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kolezko [41]
1 year ago
8

All traditional oriental medicine recognizes that an energy force called _____ flows through the body.

Physics
1 answer:
valentina_108 [34]1 year ago
7 0

All traditional oriental medicine recognizes that an energy force called Qi flows through the body.

Traditional Chinese medicine is predicated on the idea that qi, or the body's life force, travels along meridian channels and maintains equilibrium in a person's physical, mental, emotional, and spiritual well-being. The Feldenkrais technique is a type of somatic education that combines focused attention and gentle movement to improve human function and mobility. You can broaden your range of motion, boost your comfort level, and enhance your flexibility and coordination by using this method. This could involve your food, your workout routine, psychotherapy, relationship counseling, and more. therapies that are complementary or alternative, including acupuncture, chiropractic adjustments, homeopathy, massage, naturopathy, and others.

Learn more about force here-

brainly.com/question/13191643

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Velocity is the rate of motion in a specific direction. ... My velocity is 30 kilometers per hour that-a-way. Average speed is described as a measure of distance divided by time. Velocity can be constant, or it can change (acceleration).

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Velocity is the rate of motion in a specific direction. ... My velocity is 30 kilometers per hour that-a-way. Average speed is described as a measure of distance divided by time. Velocity can be constant, or it can change (acceleration).

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A pitcher throws a 0.140 kg baseball, and it approaches the bat at a speed of 35.0 m/s. The bat does Wnc = 75.0 J of work on the
Eva8 [605]

Answer:

The speed of the ball is 42.5 m/s

Explanation:

The initial kinetic energy of the ball is:

K_1=\frac{1}{2} m v_0^2=\frac{1}{2}*0.140 kg*(35.0 m/s)^2= 85.75 J

The speed of the ball after leaving the bat is:

K_2=K_1+W_{nc}\\ \frac{1}{2}mV^2= 85.75 J + 75 J\\ (\frac{1}{2}mV^2)2=( 160.75 J)2\\ mV^2= 321.5 J\\ V^2= \frac{321.5 J}{0.140kg} \\ V=\sqrt{\frac{321.5 J}{0.140kg}}

V=47.92 m/s

Using kinematic equation we can find the speed of the ball after being 25 m above the point of collision:

V_f^2-V^2=-2gh

V_f^2-(47.92 m/s)^2=-2*9.81m/s^2*25m

V_f^2=-2*9.81m/s^2*25m+(47.92 m/s)^2

V_f=\sqrt{-2*9.81m/s^2*25m+(47.92 m/s)^2}

V_f=42.5m/s

3 0
2 years ago
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