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gregori [183]
3 years ago
8

Why is the rotation system used in volleyball? Physics was the closest thing to PE

Physics
2 answers:
iris [78.8K]3 years ago
5 0
Eso es por que.. lo que la persona dijo
tankabanditka [31]3 years ago
3 0

Answer:

A rotation occurs after every side out, which is when the receiving team gains the right to serve by winning a rally. ... The new serving team will rotate clockwise one spot. The purpose of this is to rotate all the players through the serving position. If you continue winning points, you stay in position.

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A radioactive isotope has a half-life of 5,000 years. A rock originally contained 16 grams of radioactive isotope, and now conta
mixas84 [53]

Answer:

10,000 years

Explanation:

Given that,

Half life of the radioactive isotope, t_{\frac{1}{2}} = 5000 years

Initial composition, N_0 = 16 grams

Final composition, N = 4 grams

We know that,

N=N_0\times (\dfrac{1}{2})^\frac{t}{t_\frac{1}{2}}}\\\\4=16(0.5)^\frac{t}{5000}}\\\\\dfrac{4}{16}=(0.5)^{\frac{t}{5000}}\\\\0.25=(0.5)^\frac{t}{5000}\\\\0.5^2=(0.5)^\frac{t}{5000}\\\\2=\dfrac{t}{5000}\\\\t=2\times 5000\\\\t=10000\ yrs

Therefore, the age of the rock is 10,000 years.

6 0
3 years ago
BRAINLIEST!
Vlada [557]

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

       

8 0
3 years ago
A ball is thrown straight up from the edge of the roof of a building. A second ball is dropped from the roof a time of 1.19s lat
victus00 [196]

Answer / Explanation

It is worthy to note that the question is incomplete. There is a part of the question that gave us the vale of V₀.

So for proper understanding, the two parts of the question will be highlighted.

A ball is thrown straight up from the edge of the roof of a building. A second ball is dropped from the roof a time of 1.19s later. You may ignore air resistance.

a) What must the height of the building be for both balls to reach the ground at the same time if (i) V₀ is 6.0 m/s and (ii) V₀ is 9.5 m/s?

b) If Vo is greater than some value Vmax, a value of h does not exist that allows both balls to hit the ground at the same time.  

Solve for Vmax

Step Process

a)  Where h = 1/2g [ (1/2g - V₀)² ] / [(g - V₀)²]

Where V₀ = 6m/s,

We have,

           h = 4.9 [ ( 4.9 - 6)²] / [( 9.8 - 6)²]

                 = 0.411 m

Where V₀ = 9.5m/s

We have,

     h = 4.9 [ ( 4.9 - 9.5)²] / [( 9.8 - 9.5)²]

                 = 1152 m

b)  From the expression above, we got to realise that h is a function of V₀, therefore, the denominator can not be zero.

Consequentially, as V₀ approaches 9.8m/s, h approaches infinity.

Therefore Vₙ = V₀max = 9.8 m/s

4 0
3 years ago
Long, long ago, on a planet far, far away, a physics experiment was carried out. First, a 0.210-kg ball with zero net charge was
tigry1 [53]

Answer:

\Delta V=316167V

Explanation:

The difference of electric potential between two points is given by the formula \Delta V=Ed, where <em>d</em> is the distance between them and<em> E</em> the electric field in that region, assuming it's constant.

The electric field formula is E=\frac{F}{q}, where <em>F </em>is the force experimented by a charge <em>q </em>placed in it.

Putting this together we have \Delta V=\frac{Fd}{q}, so we need to obtain the electric force the charged ball is experimenting.

On the second drop, the ball takes more time to reach the ground, this means that the electric force is opposite to its weight <em>W</em>, giving a net force N=W-F. On the first drop only <em>W</em> acts, while on the second drop is <em>N</em> that acts.

Using the equation for accelerated motion (departing from rest) d=\frac{at^2}{2}, so we can get the accelerations for each drop (1 and 2) and relate them to the forces by writting:

a_1=\frac{2d}{t_1^2}

a_2=\frac{2d}{t_2^2}

These relate with the forces by Newton's 2nd Law:

W=ma_1

N=ma_2

Putting all together:

N=W-F=ma_1-F=ma_2

Which means:

F=ma_1-ma_2=m(a_1-a_2)=m(\frac{2d}{t_1^2}-\frac{2d}{t_2^2})=2md(\frac{1}{t_1^2}-\frac{1}{t_2^2})

And finally we substitute:

\Delta V=\frac{Fd}{q}=\frac{2md^2}{q}(\frac{1}{t_1^2}-\frac{1}{t_2^2})

Which for our values means:

\Delta V=\frac{2(0.21Kg)(1m)^2}{7.7\times10^{-6}C}(\frac{1}{(0.35s)^2}-\frac{1}{(0.65s)^2})=316167V

7 0
3 years ago
You have a small piece of iron at 25 °C and place it into a large container of water at 75 °C. Which of these could be the tempe
Ivanshal [37]

50 degrees because the would most likely equal out

8 0
3 years ago
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