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3 years ago

A waist-to-hip ratio of 1.0 is considered healthy. True False

Physics

2 answers:

rjkz [21]3 years ago

8 0

Answer: False

Explanation:

The waist to hip ratio can be define as the dimensionless ration which measures the circumference of waist to hip. It is obtain by dividing the circumference of the waist to that of the circumference of hip. The waist to hip ratio more than 1 can be considered as risk in development of heart disease. A healthy women is expected to have a waist-hip ratio under .80 and for healthy males it should be equal to and less than .90.

On the basis of the above description, the statement, A waist-to-hip ratio of 1.0 is considered healthy is false.

laila [671]3 years ago

7 0

false a waist to hip ratio that is healthy would be .80 or less is considered healthy for most individuals hope this helps

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Statistical time division multiplexing does not require the capacity of the circuit to be equal to the sum of the combined circu

aleksklad [387]

Answer:

The answer is True

Explanation:

Statistical Multiplexing is considered an example of communication link sharing which makes it comparable to DBA (Dynamic Bandwidth Allocation). Here, communication channels are broken down into data streams to optimize the communication process.

In Statistical Time-division Multiplexing, time slots are allocated to data streams for communication optimization. This method makes sure that no time slot or bandwidth is wasted.

Hence, the sum of combined circuits must not be equal to the capacity of the circuit to work effectively.

7 0

3 years ago

If 270 watts of power is used in 42 seconds, how much work was done

navik [9.2K]

Answer: W = 11340J

Explanation:

Hey there! I will give the following steps, if you have any questions feel free to ask me in the comments below.

So this is the Formula: Power = Work / Time.

Step 1: Find the Formula

P = W / T

Step 2: Make W the subject of the equation.

W = PT

Step 3: Given.

P = 270 Watts, T = 42 seconds

Step 4: Substitute these values into equation 2 .

W = 270(42)

Step 5: Simplify.

W = 11340J

The amount of work done was 11340.

~I hope I helped you! :)~

4 0

3 years ago

For an object starting from rest and accelerating with constant acceleration, distance traveled is proportional to the square of

natali 33 [55]

The problem states that the distance travelled (d) is directly proportional to the square of time (t^2), therefore we can write this in the form of:

d = k t^2

where k is the constant of proportionality in furlongs / s^2

Using the 1st condition where d = 2 furlongs, t = 2 s, we calculate for the value of k:

2 = k (2)^2

k = 2 / 4

k = 0.5 furlongs / s^2

The equation becomes:

d = 0.5 t^2

Now solving for d when t = 4:

d = 0.5 (4)^2

d = 0.5 * 16

d = 8 furlongs

It traveled 8 furlongs for the first 4.0 seconds.

8 0

3 years ago

Using a power supply, I place +1.00µC on one plate of a capacitor. A capacitor is a device for storing charge. I, then, connect

Paraphin [41]

Answer:

$N=62421972534 electrons$

Explanation:

The charge transferred charge Q= −10nC = -10*10-9 C

Charge of a electron : $Q_{e}= -1,602*10^{-19}C$

Number of electrons transferred: $N=\frac{Q}{Q_{e} }=(-10*10^{-9} )/(-1,602*10^{-19})=62421972534 electrons$

5 0

3 years ago

I need both parts please (a) Given a material with an attenuation coefficient (a) of 0.6/cm, what is the intensity of a beam (wi

Masteriza [31]

Answer:

After it has traveled through 1 cm : $I(1 \ cm) = 0.5488 I_0$

After it has traveled through 2 cm : $I(2 \ cm) = 0.3012 I_0$

After it has traveled through 1 cm : $od( 1\ cm) = 0.2606$

After it has traveled through 2 cm : $od( 2\ cm) = 0.5211$

Explanation:

For this problem, we can use the Beer-Lambert law. For constant attenuation coefficient $\mu$ the formula is:

$I(x) = I_0 e^{-\mu x}$

where I is the intensity of the beam, $I_0$ is the incident intensity and x is the length of the material traveled.

For our problem, after travelling 1 cm:

$I(1 \ cm) = I_0 e^{- 0.6 \frac{1}{cm} \ 1 cm}$

$I(1 \ cm) = I_0 e^{- 0.6}$

$I(1 \ cm) = 0.5488 \ I_0$

After travelling 2 cm:

$I(2 \ cm) = I_0 e^{- 0.6 \frac{1}{cm} \ 2 cm}$

$I(2 \ cm) = I_0 e^{- 1.2}$

$I(2 \ cm) = 0.3012 \ I_0$

The optical density od is given by:

$od(x) = - log_{10} ( \frac{I(x)}{I_0} )$ .

So, after travelling 1 cm:

$od( 1\ cm) = - log_{10} ( \frac{0.5488 \ I_0}{I_0} )$

$od( 1\ cm) = - log_{10} ( 0.5488 )$

$od( 1\ cm) = - ( - 0.2606)$

$od( 1\ cm) = 0.2606$

After travelling 2 cm:

$od( 2\ cm) = - log_{10} ( \frac{0.3012 \ I_0}{I_0} )$

$od( 2\ cm) = - log_{10} ( 0.3012 )$

$od( 2\ cm) = - ( - 0.5211)$

$od( 2\ cm) = 0.5211$

3 0

3 years ago