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lisabon 2012 [21]
3 years ago
8

BRAINLIEST!

Physics
1 answer:
Vlada [557]3 years ago
8 0

The two tests used for measuring each of the following fitness areas are as follows:

     a) Cardiorespiratory fitness → The 2.4 km run test; Astrand treadmill test

     b) Muscle strength → Muscle Testing Scale,  Bench/Leg press

     c) Body Composition → DEXA Scan, Bioelectrical Impedance Analysis

      Cardiorespiratory fitness is the ability of the circulatory as well as the respiratory systems to deliver oxygen to the mitochondria of skeletal muscle for energy generation during physical activity.

     

      Cardiorespiratory fitness is a vital indicator of both physical and mental wellbeing. It can lower the chance of developing heart disease and type 2 diabetes. Cardiorespiratory fitness improves lung and heart health while also increasing emotions of well-being.

  • The astrand treadmill test is used widely for testing the fitness of cardiorespiratory. The usefulness of this test is used to assess a client's aerobic fitness.
  • The 2.4 km run test is an uncomplicated running test that needs the use of a stopwatch and a running track. It is widely used for aerobic fitness

      Muscle strength refers to the capacity to move and lift items.  It is determined by the amount of power you can exert and the amount of weight you can lift within a short amount of time.

   

      The test used to determine the muscle strength can be:

  • Muscle Testing Scale;
  • Bench/Leg Press

     

     Muscular strength testing is used to examine a complaint of weakness, which is common when there is a suspectable neurologic illness or muscle imbalance.

 

      Bench presses are weight-training exercises that can be added to your routine to help increase upper body strength and improve muscular endurance.

      Body Composition relates to the bones, muscles, and body fat in the body. Body physique composition is used by medical professionals to determine if you are at a healthy weight for your specific body.

       The test used to determine the body Composition can be:

  • DEXA Scan
  • Bioelectrical Impedance Analysis  

        Dexa Scan scans your body using x-ray technology to offer a thorough evaluation of the amount of muscle mass and fat mass your body contains, as well as where fat and muscle are deposited or stored on your body.

        Bioelectrical Impedance Analysis evaluates fat-free mass by passing a low electric current across the body. Because electricity can only move by water, and each body tissue includes variable quantities of water, the conductivity of each tissue type influences the flow of the electric current. 

Therefore, from the above explanations, we can conclude that we've clearly understood the various test for each fitness area.

Learn more about body fitness here:

brainly.com/question/10033432?referrer=searchResults

       

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A bag of sugar weighs 5.00 lb on Earth. What would it weigh in newtons on the Moon, where the free-fall acceleration is one-sixt
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Answer:

Earth: 22.246 N

Moon: 3.71 N

Jupiter: 58.72 N

Explanation:

The mass of an object will remain constant in any location, its weight however, can fluctuate depending on its location. For example, a golf ball will weigh less on the moon, but its mass will not be different if it was on earth.

To calculate anything, we need to convert to standard measurements.

5.00 lbs = 2.27 kg

On earth, gravity is measured to be 9.8 m/s², so the weight in Newtons on Earth would be: (2.27 kg) x (9.8 m/s²) = 22.246 N

Repeated on the moon where gravity is (9.8 m/s²) x (1/6) = 1.633 m/s², so the weight in Newtons on the moon would be: (2.27 kg) x (1.633 m/s²) = 3.71 N

Repeated on Jupiter where gravity is (9.8 m/s²) x (2.64) = 25.87 m/s², so the wight in Newtons on Jupiter would be: (2.27 kg) x (25.87 m/s²) = 58.72 N

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Answer:

For left = 0  N/C

For right = 0  N/C

At middle = -7.6836 * 10^{-11} \vec{i}  N/C

Explanation:

Given data :-

б =6.8 * 10^{-22} C/ m²

Considering the two thin metal plates to be non conducting sheets of charges.

Electric field is given by

E = \frac{\sigma }{2\varepsilon }

1) To the left of the plate

\vec{E}= (\frac{\sigma }{2\varepsilon })(-\vec{i})+  (\frac{\sigma }{2\varepsilon })(\vec{i})   = 0 N/C.

2) To the right of them.

\vec{E}= (\frac{\sigma }{2\varepsilon })(-\vec{i})+  (\frac{\sigma }{2\varepsilon })(\vec{i})   = 0 N/C.

3) Between them.

\vec{E}= (\frac{\sigma }{2\varepsilon })(-\vec{i})+  (\frac{\sigma }{2\varepsilon })(-\vec{i}) = (\frac{\sigma }{\varepsilon })(-\vec{i}) = -\frac{6.8 * 10^{-22} }{8.85 * 10 ^{-12} }  \vec{i} =   -7.6836 * 10^{-11} \vec{i} N/C

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