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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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Two 13 cm -long thin glass rods uniformly charged to +11nC are placed side by side, 4.0 cm apart. What are the electric field st
DedPeter [7]

Answer:

E1  = 10.15 * 10^4 N/C

E2 = 0

E3 = 10.15 *10^4 N/C

Explanation:

Given data:

Two 13 cm-long thin glass rods ( L ) = 0.13 m

charge (Q)  = +11nC

distance between thin glass rods   = 4 cm .

<u>Calculate the electric field strengths </u>

electric charge due to a single glass rod in the question ( E ) = \frac{Q}{2\pi e_{0}rL }

equation 1 can be used to determine E1, E2 and E3 because the points lie within the two rods hence the net electric field produced will be equal to the difference in electric fields produced

applying equation 1 to determine E1

E1 = \frac{Q}{2\pi e_{0}rL } ( \frac{1}{0.01} - \frac{1}{0.03} )    ( distance from 1 rod is 0.01 m and from the other rod is 0.03 )

   = \frac{11*10^{-9} }{2*3.14*8.85*10^{-12}*0.13 } ( 66.67 )

   = 10.15 * 10^4 N/C

applying equation 1 to determine E2

E2 = \frac{Q}{2\pi e_{0}rL }( \frac{1}{0.02} - \frac{1}{0.02} )

therefore E2 = 0

E1 = E3

hence E3 = 10.15*10^4 N/C

4 0
2 years ago
Suppose you were digging a well into saturated sediments. Why is the sediment’s permeability an important factor in deciding whe
zloy xaker [14]

Answer:

The importance of the sediments permeability is that if it is permeable, water will flow easily through the sediment and thereby produce a very good supply of water for the well.

Explanation:

When digging a well into saturated sediments, the possibility of the sediment with either little saturation or full saturation being able to provide steady water supply will be limited by how permeable it is. Now, the importance of the sediments permeability is that if it is permeable, water will flow easily through the sediment and thereby produce a very good supply of water for the well.

4 0
3 years ago
You hold a bucket in one hand. In the bucket is a 500 g rock. You swing the bucket so the rock moves in a vertical circle 2.2 m
slavikrds [6]

Answer:v=3.28 m/s

Explanation:

Given

mass of rock m=500 gm

diameter of circle d=2.2 m

radius r=\frac{2.2}{2}=1.1 m

At highest Point

mg+N=\frac{mv^2}{r}

At highest Point N=0 because mass is just balanced by centripetal Force

thus mg=\frac{mv^2}{r}

v=\sqrt{gr}

v=\sqrt{9.8\times 1.1}

v=\sqrt{10.78}

v=3.28 m/s

6 0
3 years ago
2 resistors of resistance 1000 ohm and 2000 ohm are joined in series with a 100V supply. A voltmeter of internal resistance 4000
Vadim26 [7]
<h2>The voltmeter reading will be 35.7 volt </h2>

Explanation:

The resistor 1000 ohm and 4000 ohm are connected in parallel .

Their combined resistance is supposed R₁

Thus \frac{1}{R_1} = \frac{1}{1000} + \frac{1}{4000}  

or R₁ = 800 ohm

Therefore the total resistance in circuit is = 2000 + 800 = 2800 ohm

Because these are in series .

We can find  current flowing through the circuit  I = \frac{V}{R} = \frac{100}{2800} = \frac{1}{28}

here R is total resistance .

The potential difference across 1000 ohm = \frac{1}{28} x 1000 = 35.7 volt

Thus voltmeter reading will be 35.7 volt

5 0
3 years ago
4. A metal fork has a density of 7.8 g/cm ³. Is this fork made up of aluminum?
Morgarella [4.7K]
No, aluminum has a density near 2.7 g/cm^3 
<span>7.8 g/cm^3 is near the density of iron (or in the case of a fork, steel).
this is it

</span>
5 0
3 years ago
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