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lana66690 [7]
3 years ago
5

List down the procedures for each swimming stroke 1.Crawl 2.Breaststoke 3.Butterflystroke 4. Backstoke​

Physics
1 answer:
Elena L [17]3 years ago
4 0

Answer: Swimming strokes are techniques that includes arm and leg movements to help push the swimmer against water and propel the swimmer forward.

Explanation:

There are different types of swimming strokes these includes:

--> FRONT CRAWL: This is the fastest of all the techniques. The procedure includes:

• the body is kept flat, facing down and in line with the water surface,

• As the swimmer proceeds with movements, the arms are alternately moved in a PULL (with your palms facing down pull in line with the body) and RECOVERY (with the hand closed to the upper thigh, lift one arm out of the water with a bent elbow) actions.

• As you finish the recovery phase, turn quickly side ways to take in some air.

• With ankles relaxed and flexible, point your toes behind you and kick up-and-down in a continuous motion from your thighs.

BUTTERFLY STROKE: The procedures for this technique includes:

• the body is kept flat, facing down and in line with the water surface.

• the arms are moved in three ways, the catch, pull and recovery movements. The Catch involves the arms being straight, shoulder width apart and palms facing down wards, press down and out against the water with both hands at the same time. The pull involves the hands being pulled towards the body in a semicircular motion. The recovery starts at the end of each pull, the arms are moved out and over the water simultaneously and is thrown forward into the starting position.

• the chin is being raised up at the recovery stage to draw in a breath while looking straight.

• With both legs together and toes pointed, kick downwards at the same time.

• the body is moved in a wave-like manner.

BREASTSTROKE: The procedure for this technique includes;

• the body is kept flat, facing down and in line with the water surface

• the arms are also moved in three ways. In the catch movements, with arms out straight and palms facing downwards, press down and out at the same time. With elevated elbows above the arms, pull hard towards the chest. Then while recovering, to reduce drag when pushing against water, the both palms are joined together Infront of the chest and pushed out until the arms are straight again.

• the head is lifted above water at the end of pulling movement to breath in air.

• bend your knees to bring your heel towards your bottom and make a circular motion outwards with your feet until they return to the starting position.

BACKSTROKE: The procedure for this technique includes

• the body kept flat while backing the water surface. But following the arm movement, it rows from side to side.

• the arms performs alternating and opposite movements. As one arm pulls backwards in the water the other arm recovers above the water.

• taking in air should be alternated with the arm movements.

• the legs are moved up and down in a quick succession to enhance movements.

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The magnitude of the resultant force on the balloon is 374.13 N.

The given forces from the image;

  • <em>Upward force = 514 N</em>
  • <em>Downward force = 267 N</em>
  • <em>Eastward force = 678 N</em>
  • <em>Westward force = 397 N</em>

The net vertical force on the balloon is calculated as follows;

F_y = 514 \ N \ \ - \ \ 267 \ N\\\\F_y = 247 \ N

The net horizontal force on the balloon is calculated as follows;

F_x = 678 \ N \ - \ 397 \ N\\\\F_x = 281 \ N

The magnitude of the resultant force on the balloon is calculated as follows;

F = \sqrt{F_y^2 + F_x^2} \\\\F = \sqrt{(247)^2 + (281)^2} \\\\F= 374.13 \ N

Thus, the magnitude of the resultant force on the balloon is 374.13 N.

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Answer:

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Explanation:

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a_{c}= \omega ^2 r

where \omega is the rotational speed and r is the radius. As the centripetal acceleration is feel like an centrifugal acceleration in the rotating frame of reference (be careful, as the rotating frame of reference is <u>NOT INERTIAL,</u> the centrifugal force is a fictitious force, the real force is the centripetal).

<h3>a. </h3>

The rotational speed  is :

2.7 \frac{m}{s^2} = \omega ^2 * 110  \ m

\omega ^2 = \frac{2.7 \frac{m}{s^2}} {110 \ m}

\omega  = \sqrt{ 0.02454 \frac{rad^2}{s^2} }

\omega  = 0.1567 \frac{rad}{s}

Knowing that there are 2\pi \ rad in a revolution and 60 seconds in a minute.

\omega  = 0.1567 \frac{rad}{s}  \frac{1 \ rev}{2\pi \ rad} \frac{60 \ s}{1 \ min}

\omega  = 1.496 \frac{rev}{min}

<h3>b. </h3>

The rotational speed needed is :

9.8 \frac{m}{s^2} = \omega ^2 * 110  \ m

\omega ^2 = \frac{9.8 \frac{m}{s^2}} {110 \ m}

\omega  = \sqrt{ 0.08909 \frac{rad^2}{s^2} }

\omega  = 0.2985 \frac{rad}{s}

Knowing that there are 2\pi \ rad in a revolution and 60 seconds in a minute.

\omega  = 0.2985 \frac{rev}{min}  \frac{1 \ rev}{2\pi \ rad} \frac{60 \ s}{1 \ min}

\omega  = 2.8502 \frac{rev}{min}

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