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fgiga [73]
3 years ago
14

The metabolic end-product(s) of the oxygen system is which of the following?

Physics
1 answer:
Dmitrij [34]3 years ago
7 0

The metabolic end-product(s) of the oxygen system is Carbon dioxide and Water

<u>Explanation:</u>

  • Our body creates energy via combustion of carbohydrates , ammino acid and fats in presence of oxygen, this is called as aerobic metabolism.
  • Oxygen system  in aerobic metabolism oxidizes carbohydrates and provide free oxygen atom .
  • These oxygen atom gets attached to carbon in the carbon dioxide molecule, which is excreted.
  • By products of such aerobic metabolism are carbon dioxide and water.
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Two students make the following claims:
antiseptic1488 [7]

Answer:

E. Student 1 is correct, because as θ is increased, h is the same.

Explanation:

Here we have the object of a certain mass falling under gravity so the force acting on the it will depend on mass of the object and the acceleration due to gravity.

Mathematically:

F=m.g

As we know that the work done is evaluated as the force applied on a body and the displacement of the body in the direction of the force.

And for work we have:

W=F.s\cos\theta

where:

s= displacement of the object

\theta= angle between the force and displacement vectors

Given that the height of the object is same in each trail of falling object under the gravity be it a free-fall or the incline plane.

  • In case of free-fall the angle between the force is and the displacement is zero.
  • In case when the body moves along the inclined plane the force applied by the gravity is same because it depends upon the mass of the object. And the net displacement in the direction of the gravitational force is the height of the object which is constant in both the cases.

So, the work done by the gravitational force is same in the two cases.

6 0
3 years ago
Please help easy 8th grade question
Tamiku [17]

Answer:

whats the question?

Explanation:

7 0
3 years ago
HELP ASAP WITH number 6 plz
Alika [10]
The answer would be c
6 0
3 years ago
A cannon is fired horizontally at 243 m/s off of a 62 meter tall, shear vertical cliff. How far in meters from the base of the c
Alekssandra [29.7K]

Answer:

865.08 m

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 243 m/s

Height (h) of the cliff = 62 m

Horizontal distance (s) =?

Next, we shall determine the time taken for the cannon to get to the ground. This can be obtained as follow:

Height (h) of the cliff = 62 m

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) =?

h = ½gt²

62 = ½ × 9.8 × t²

62 = 4.9 × t²

Divide both side by 4.9

t² = 62/4.9

Take the square root of both side.

t = √(62/4.9)

t = 3.56 s

Finally, we shall determine the horizontal distance travelled by the cannon ball as shown below:

Initial velocity (u) = 243 m/s

Time (t) = 3.56 s

Horizontal distance (s) =?

s = ut

s = 243 × 3.56 s

s = 865.08 m

Thus, the cannon ball will impact the ground 865.08 m from the base of the cliff.

6 0
3 years ago
PLS HELP ME OUT!!
MaRussiya [10]
In order to solve this problem, we must first find out the value of each line on the number line. However, we can make this problem more simple by ignoring every interval except for the ones between 0 and 6. There are three total intervals in between 0 and 6 (including 6 and excluding 0). Therefore, we can do 6/2, and get an interval value of 2. This means that each line adds a value of 2. Since the car is only one line past zero, we only have to add one value of 2. Since 0 + 2 = 2, our final answer is C. 2.

Hope this helps!
4 0
3 years ago
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