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lara31 [8.8K]
3 years ago
7

10 basic rules of badminton?​

Physics
2 answers:
saw5 [17]3 years ago
8 0

Answer:

The 10 rules of badminton are as follows:

1. A game starts with a coin toss. Whoever wins the toss gets to decide whether they would serve or receive first OR what side of the court they want to be on. The side losing the toss shall then exercise the remaining choice.

2. At no time during the game should the player touch the net, with his racquet or his body.

3. The shuttlecock should not be carried on or come to rest on the racquet.

4. A player should not reach over the net to hit the shuttlecock.

5. A serve must carry cross court (diagonally) to be valid.

6. During the serve, a player should not touch any of the lines of the court, until the server strikes the shuttlecock. During the serve the shuttlecock should always be hit from below the waist.

7. A point is added to a player's score as and when he wins a rally.

8. A player wins a rally when he strikes the shuttlecock and it touches the floor of the opponent's side of the court or when the opponent commits a fault. The most common type of fault is when a player fails to hit the shuttlecock over the net or it lands outside the boundary of the court.

9. Each side can strike the shuttlecock only once before it passes over the net. Once hit, a player can't strike the shuttlecock in a new movement or shot.

10. The shuttlecock hitting the ceiling, is counted as a fault.

Explanation:

Paladinen [302]3 years ago
3 0

Explanation:

Rules

A match consists of the best of three games of 21 points.

The player/pair winning a rally adds a point to its score.

At 20-all, the player/pair which first gains a 2-point lead wins that game.

At 29-all, the side scoring the 30th point wins that game.

The player/pair winning a game serves first in the next game.

A badminton match can be played by two opposing players (singles) or four opposing players (doubles).

A competitive match must be played indoors utilising the official court dimensions.

A point is scored when the shuttlecock lands inside the opponent's court or if a returned shuttlecock hits the net or lands outside of the court the player will lose the point.

At the start of the rally, the server and receiver stand in diagonally opposite service courts.

A legal serve must be hit diagonally over the net and across the court.

A badminton serve must be hit underarm and below the server's waist height with the racquet shaft pointing downwards, the shuttlecock is not allowed to bounce. After a point is won, the players will move to the opposite serving stations for the next point.

The rules do not allow second serves.

During a point a player can return the shuttlecock from inside and outside of the court.

A player is not able to touch the net with any part of their body or racket.

A player must not deliberately distract their opponent.

A player is not able to hit the shuttlecock twice.

A 'let' may be called by the referee if an unforeseen or accidental issue arises.

A game must include two rest periods. These are a 90-second rest after the first game and a 5-minute rest after the second game.

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The strength of the gravitational force between two objects depends on two factors, mass and distance. the force of gravity the masses exert on each other. If one of the masses is doubled, the force of gravity between the objects is doubled. increases, the force of gravity decreases

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3 years ago
A 40.0 kg beam is attached to a wall with a hi.nge and its far end is supported by a cable. The angle between the beam and the c
GalinKa [24]

The magnitude of the force that the beam exerts on the hi.nge will be,261.12N.

To find the answer, we need to know about the tension.

<h3>How to find the magnitude of the force that the beam exerts on the hi.nge?</h3>
  • Let's draw the free body diagram of the system using the given data.
  • From the diagram, we have to find the magnitude of the force that the beam exerts on the hi.nge.
  • For that, it is given that the horizontal component of force is equal to the 86.62N, which is same as that of the horizontal component of normal reaction that exerts by the beam on the hi.nge.

                           N_x=86.62N

  • We have to find the vertical component of normal reaction that exerts by the beam on the hi.nge. For this, we have to equate the total force in the vertical direction.

                           N_y=F_V=mg-Tsin59\\

  • To find Ny, we need to find the tension T.
  • For this, we can equate the net horizontal force.

                           F_H=N_x=Tcos59\\\\T=\frac{F_H}{cos59} =\frac{86.62}{0.51}= 169.84N

  • Thus, the vertical component of normal reaction that exerts by the beam on the hi.nge become,

                    N_y= (40*9.8)-(169.8*sin59)=246.4N

  • Thus, the magnitude of the force that the beam exerts on the hi.nge will be,

                 N=\sqrt{N_x^2+N_y^2} =\sqrt{(86.62)^2+(246.4)^2}=261.12N

Thus, we can conclude that, the magnitude of the force that the beam exerts on the hi.nge is 261.12N.

Learn more about the tension here:

brainly.com/question/28106871

#SPJ1

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Which most accurately describes the path that sound​ travels??
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A vertical spring stretches 4.0 cm when a 12-g object is hung from it. The object is replaced with a block of mass 28 g that osc
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Answer:

Time period of the osculation will be 0.0671 sec  

Explanation:

It is given a vertical spring is stretched by 4 cm

So change in length of the spring x = 4 cm = 0.04 m

Mass which is hung from it m = 12 gram = 0.012 kg

Sprig force will be equal to weight of the mass

So kx=mg

k\times 0.04=0.012\times 9.8

k = 244.7 N/m

Now new mass is m = 28 gram = 0.028 kg

So time period with new mass will be

T=2\pi \sqrt{\frac{m}{k}}

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A car speeds up from 18.54 m/s to<br> 29.52 m/s in 13.84 s.<br> The acceleration of the car is:
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Answer:

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Explanation:

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A = 10.98m/s / 13.84s

A = .7934m/s^{2}

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