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Oksana_A [137]
3 years ago
5

What is an inexpensive, portable, and common way to assess body fat in the fitness industry?

Physics
1 answer:
Musya8 [376]3 years ago
5 0

Answer: Skinfold testing

Explanation:

Skinfold testing, is also referred to as calliper testing and it's used to know the body fat percentage. Skinfold testing is an inexpensive, portable, and common way to assess body fat in the fitness industry.

It is typically done with the use of caliper tapes, marker pens which makes it cheap. Skinfold testing isn't usually accurate which is as a result of human errors.

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A uniform line charge of density λ lies on the x axis between x = 0 and x = L. Its total charge is 7 nC. The electric field at x
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Answer:

The electric field at x = 3L is 166.67 N/C

Solution:

As per the question:

The uniform line charge density on the x-axis for x, 0< x< L is \lambda

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Now, we know that:

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Also the line charge density:

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Thus

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Now, for small element:

d\vec{E} = K\frac{dq}{x^{2}}

d\vec{E} = K\frac{\lambda }{x^{2}}dx

Integrating both the sides from x = L to x = 2L

\int_{0}^{E}d\vec{E_{2L}} = K\lambda \int_{L}^{2L}\frac{1}{x^{2}}dx

\vec{E_{2L}} = K\lambda[\frac{- 1}{x}]_{L}^{2L}] = K\frac{Q}{L}[frac{1}{2L}]

\vec{E_{2L}} = (9\times 10^{9})\frac{7\times 10^{- 9}}{L}[frac{1}{2L}] = \frac{63}{L^{2}}

Similarly,

For the field in between the range 2L< x < 3L:

\int_{0}^{E}d\vec{E} = K\lambda \int_{2L}^{3L}\frac{1}{x^{2}}dx

\vec{E} = K\lambda[\frac{- 1}{x}]_{2L}^{3L}] = K\frac{Q}{L}[frac{1}{6L}]

\vec{E} = (9\times 10^{9})\frac{7\times 10^{- 9}}{L}[frac{1}{6L}] = \frac{63}{6L^{2}}

Now,

If at x = 2L,

\vec{E_{2L}} = 500 N/C

Then at x = 3L:

\frac{\vec{E_{2L}}}{3} = \frac{500}{3} = 166.67 N/C

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Explanation:

I hope this helps a little! :)

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