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2 years ago

10 basic rules of badminton?

Physics

2 answers:

saw5 [17]2 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]2 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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3 years ago

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<h3><u>Answer;</u></h3>

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3 years ago

Consider a cyclotron in which a beam of particles of positive charge q and mass m is moving along a circular path restricted by

Ulleksa [173]

A) $v=\sqrt{\frac{2qV}{m}}$

B) $r=\frac{mv}{qB}$

C) $T=\frac{2\pi m}{qB}$

D) $\omega=\frac{qB}{m}$

E) $r=\frac{\sqrt{2mK}}{qB}$

Explanation:

When the particle is accelerated by a potential difference V, the change (decrease) in electric potential energy of the particle is given by:

$\Delta U = qV$

where

q is the charge of the particle (positive)

On the other hand, the change (increase) in the kinetic energy of the particle is (assuming it starts from rest):

$\Delta K=\frac{1}{2}mv^2$

where

m is the mass of the particle

v is its final speed

According to the law of conservation of energy, the change (decrease) in electric potential energy is equal to the increase in kinetic energy, so:

$qV=\frac{1}{2}mv^2$

And solving for v, we find the speed v at which the particle enters the cyclotron:

$v=\sqrt{\frac{2qV}{m}}$

When the particle enters the region of magnetic field in the cyclotron, the magnetic force acting on the particle (acting perpendicular to the motion of the particle) is

$F=qvB$

where B is the strength of the magnetic field.

This force acts as centripetal force, so we can write:

$F=m\frac{v^2}{r}$

where r is the radius of the orbit.

Since the two forces are equal, we can equate them:

$qvB=m\frac{v^2}{r}$

And solving for r, we find the radius of the orbit:

$r=\frac{mv}{qB}$ (1)

The period of revolution of a particle in circular motion is the time taken by the particle to complete one revolution.

It can be calculated as the ratio between the length of the circumference ( $2\pi r$ ) and the velocity of the particle (v):

$T=\frac{2\pi r}{v}$ (2)

From eq.(1), we can rewrite the velocity of the particle as

$v=\frac{qBr}{m}$

Substituting into(2), we can rewrite the period of revolution of the particle as:

$T=\frac{2\pi r}{(\frac{qBr}{m})}=\frac{2\pi m}{qB}$

And we see that this period is indepedent on the velocity.

The angular frequency of a particle in circular motion is related to the period by the formula

$\omega=\frac{2\pi}{T}$ (3)

where T is the period.

The period has been found in part C:

$T=\frac{2\pi m}{qB}$

Therefore, substituting into (3), we find an expression for the angular frequency of motion:

$\omega=\frac{2\pi}{(\frac{2\pi m}{qB})}=\frac{qB}{m}$

And we see that also the angular frequency does not depend on the velocity.

For this part, we use again the relationship found in part B:

$v=\frac{qBr}{m}$

which can be rewritten as

$r=\frac{mv}{qB}$ (4)

The kinetic energy of the particle is written as

$K=\frac{1}{2}mv^2$

So, from this we can find another expression for the velocity:

$v=\sqrt{\frac{2K}{m}}$

And substitutin into (4), we find:

$r=\frac{\sqrt{2mK}}{qB}$

So, this is the radius of the cyclotron that we must have in order to accelerate the particles at a kinetic energy of K.

Note that for a cyclotron, the acceleration of the particles is achevied in the gap between the dees, where an electric field is applied (in fact, the magnetic field does zero work on the particle, so it does not provide acceleration).

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3 years ago

What are some common characteristics that infectious agents like viruses bacteria fungi and parasites have in common and what ar

CaHeK987 [17]

Answer: Some can and can not kill you

Explanation:

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3 years ago

You toss a rock up vertically at an initial speed of 39 feet per second and release it at an initial height of 6 feet. The rock

3241004551 [841]

Answer:

2.583 s, 29.77 ft and 1.219 s

Explanation:

Using equation of motion and taken the motion upward as positive, also a = g ( acceleration due to gravity) = - 32 fts⁻², V= 39 fts⁻¹ V₁ is final velocity, y is the distance in ft from the ground

H = 6 ft, the height from which it is tossed

V₁ = V + gt = V - gt

at maximum height the body came to rest momentarily V₁ = 0

0 = V - gt

-V = -gt

- 39 / -32 = t

t time to reach maximum height = 1.219 s

To Maximum height reached can be calculated with the formula

V₁² = V² + 2g( y - H) where H is the initial height reached by the tossed rock

where V₁ is the final velocity at maximum height which = 0

0 = V² - 2g(y-H) where y is the distance traveled from the ground

-V² = -2g(y-H)

₋V² / -2g = y-H

(V²/2g) + H = y in ft

(39² / (2 × 32)) + 6

y = 29.77 ft

The total time it will be in air can be calculated with the formula below

y = H + Vt - 0.5gt² from y-H = ut + 0.5at²

0.5gt² - Vt - H = 0 since the body returned to the ground ( y = 0)

0.5gt² - Vt - H = 0

using quadratic formula

- (-V)² ± √ ((-V²) - 4 × 0.5g × -H) / (2 × 0.5 × g)

(V ± √ (V² + 2gH)) ÷ g

substitute the values into the expression

t = (39 + √(39² + (2×-32× 6)))/ 32 or (39 - √ (39² + (2 × -32×6))/ 32

t = (39 + √(1521 +384))/32 = (39 + √1905) / 32 = 2.583 s

t = (39 - √1905) / 32 = -0.15 s

The will remain in air (V ± √ (V² + 2gH)) / g seconds. It will reach a maximum height of (V²/2g) + H feet after V/g seconds

8 0

3 years ago

10 basic rules of badminton?​

10 basic rules of badminton?