What’s important about physical health

Physics

1 answer:

coldgirl [10]2 years ago

4 0

Answer:

Physical activity or exercise can improve your health and reduce the risk of developing several diseases like type 2 diabetes, cancer and cardiovascular disease. Physical activity and exercise can have immediate and long-term health benefits. Most importantly, regular activity can improve your quality of life.

Explanation:

<h2>hope it is helpful for you </h2><h2>keep smiling </h2>

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1. What is the intensity of a sound when an observer is 10.0 m from the speaker

Nana76 [90]

Answer:

About 133 db.

Explanation:

Sound Intensity Level in db (SIL db) is equal to 10log (base 10) times the ratio of the sound intensity at 200 watts (I) relative to the sound intensity of the reference sound intensity (I sub 0), which by default is equal to 10⁻¹² W/m² or 0 dB.

I = 200 w / 10 m^2 = 20 w per square meter

I sub 0 = 10^-12 w per square meter

SIL = 10log ( I / I sub o) = 20 / 10^-12 = 10log ( 20^12) = 10 ( 13.3 ) = 133 db

Hope I typed this part correctly. Hard to get it in without being able to do exponents, etc. :D

8 0

3 years ago

Problem: The frequency of an FM radio station is 89.3 MHz. Calculate its period. Part B: From the Library, select the general eq

vekshin1

Answer:

Time period, $T=1.11\times 10^{-8}\ s$

Explanation:

We have,

The frequency of an FM radio station is 89.3 MHz.

It is required to find the period of the wave.

The reciprocal of frequency is called time period of a wave. It can be given by :

$T=\dfrac{1}{f}\\\\T=\dfrac{1}{89.3\times 10^{6}\ Hz}\\\\T=1.11\times 10^{-8}\ s$

So, the period of the wave is $1.11\times 10^{-8}\ s$ .

5 0

2 years ago

During a rainy day, as a result of colliding clouds an observer saw lighting and a heard thunder sound. The time between seeing

hodyreva [135]

Answer:

d = 1700 meters

Explanation:

During a rainy day, as a result of colliding clouds an observer saw lighting and a heard thunder sound. The time between seeing the lighting and hearing the sound was 5 second, t = 5 seconds

Speed of sound, v = 340 m/s (say)

Let d is the distance of the colliding cloud from the observer. The distance covered by the object. It is given by :

$d=v\times t$