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GREYUIT [131]
3 years ago
9

To be effective, an exercise program must be?

Physics
2 answers:
Diano4ka-milaya [45]3 years ago
7 0

A.

uniquely tailored to every person

ExtremeBDS [4]3 years ago
3 0

To be effective, an exercise program must have an aerobic form, portion for strength enhancement, and a stretching part. These three things are essential because they each target specific improvements in your body. For example, aerobics can help you maintain your body’s fitness or make it better. This usually targets your heart rate and ensures that you burn fat while doing so. Second is strength enhancement; this will make sure that your body becomes better – not just in a feeble state. Lastly is stretching, your muscles are like rubber bands. You cannot end or start your exercise program without stretching simply because they can damage your muscles as well. Aside from this, stretching can stop you from shocking your body into a physical activity, which may cause you to lose consciousness or have undue stress and fatigue. 

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The sun is 1.5 × 108 km from Earth. The index of refraction for water is 1.349. How much longer would it take light from the sun
tankabanditka [31]

Answer:

175s

Explanation:

time it takes sunlight to reach the earth in  vacuum

C=light speed=299792458m/s

X=1.5x10^8km=1.5x10^11m

c=X/t

T1=X/c

T1=1.5X10^11/299792458=500.34s

time it takes sunlight to reach the earth in  water:

First we calculate the speed of light in water taking into account the refractive index

Cw=299792458m/s/1.349=222233104.5m/s

T2=1.5x10^11/222233104.5m/s=675s

additional time it would take for the light to reach the earth

ΔT=T2-T1=675-500=175s

4 0
3 years ago
Una manguera de agua de 1.3 cm de diametro es utilizada para llenar una cubeta de 24 Litros. Si la cubeta se llena en 48 s. A) ¿
Viefleur [7K]

Answer:

We must translate this:

a 1.3 cm diameter water hose is used to fill a 24-liter bucket. If the bucket is filled in 48 s.  

A) What is the speed with which the water leaves the hose?

B) if the diameter of the hose is reduced to 0.63 cm and assuming the same flow, what will be the speed of the water leaving the hose?

A) If the velocity of the water is Xcm/s

and the radius of the hose is equal to half its diameter, so it is 1.3cm/2

Then in one second we can considerate that a cylinder of:

V = pi*(1.3cm/2)^2*X cm^3 of water.

So we have that quantity in one second of flow.

where pi = 3.14

then in 48 seconds, the amount of water in the bucket is:

V = 48*pi*(1.3/2)^2*X = 24 L = 24,000 cm^3

Now we need to solve this for X.

48*3.14*(1.3/2)^2*X = 24000

63.679*x = 24000

x = 24000/63.679 = 376.89

So the velocity of the water is 376 cm per second.

B) if the diameter is 0.64cm, we have the equation:

48*3.14*(0.63/2)^2*x = 24000

14.955*X = 24000

X = 24000/14.955 = 1604.814 cm/s

6 0
3 years ago
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Veseljchak [2.6K]
The gravity is pushing rhe boat down
3 0
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Alenkinab [10]

Explanation:

given

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