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:
Elliptical orbit.<<<<<<<<<<
Answer:
Explanation:
Momentum is a concept and is defined as,
Momentum = mass × velocity
So to calculate the momentum of the car
momentum of the car = mass of the car × velocity of the car
So we get,
momentum of the car = 1800 × 30
= 54000 Ns
The highest elevation reached by the ball in its trajectory is 16.4 m.
To find the answer, we need to know about the maximum height reached in a projectile.
What's the mathematical expression of the maximum height reached in a projectile motion?
- The maximum height= U²× sin²(θ)/g
- U= initial velocity, θ= angle of projectile with horizontal and g= acceleration due to gravity
What's the maximum height reached by a block that is thrown with an initial velocity of 30.0 m/s at an angle of 25° above the horizontal?
- Here, U = 30.0 m/s and θ= 25°
- Maximum height= 30²× sin²(25)/9.8
= 16.4m
Thus, we can conclude that the highest elevation reached by the ball in its trajectory is 16.4 m.
Learn more about the projectile motion here:
brainly.com/question/24216590
#SPJ4
You said that she's losing 1.9 m/s of her speed every second.
So it'll take
(6 m/s) / (1.9 m/s²) = 3.158 seconds (rounded)
to lose all of her initial speed, and stop.