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S_A_V [24]
3 years ago
13

The stimuli for kinesthesis is the __________ energy of joint and muscle movement. A. thermal B. electrical C. mechanical D. che

mical Please select the best answer from the choices provided A B C D
Physics
2 answers:
Irina18 [472]3 years ago
8 0

Answer: C. mechanical

Kinesthetic stimuli is the stimuli which stimulate the awareness in the body regarding the movement of muscles by providing information to the receptors. It controls the activity of muscle, bones, tendons, joints and other body parts. The kinesthetic stimuli by stimulating the receptors present in the muscles and joints generates mechanical energy which facilitates movement of muscles.

lilavasa [31]3 years ago
4 0
The answer to this question is The first option, Or what I should say "A.Thermal"

Your welcome!
Whit
2 years ago
Totally wrong answer but thanks for trying
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True or false cells of humans and plants are same.​
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Explain two reasons why astronomers are continually building larger and larger telescopes. Explain two reasons why astronomers a
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Answer:

* Larger mirrors collect more light and therefore fainter and more distant objects can have enough intensity to be detected

* arger mirrors decreases the angle of dispersion giving a better resolution of the bodies

Explanation:

Refracting telescopes get bigger every day for two main reasons.

* Larger mirrors collect more light and therefore fainter and more distant objects can have enough intensity to be detected

* the diffraction process for circular apertures is given by

               θ = 1.22 λ / D

where d is the diameter of the mirror, therefore having larger mirrors decreases the angle of dispersion giving a better resolution of the bodies

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3 years ago
A 20.0 cm tall object is placed 50.0 cm in front of a convex mirror with a radius of curvature of 34.0 cm. Where will the image
Neko [114]

Answer:

1.Theimage will be located at -0.13m or -13 cm

2.The height of the image will be 0.052m or 5.2cm

Explanation:

Given that;

Height of object, h=20 cm = 0.2m

Object distance in front of convex mirror, o,= 50 cm =0.5m

Radius of curvature, r, =34 cm =0.34m

Let;

Image distance, i,=?

Image height, h'=?

You know that focal length,f, is half the radius of curvature,hence

f=r/2 = 0.34/2 = 0.17m ( this length is inside the mirror, in a virtual side, thus its is negative)

f= -0.17m

Apply the relationship that involves the focal length;

=\frac{1}{o} +\frac{1}{i} =\frac{1}{f}

=\frac{1}{0.5} +\frac{1}{i} =-\frac{1}{0.17}

Re-arrange to get i

\frac{1}{i} =-2-5.88\\\\\\\frac{1}{i} =-7.88\\\\i=-0.13m

This is a virtual image formed at a negative distance produced through extension of drawing rays behind the mirror if you use rays to locate the image behind the mirror

Apply the magnification formula

magnification, m=height of image÷height of object

m=\frac{h'}{h} =-\frac{i}{o}

substitute the values to get the height of image h'

\frac{h'}{0.20} =-\frac{-0.13}{0.5} \\\\\\h'=\frac{0.13*0.20}{0.5} \\\\\\h'=\frac{0.025}{0.5} =0.052m\\\\\\h'=5.2cm

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4 years ago
A type of star called a red dwarf gets its name because it is a relatively small star that appears redder than the Sun when view
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