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2 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

2 answers:

Irina18 [472]2 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]2 years ago

4 0

The answer to this question is The first option, Or what I should say "A.Thermal"

Your welcome!

Guest

1 year ago

Totally wrong answer but thanks for trying

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hammer [34]

Answer:

I. Electromagnetic waves

Explanation:

7 0

2 years ago

Read 2 more answers

A stunt cyclist needs to make a calculation for an upcoming cycle jump. The cyclist is traveling 100 ft/sec toward an inclined r

Elena-2011 [213]

Answer: $A=23.57^{\circ}$

Explanation:

Given

velocity=100 ft/s

height of landing zone=10 ft

Equation of $x(t)=100 t\cos (A)$

$y(t)=-16t^2+100t\sin (A)+10$

Maximum height=35 feet

at maximum height

$\frac{\mathrm{d} Y}{\mathrm{d} t}=0$

$\frac{\mathrm{d} x}{\mathrm{d} t}=-32t+100\sin A$

$t=3.125\sin A$

At $t=3.125\sin A$

$Y(t)=35=-16\times (3.125\sin A)^2+100\times 3.125\sin A\times \sin A+10$

$312.5\sin^2 A-156.25\sin^2 A=25$

$sin^2 A=0.16$

$A=23.57^{\circ}$

8 0

2 years ago

To measure the coefficient of kinetic friction by sliding a block down an inclined plane the block must be in equilibrium.

lozanna [386]

Answer:

a)

Explanation:

A block sliding down an inclined plane, is subject to two external forces along the slide.
One is the component of gravity (the weight) parallel to the incline.
If the inclined plane makes an angle θ with the horizontal, this component (projection of the downward gravity along the incline, can be written as follows:

$F_{gp} = m*g* sin \theta (1)$

(taking as positive the direction of the movement of the block)

The other force, is the friction force, that adopts any value needed to meet the Newton's 2nd Law.
When θ is so large, than the block moves downward along the incline, the friction force can be expressed as follows:

$F_{f} = \mu_{k} * N (2)$

The normal force, adopts the value needed to prevent any vertical movement through the surface of the incline:

$N = m*g* cos \theta (3)$

In equilibrium, both forces, as defined in (1), (2) and (3) must be equal in magnitude, as follows:

$m*g* sin \theta = \mu_{k} * m*g* cos \theta$

As the block is moving, if the net force is 0, according to Newton's 2nd Law, the block must be moving at constant speed.
In this condition, the friction coefficient is the kinetic one (μk), which can be calculated as follows:

$\mu_{k} = tg \theta$

8 0

1 year ago

You set a tuning fork into vibration at a frequency of 723 Hz and then drop it off the roof of the Physics building where the ac

zaharov [31]

Answer:

Explanation:

Given

Original Frequency $f=723\ Hz$

apparent Frequency $f'=697\ Hz$

There is change in frequency whenever source move relative to the observer.

From Doppler effect we can write as

$f'=f\cdot \frac{v-v_o}{v+v_s}$

where

$f'=$ apparent frequency

v=velocity of sound in the given media

$v_s=$ velocity of source

$v_0=$ velocity of observer

here $v_0=0$

$697=723\cdot (\frac{343-0}{343+v_s})$

$v_s=(\frac{f}{f'}-1)v$

$v_s=(\frac{723}{697}-1)\cdot 343$

$v_s=12.79\approx 12.8\ m/s$

i.e.fork acquired a velocity of $12.8 m/s$

distance traveled by fork is given by

$v^2-u^2=2as$

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

$v_s^2-0=2\times 9.8\times s$

$s=\frac{12.8^2}{2\times 9.8}$

$s=8.35\ m$

5 0

1 year ago

A 4500-kg spaceship is in a circular orbit 190 km above the surface of Earth. It needs to be moved into a higher circular orbit

Maslowich

Answer:

Explanation:

Total energy of a satellite in an orbit , h height away

= - GMm /2 ( R + h )

When h = 380 km

Total energy of a satellite = $\frac{6.67\times10^{-11}\times5.97\times10^{24}\times 4500}{2\times(6378+380)\times10^3}$

= - 13.25 x 10¹⁰ J

When h = 190 km

Total energy of a satellite =

$\frac{6.67\times10^{-11}\times5.97\times10^{24}\times 4500}{2\times(6378+190)\times10^3}$

= - 13.63 x 10¹⁰ J

Diff

= 38 x 10⁸ J Energy will be required.

8 0

2 years ago