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s344n2d4d5 [400]
2 years ago
15

Please explain in detail why it is possible to estimate, to a good approximation, the energy derived from a sugar cube in the bo

dy, by doing laboratory experiments. Obviously the mechanisms are completely different and your answer should address this issue.
Physics
1 answer:
Kaylis [27]2 years ago
8 0
The reactions that are carried out on sugar within are body are very similar to the reaction of combustion. The reactants used and the product formed are the same in the aerobic respiration reaction. Therefore, it is possible to estimate the energy released by the sugar in the body by finding out the energy released when the sugar is burned. Despite the products arising from different pathways with different mechanisms, the reactants and products being the same means that the energy released will be similar.
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1| Page
andreev551 [17]

Answer:

Polarization occurs when an electric field distorts the negative cloud of electrons around positive atomic nuclei in a direction opposite the field. Polarization P in its quantitative meaning is the amount of dipole moment p per unit volume V of a polarized material, P = p/V.

Explanation:

8 0
2 years ago
Read 2 more answers
A charge of 5.67 x 10^-18 C is placed 3.5 x 10^-6 m away from another charge of -
Fudgin [204]
+ 1.58 e -15

Please hit thanks button! :)
5 0
1 year ago
A closed container initially holds 50 monatomic Aparticles that have a combined energy of 480 units. After 100 monatomic B parti
Molodets [167]

Answer:

"8 units" is the appropriate answer.

Explanation:

According to the question,

Throughout equilibrium all particles are of equivalent intensity, and as such the integrated platform's total energy has been uniformly divided across all individuals.

Now,

The total energy will be:

= 480+720

= 1200 \ units

The total number of particles will be:

= 50+100

= 150

hence,

Energy of each A particle or each B particle will be:

= \frac{1200}{150}

= 8 \ units

5 0
2 years ago
The acceleration due to gravity on the surface of Mars is about one-third the acceleration due to gravity on Earth’s surface.
aksik [14]

Answer:

one-third of its weight on Earth's surface

Explanation:

Weight of an object is = W = m*g

Gravity on Earth = g₁ = 9.8 m/s

Gravity on Mars = g₂ = \frac{1}{3} g₁

Weight of probe on earth = w₁ = m * g₁

Weight of probe on Mars = w₂ = m * g₂ -------- ( 1 )

As g₂ = g₁/3 --------- ( 2 )

Put equation (2) in equation (1)

so

Weight of probe on Mars = w₂ = m * g₁ /3

Weight of probe on Mars = \frac{1}{3}  m * g₁ = \frac{1}{3} w₁

⇒Weight of probe on Mars =\frac{1}{3} Weight of probe on earth

6 0
3 years ago
Three wires meet at a junction. Wire 1 has a current of 0.40 A into the junction. The current of wire 2 is 0.57 A out of the jun
AlekseyPX

Answer:

a. 1.56 × 10¹⁸ electrons per second

b. The electrons in wire 3 flow into the junction.

Explanation:

Here is the complete question

Three wires meet at a junction. Wire 1 has a current of 0.40 A into the junction. The current of wire 2 is 0.65 A out of the junction. (a) How many electrons per second move past a point in wire 3? (b) In which direction do the electrons move in wire 3 -- into or out of the junction?

Solution

(a) How many electrons per second move past a point in wire 3?

Using Kirchhoff's current law, at the junction, i₁ + i₂ + i₃ = 0 where i₁ = current in wire 1 = 0.40 A, i₂ = current in wire 2 = 0.65 A and  i₃ = = current in wire 3,

So, i₃ = -(i₁ + i₂)

taking current flowing into the junction as positive and those leaving as negative, i₁ = + 0.40 A and i₂ = -0.65 A

So, i₃ = -(i₁ + i₂)

i₃ = -(0.40 A + (-0.65 A))

i₃ = -(0.40 A - 0.65 A)

i₃ = -(-0.25 A)

i₃ = 0.25 A

Since i₃ = 0.25 C/s and we have e = 1.602 × 10⁻¹⁹ C per electron, then the number of electrons flowing in wire 3 per second is i₃/e = 0.25 C/s ÷ 1.602 × 10⁻¹⁹ C per electron = 0.1561  × 10¹⁹ electrons per second = 1.561  × 10¹⁸ electrons per second ≅ 1.56 × 10¹⁸ electrons per second

(b) In which direction do the electrons move -- into or out of the junction?

Given that i₃ = + 0.25 A and that positive flows into the junction, thus, the electrons in wire 3 flow into the junction.

8 0
3 years ago
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