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3 years ago

6

All of the following are locations for kinesthetic system receptors except __________. A. muscles B. semicircular canals C. join

ts D. tendons Please select the best answer from the choices provided A B C D

2 answers:

jasenka [17]3 years ago

5 0

Answer:

I think it's B

Explanation:

Because hese receptors are found in muscles, tendons, and joints.

spin [16.1K]3 years ago

3 0

Answer:

B semicircular canals

Explanation:

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A school bus takes 0.530 h to reach the school from your house. If the average velocity of the bus is 19.0 km/h to the east, wha

kifflom [539]

Answer:

$x_{distance}=10.07km$

Explanation:

Given data

time=0.530 h

Average velocity Vavg=19.0 km/s

To find

Displacement Δx

Solution

The Formula for average velocity is given as

$V_{avg}=(x_{distance} )/(t_{time} )\\ x_{distance}=V_{avg}*(t_{time} )\\x_{distance}=(19.0km/h)*(0.530h)\\x_{distance}=10.07km$

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4 years ago

A truck with a mass of 1370 kg and moving with a speed of 12.0 m/s rear-ends a 593 kg car stopped at an intersection. The collis

Elza [17]

Answer:

speed of car after collision, v2 =16.1 m/s and of the truck, v1 = 4.6 m/s

Explanation:

Given:

mass of truck M = 1370 kg

speed of truck = 12.0 m/s

mass of car m = 593 kg

collision is elastic therefore,

Applying law of momentum conservation we have

momentum before collision = momentum after collision

1370×12 + 0( initially car is at rest) = 1370×v1+ 593×v2 ....(i)

Also for a collision to be elastic,

velocity of approach = velocity of separation

12 -0 = v2-v1 ....(ii)

using (i) and (ii) we have

So speed of car after collision, v2 =16.1 m/s and of the truck, v1 = 4.6 m/s

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3 years ago

Which factor causes global wind patterns?

Kay [80]

Answer:

B . unequal heating of the Earth's surface by the sun

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3 years ago

Read 2 more answers

A sinusoidal electromagnetic wave is propagating in vacuum. At a given point P and at a particular time, the electric field is i

olasank [31]

Answer:

a

The direction of the wave propagation is the negative z -axis

b

The amplitude of electric and magnetic field are $A_E= 3.35*10^5 V/m$ ,

$A_M= 1.12 *10^{-3} T$ respectively

Explanation:

According to right hand rule, your finger (direction of electric field) would be pointing in the positive x-axis i.e towards your right let your palms be face toward the direction of the magnetic field i.e negative y-axis (toward the ground ) Then anywhere your thumb stretched out is facing is the direction of propagation of the wave here in this case is the negative z -axis

The Intensity of the wave is mathematically represented as

$I = \frac{1}{2} c \epsilon _O E_{rms}^2$

Given that $I = 7.43 \frac{kW}{cm^2} = 7.43 \frac{*10^3}{*10^-{4} }= 7.43*10^7 \frac{W}{m^2}$

Making $E_{rms}$ the subject we have

$E_{rms} = \sqrt{\frac{I}{0.5*c*\epsilon_o} }$

Substituting values as given on the question

$E_{rms} = \sqrt{\frac{7.43 *10^7[\frac{W}{m^2} ]}{0.5 * 3.08*10^8 *8.85*10^{-12}} }$

$= 2.37*10^5 \ V/m$

The amplitude of the electric field is mathematically represented as

$A_E = \sqrt{2} * E_{rms}$

$= \sqrt{2} * 2.37*10^5$

$A_E= 3.35*10^5 V/m$

The amplitude of the magnetic field is mathematically represented as

$A_M = \frac{A_E}{c}$

Substituting value

$A_M = \frac{3.35 *10^5}{3.0*10^8}$

$A_M= 1.12 *10^{-3} T$

7 0

3 years ago

A person's prescription for her new bifocal glasses calls for a refractive power of -0.450 diopters in the distance-vision part,

Angelina_Jolie [31]

Answer:

Far point of the eye is 22.24 m

Far point of the eye is 0.4 m

Explanation:

$\frac{1}{f}=-0.045$

Object distance = u

Image distance = v

Lens equation

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}-\frac{1}{u}=\frac{1}{v}\\\Rightarrow \frac{1}{v}=-0.045-\frac{1}{\infty}\\\Rightarrow \frac{1}{v}=\frac{1}{-0.045}\\\Rightarrow v=-22.22\ m$

Far point

$|v|+\text{Position from eye}\\ =|-22.22|+0.02\\ =22.24\ m$

Far point of the eye is 22.24 m

Object distance = u = 0.25-0.02 = 0.23 m

$\frac{1}{f}=1.75$

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}-\frac{1}{u}=\frac{1}{v}\\\Rightarrow \frac{1}{v}=1.75-\frac{1}{0.23}\\\Rightarrow v=-0.38\ m$

Near point

$|v|+\text{Position from eye}\\= |-0.38|+0.02\\ =0.4\ m$

Far point of the eye is 0.4 m

7 0

3 years ago