The purpose of the scapula to move during arm elevation is increase the range of elevation of the arm.
<h3>What is the importance of movement of the scapula during arm elevation?</h3>
The scapula is an important bone which is found in the shoulder and back region of the body.
The scapula enables and increases the range of motion of the arm with its motions.
During arm elevation, the scapula undergoes an upward rotational motion.
Therefore, the purpose of the scapula to move during arm elevation is increase the range of elevation of the arm.
Learn more about scapula motion at: brainly.com/question/5133017
#SPJ12
Answer:
Net charge contained in the cubeq= 3.536×10^-6C
Explanation:
Formular for total flux in a cube is given as:
Total flux= E300Acos(180) + E200Acos(0)
Where A is crossectional area
Total flux= A(E200-E300)
Total flux= q/Eo
q= Eo×total flux
q=(8.84×10^-12)×(100)^2×(100-60)
q= 3.536×10^-6C
Answer:
The magnetic field is
Explanation:
From the question we are told that
The mass of the metal rod is 
The current on the rod is 
The distance of separation(equivalent to length of the rod ) is 
The coefficient of kinetic friction is 
The kinetic frictional force is 
The constant speed is 
Generally the magnetic force on the rod is mathematically represented as

For the rod to move with a constant velocity the magnetic force must be equal to the kinetic frictional force so

=> 
=> 
=> 
<h2><u>Q</u><u>u</u><u>e</u><u>s</u><u>t</u><u>i</u><u>o</u><u>n</u>:-</h2>
The speed of a wave is 40 m/s. If the wavelength is 80 centimeters, what is the frequency of the wave ?
<h2><u>A</u><u>n</u><u>s</u><u>w</u><u>e</u><u>r</u>:-</h2>
<h3>Given:-</h3>
Velocity (V) = 40 m/s
Wavelength
= 80 cm = 0.8 m
<h3>To Find:-</h3>
The frequency (F) of the wave.
<h2>Solution:-</h2>
We know,

40 = F × 0.8
F = 
F = 50
<h3>The frequency of the wave is <u>5</u><u>0</u><u> </u><u>H</u><u>z</u>. [Answer]</h3>