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3 years ago

How could losing weight be considered an energy transformation?

Physics

2 answers:

Tanya [424]3 years ago

8 0

Answer:

“So while yes, it is true that fat doesn't transform into energy… it is the transformation of fat into carbon dioxide that creates energy.” She says when our bodies give the signal that it needs more energy, lipase in fat cells helps release the fat into circulation

Explanation:

natita [175]3 years ago

8 0

Loosing weight uses energy because your body is creating heat. Heat =energy. Every living thing has energy in it! And when your body is using energy to loose weight, it’s using heat! Without heat, we’ll there’d end no energy. Without energy, there’s no life

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Page 40-44 earth science regents<br> just post the picture of the pages please

Anna007 [38]

I don’t know what book you’re talking about so I can’t help but have a look online, you may be able to find it if you search up the book name and look around a few websites

8 0

3 years ago

The diffusion rate for a solute is 4.0 x 10^-11 kg/s in a solvent- filled channel that has a cross-sectional area of 0.50 cm^2 a

zlopas [31]

Answer:

$s = 9.6\times 10^{-12}kg/s$

Explanation:

Given:

Solute Diffusion rate = 4.0 × 10⁻¹¹ kg/s

Area of cross-section = 0.50 cm²

Length of channel =0.25 cm

Now for the new channel

Area of cross-section = 0.30 cm²

Length of channel =0.10 cm

let the Solute Diffusion rate of new channel = s

now equating the diffusion rate per unit volume for both the channels

$\frac{4\times 10^{-11}}{0.50\times 0.25}=\frac{s}{0.30\times 0.10}$

thus,

$s = 9.6\times 10^{-12}kg/s$

7 0

4 years ago

(1) Explain : Market and its characteristics.

AlekseyPX

Answer:

market is a place where we sell or buy things ( goods)

it's characteristics are

buying and selling goods

perfect competitions

market doesn't refer only a fix place

6 0

3 years ago

A car travels around a level, circular track that is 750m across. What coefficient of friction is required to ensure the car can

Crank

The coefficient of friction must be 0.196

Explanation:

For a car moving on a circular track, the frictional force provides the centripetal force needed to keep the car in circular motion. Therefore, we can write:

$\mu mg = m\frac{v^2}{r}$

where the term on the left is the frictional force acting between the tires of the car and the road, while the term on the right is the centripetal force. The various terms are:

$\mu$ is the coefficient of friction between the tires and the road