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Butoxors [25]
2 years ago
15

A movement that keeps the distal end of the body segment fixed in one location describes what type of kinetic chain movement? Op

en kinetic chain movement Lateral kinetic chain movement Dynamic kinetic chain movement Closed kinetic chain movement
Physics
1 answer:
Greeley [361]2 years ago
5 0

A movement which keeps the distal end of the body segment fixed in one location describes what type of kinetic chain movement is known as a <u>closed kinetic chain movement.</u>

<h3>What is a closed kinetic chain movement?</h3>

A closed kinetic chain movement is defined as that position where the most distal aspects of a particular extremity are fixed to the earth or another solid object.

In conclusion; a movement which keeps the distal end of the body segment fixed in one location describes what type of kinetic chain movement is known as a closed kinetic chain movement.

Learn more about closed kinetic chain motion:

brainly.com/question/12405822

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Which of the following causes tornados A. Dry climates B.Sationary fronts B. Condensation D. Wind shear
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compare the magnitude and range of the four basic forces—gravitational, electromagnetic, weak nuclear and strong nuclear
Andrews [41]

Answer:

The four fundamental forces magnitude and range are in order of Strong Nuclear Force > electromagnetic force > weak nuclear force > gravitational force.

Explanation:

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6 0
3 years ago
A girl of mass 50.6 kg stands on the edge of a frictionless merry-go-round of mass 827 kg and radius 3.72 m that is not moving.
GaryK [48]

Answer:

(a). The value of angular speed of the merry-go-round \omega = - 5.82 × 10^{-3} \frac{rad}{s}

(b). The linear speed of the girl after the rock is thrown V = -1.89 × 10^{-2} \frac{m}{s}

Explanation:

Given data

Mass of the girl m_{g} = 50.6 kg

Mass of merry-go-round m_{m} = 827 kg

Radius r = 3.72 m

The speed of the rock relative to the ground V_{r} = 7.82 \frac{m}{s}

(a). The angular speed of the merry-go-round is given by

\omega = - [\frac{m_{r}v_{r}  }{r} ] \frac{2}{m_{m} + 2m_{g} }

Put all the values in above formula

\omega = \frac{(1.13)(7.82)}{3.27} \frac{2}{827 + (2)50.6}

\omega = - 5.82 × 10^{-3} \frac{rad}{s}

This is the value of angular speed of the merry-go-round.

(b). The liner speed of the girl is given by

⇒ V = r × \omega

⇒ V = - 3.72 × 5.82 × 10^{-3}

⇒ V = -1.89 × 10^{-2} \frac{m}{s}

This is the linear speed of the girl after the rock is thrown.

7 0
3 years ago
An undersea research chamber is spherical with an external diameter of 5.50 m . The mass of the chamber, when occupied, is 87600
arsen [322]

Answer:

A)

8.75\times10^{5} N

B)

1.7\times10^{4} N

Explanation:

d = diameter of the spherical chamber = 5.50 m

Volume of the spherical chamber is given as

V = \frac{4\pi (\frac{d}{2} )^{3}}{3}\\ V = \frac{4(3.14) (\frac{5.50}{2} )^{3}}{3}\\V = 87.07 m^{3}

\rho = density of the seawater = 1025 kg m⁻³

Part A)

Force of buoyancy by seawater on the chamber is given as

F_{b} = \rho V g \\F_{b} = (1025) (87.07) (9.8)\\F_{b} = 8.75\times10^{5} N

Part B)

T = tension force in the cable in upward direction

m = mass of the chamber = 87600 kg

Weight of the chamber is given as

F_{g} = mg \\

Using equilibrium of force in the vertical direction, we have

T + F_{g} = F_{b}\\T + m g = F_{b}\\T + (87600) (9.8) = 8.75\times10^{5}\\T = 16520 N \\T = 1.7\times10^{4} N

3 0
3 years ago
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