ago
Standing still, Bruce, the quarterback, gets tackled by Biff, the 90.0-kg tackle, who is traveling at 7.0 m/s. Upon collision, B
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\\ \sf\longmapsto \Delta P=P

\\ \sf\longmapsto m1v1=m2v2

\\ \sf\longmapsto 90(7)=m2(10)

\\ \sf\longmapsto 10m2=630

\\ \sf\longmapsto m2=\dfrac{630}{10}

\\ \sf\longmapsto m2=63kg

Bruces mass is 63kg

3 0
2 years ago
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