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nikdorinn [45]
3 years ago
14

List five rules of golf etiquette. Explain the correct stance, grip, orientation, and parts of a swing.

Physics
2 answers:
Verizon [17]3 years ago
8 0
- keep you eyes on the ball, after you hit the ball you look up, you have to see the club hit the ball

- your grip it really matters on what your comfortable with but first you put your left hand on the club then put your right hand below your left hand, then put your right pinky on top of your index finger, but you can inter lock your pinky and index finger

- DON'T HIT AT THE BALL HIT THREW THE BALL

- parts of the swing, can be pretty tricky to explain, but I'll try my best to explain it, okay so a golf swing is like a pendulum, you swing right to left. Once you lift the club to the right you still look down at the ball. After swimming to the right your have to hit the ball with your right hand strength, in order the ball to fly.

- the way you do your grip defines if the ball will go straight or not. But also the way you swing towards the ball also defines how far the ball is going to go.
Makovka662 [10]3 years ago
8 0

Answer:

1: Wear proper uniform as per the requirement.

2: One should be agree to rules before start of game.

3: Not to make too much noise on ground.

4: Respect the decision.

5: Balls.

Explanation:

* For correct posture, one should stand with a larger open feet, this will help to swing easily and a greater force will be applied.

* Tight grip is necessary to hold the stick and for proper shots. The grip should be tight at the end of the stick.

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The buoyant force must be greater than water.
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Toon Train is traveling at the speed of 10 m/s at the top of a hill. Five seconds later it reaches the bottom of the hill and is
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Answer:

the rate of acceleration of the train is 4 m/s²

Explanation:

Given;

initial velocity of the train, u = 10 m/s

change in time of motion, dt = 5 s

final velocity of the train, v = 30 m/s

The rate of acceleration of the train is calculated as;

a = \frac{dv}{dt} = \frac{v-u}{dt} = \frac{30-10}{5} = \frac{20}{5} = 4 \ m/s^2

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A 1.5 kg cart is attached to a spring with spring constant of 5 N/m. The cart & spring is pulled to stretch the spring by 3
vaieri [72.5K]

22.5 J

Explanation:

Given:

x = 3 m

k = 5\:\text{N/m}

The spring potential energy PE_s is

PE_s = \frac{1}{2}kx^2 = \frac{1}{2}(5\:\text{N/m})(3\:\text{m})^2

\:\:\:\:\:\:\:=22.5\:\text{J}

3 0
2 years ago
In a ballistics test, a 24 g bullet traveling horizontally at 1200 m/s goes through a 31-cm-thick 320 kg stationary target and e
Zanzabum

Answer:

The  velocity is  v_t  =  0.02175 \  m/s

Explanation:

From the question we are told that

   The  mass of the bullet is  m_b  =  0.024 \  kg

    The initial speed of the bullet is  u_b  =  1200 \  m/s

   The mass of the target is  m_t  =  320 \  kg

    The  initial velocity of target is  u_t  =  0  \ m/s

    The  final velocity of the bullet is  is  v_b  =  910 \  m/s

   

Generally according to the law of momentum conservation we have that

      m_b *  u_b  +  m_t *  u_t  =  m_b *  v_b  +  m_t  *  v_t

=>   0.024  *  1200  +  320 *  0  =  0.024 *  910   +  320  *  v_t

=>    v_t  =  0.02175 \  m/s

3 0
3 years ago
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