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TiliK225 [7]
3 years ago
12

Which part of fitness will help you hit a baseball farther?

Physics
2 answers:
Dmitry [639]3 years ago
6 0

Answer:

Flexibility! :)

Explanation:

Agility: The ability to rapidly and accurately change the direction of the body. It is important in sports such as tennis which requires players to change direction quickly to hit the ball.

Flexibility: The muscles’ ability to move a joint through a full range of motion, and staying flexible is important to health and performance. As the body ages, the muscles, tendons, and ligaments stiffen, lose elasticity, and become less flexible.

Power: The ability to move the body parts swiftly while applying the maximum force of the muscles.

Body Composition: The combination of fat mass and fat-free mass, including bones, muscles, organs, and water.

Nesterboy [21]3 years ago
5 0

Answer:

The Answer Is Either B,Flexibility, Or C, Power. I Personally Think It's C, Power.

Explanation:

Strong core muscles will create the power needed to hit the ball farther. Most golfers want increased distance on their shots, which is associated with more club head speed. If a golfer increases their ability to generate more force, they'll generate more speed, and the ball will travel farther.

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A horizontal force of 20 N is applied to the object. Will the force applied make the object move?
goldenfox [79]

It depends on two factors: The direction of application of force and the coefficient of friction, \mu, between the object and the surface on which the object is placed.

If the direction of application of force is vertically downwards, then the object will not move for any value of \mu.

If the coefficient of friction is zero, \mu=0, then any value of force that is not in a vertically downward direction, the object will move.

If the direction of application of force is not vertically downwards, then it depends on the value of frictional force.

8 0
3 years ago
You are flying your personal rocketcraft at 0.9c from Star A toward Star B. The distance between the stars, in the stars' refere
Alexxx [7]

Answer:

See explanation.

Explanation:

If both stars explode in simultaneously in the <em>your </em>frame of reference then obviously you will see the two flashes simultaneously, and therefore, the time difference between the events would be zero.

If however, the stars exploded simultaneously in their frame of reference, then you would not observe the flashes simultaneously.  Then the time difference between the events will not be zero, rather, you will observe star B exploding first and star A after.

7 0
3 years ago
Which of the following is the water cycle process where the extra water that plants release is evaporated from their leaves?
ddd [48]
B evaporation since the water is going up
6 0
3 years ago
Which of these describes static electricity?
jarptica [38.1K]
The correct answer Temporary charge
5 0
3 years ago
Read 2 more answers
When switched on, the grinding machine accelerates from rest to its operating speed of 3550 rev/min in 10 seconds. When switched
Paraphin [41]

Answer:

During start total turns

N_1 = 296 turn

After half of the time total turns

N_3 = \frac{29.6 + 0}{2}(5) = 74 turns

Total number of turns during it stop

N_2 = \frac{59.17 + 0}{2}(31) = 917 turn

After half of the time total turns

N_4 = 688 turns

Explanation:

Initially the machine is at rest and then starts rotating with speed 3550 rev/min

now we will have

f = \frac{3550}{60} = 59.17 rev/s

now we know that it took 10 s to reach the speed

so angular acceleration is given as

\alpha = \frac{\omega_f - \omega_i}{t}

\alpha = \frac{59.17- 0}{10} = 5.917 rev/s^2

now it stops in 31 s so the angular deceleration is given as

\alpha_2 = -\frac{59.17}{31} = -1.91 rev/s^2

now initially number of turn to reach the given speed

N_1 = \frac{59.17 + 0}{2}(10) = 296 turn

number of turns during it stop

N_2 = \frac{59.17 + 0}{2}(31) = 917 turn

Now during startup speed after t = 5 s is given as

\omega_1 = (5.917)(5) = 29.6 rev/s

N_3 = \frac{29.6 + 0}{2}(5) = 74 turns

now during it stop the speed after half the time is given as

\omega_2 = 59.17 - (1.91)(15.5) = 29.56 rev/s

now the number of turns is given as

N_4 = \frac{59.17 + 29.56}{2}(15.5)

N_4 = 688 turns

5 0
3 years ago
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