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What is the purpose of the scapula to move during arm elevation?

Inessa [10]

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

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3 0

1 year ago

To do your homework correctly you should (5 points)

pychu [463]

Answer:

C according to me

Explanation:

ITS HARD

4 0

3 years ago

Read 2 more answers

how does spatial pattern of heights illustrate the relationship between temperature density and the rate of vertical pressure ch

Anika [276]

The rate of change of vertical pressure is directly proportional to density and also directly proportional to temperature.

Generally, the relationship between temperature, density and rate of vertical pressure is given as;

$\rho = \frac{PM}{RT}$

$\frac{dP}{dz} = -\rho g\\\\$

where;

<em>ρ is density</em>
<em>T is temperature</em>
<em>dP is rate of change of vertical pressure</em>

Thus, from the formula above, we can conclude the following relationship between temperature, density and the rate of vertical pressure change in spatial pattern of heights.

The rate of change of vertical pressure is directly proportional to density and also directly proportional to temperature.

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5 0

2 years ago

A charge of 32.0 nC is placed in a uniform electric field that is directed vertically upward and has a magnitude of 4.30x 104 V/

hodyreva [135]

A) The work done by the electric field is zero

B) The work done by the electric field is $9.1\cdot 10^{-4} J$

C) The work done by the electric field is $-2.4\cdot 10^{-3} J$

Explanation:

A)

The electric field applies a force on the charged particle: the direction of the force is the same as that of the electric field (for a positive charge).

The work done by a force is given by the equation

$W=Fd cos \theta$

where

F is the magnitude of the force

d is the displacement of the particle

$\theta$ is the angle between the direction of the force and the direction of the displacement

In this problem, we have:

The force is directed vertically upward (because the field is directed vertically upward)
The charge moves to the right, so its displacement is to the right

This means that force and displacement are perpendicular to each other, so

$\theta=90^{\circ}$

and $cos 90^{\circ}=0$ : therefore, the work done on the charge by the electric field is zero.

B)

In this case, the charge move upward (same direction as the electric field), so

$\theta=0^{\circ}$

and

$cos 0^{\circ}=1$

Therefore, the work done by the electric force is

$W=Fd$

and we have:

$F=qE$ is the magnitude of the electric force. Since

$E=4.30\cdot 10^4 V/m$ is the magnitude of the electric field

$q=32.0 nC = 32.0\cdot 10^{-9}C$ is the charge

The electric force is

$F=(32.0\cdot 10^{-9})(4.30\cdot 10^4)=1.38\cdot 10^{-3} N$

The displacement of the particle is

d = 0.660 m

Therefore, the work done is

$W=Fd=(1.38\cdot 10^{-3})(0.660)=9.1\cdot 10^{-4} J$

C)

In this case, the angle between the direction of the field (upward) and the displacement (45.0° downward from the horizontal) is

$\theta=90^{\circ}+45^{\circ}=135^{\circ}$

Moreover, we have:

$F=1.38\cdot 10^{-3} N$ (electric force calculated in part b)

While the displacement of the charge is

d = 2.50 m

Therefore, we can now calculate the work done by the electric force:

$W=Fdcos \theta = (1.38\cdot 10^{-3})(2.50)(cos 135.0^{\circ})=-2.4\cdot 10^{-3} J$

And the work is negative because the electric force is opposite direction to the displacement of the charge.

Learn more about work and electric force:

brainly.com/question/6763771

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5 0

3 years ago

At this radius, what is the magnitude of the net force that maintains circular motion exerted on the pilot by the seat belts, th

Ainat [17]

Answer:

Fc=5253 N

Explanation:

Answer:

Fc=5253 N

Explanation:

sequel to the question given, this question would have taken precedence:

"The 86.0 kg pilot does not want the centripetal acceleration to exceed 6.23 times free-fall acceleration. a) Find the minimum radius of the plane’s path. Answer in units of m."

so we derive centripetal acceleration first

ac (centripetal acceleration) = v^2/r

make r the subject of the equation

r= v^2/ac

ac is 6.23*g which is 9.81

v is 101m/s

substituing the parameters into the equation, to get the radius

(101^2)/(6.23*9.81) = 167m

Now for part

( b) there are two forces namely, the centripetal and the weight of the pilot, but the seat is exerting the same force back due to newtons third law.

he net force that maintains circular motion exerted on the pilot by the seat belts, the friction against the seat, and so forth is the centripetal force.

Fc (Centripetal Force) = m*v^2/r

So (86kg* 101^2)/(167) =

Fc=5253 N

4 0

3 years ago