All categories

3 years ago

What’s the most important benefit of maintaining a neutral posture

1 answer:

7 0

Neutral posture is essential for optimal wellbeing and functioning of the body. Holding the weight of the body The most important function of a neutral posture is to maintain the body in an upright position, supporting the body against gravity

You might be interested in

Explain the difference between the connection of a cumulative compound and a differential compound motor

Alexandra [31]

Answer:

Explanation:

A motor is a device that directs current in electrical energy form to mechanical energy, which is known as direct current (DC) motors.

DC motors are of three types: (a) The series motor, (b) The shunt motor, and (c) the compound motor. Our main focus here is the Compound motor, which is further sub-divided into:

i) The cumulative compound motors

ii) The differential compound motors

The difference between these two are:

Cumulative compound motors Differential compound motors

In cumulative compound motors, In differential compound motors,

both the series and shunt windings both series and shunt are

are connected in a way that, connected in a way that the

production of fluxes through them production of fluxes via them

assist each other i.e. they aid each always opposes each other i.e.

other in the production of magnetism they oppose each other in the

production of magnetism.

6 0

3 years ago

Statement 1: All balls hit the ground at the same time. Statement 2: All balls hit the ground with the same force. Statement 3:

laiz [17]

Answer:

Statement 1: All balls hit the ground at the same time

Explanation:

When there is no resistance of air, the acceleration due to gravity experienced by all the bodies are same. So for falling bodies, neglecting the air resistance, the falling object will be weightless and therefore all the objects will hit the ground at the same time when there is nor air resistance and the objects are considered to be falling in vacuum.

7 0

3 years ago

use engineering judgement to estimate the size of cooling and heating equipment that is needed for a three story, 31,000 square

Blizzard [7]

Answer: operating cost = 22820.736 $

Energy = 32.291 KBtu/sf.

Explanation:

Total heating load is given as = 31000 * 31.25

= 968.75* 103 Btu

From the cooling capacity application;

If 1000ftsquare = 2.8 TR

Therefore 30000ftsquare = x

Where x is the total cooling load,

Therefore x = (30000 * 2.8) / 1000

x = 84TR.

Therefore, the total cooling load, x = 84TR

Using conversion factor;

i.e. converting Btu/hr to MBH

1 MBH = 1000 Btu per hour

we have 968.75* 103 Btu/hr = 968.75 MBH

let us proceed.

Estimating the “Annual operating cost” we first calculate the maximum operating cost (Avg/Annum)

For cooling load:

Max. operatn. Cost = 14 * 84 = 1176$

For heating load:

The average heating loading hours = 1000 hrs.

Conversion to Btu/yr gives = 968.75* 103 * 1000

= 968.75* 106 Btu/yr.

Conversion of 968.75* 103Btu to Kwh gives

1Btu = 0.000293

968.75* 103 Btu = 283.84 Kwh

Therefore, average cost gives;

(8.04/100) $ = 1Kwh

X = 283.84*103

X = 22820.736 $

Operating cost = 22820.736 $

Energy use in KBtu/sf is given as

E = 968.75* 103 / 30000

E = 32.291 KBtu/sf.

8 0

4 years ago

A solid shaft and a hollow shaft of the same material have same length and outer radius R. The inner radius of the hollow shaft

alexandr402 [8]

Answer with Explanation:

By the equation or Torque we have

$\frac{T}{I_{p}}=\frac{\tau }{r}=\frac{G\theta }{L}$

where

T is the torque applied on the shaft

$I_{p}$ is the polar moment of inertia of the shaft

$\tau$ is the shear stress developed at a distance 'r' from the center of the shaft

$\theta$ is the angle of twist of the shaft

'G' is the modulus of rigidity of the shaft

We know that for solid shaft $I_{p}=\frac{\pi R^4}{2}$

For a hollow shaft $I_{p}=\frac{\pi (R_o^4-R_i^4)}{2}$

Since the two shafts are subjected to same torque from the relation of Torque we have

1) For solid shaft

$\frac{2T}{\pi R^4}\times r=\tau _{solid}$

2) For hollow shaft we have

$\tau _{hollow}=\frac{2T}{\pi (R^4-0.7R^4)}\times r=\frac{2T}{\pi 0.76R^4}$

Comparing the above 2 relations we see

$\frac{\tau _{solid}}{\tau _{hollow}}=0.76$

Similarly for angle of twist we can see

$\frac{\theta _{solid}}{\theta _{hollow}}=\frac{\frac{LT}{I_{solid}}}{\frac{LT}{I_{hollow}}}=\frac{I_{hollow}}{I_{solid}}=1.316$

Part b)

Strength of solid shaft = $\tau _{max}=\frac{T\times R}{I_{solid}}$

Weight of solid shaft = $\rho \times \pi R^2\times L$

Strength per unit weight of solid shaft = $\frac{\tau _{max}}{W}=\frac{T\times R}{I_{solid}}\times \frac{1}{\rho \times \pi R^2\times L}=\frac{2T}{\rho \pi ^2R^5L}$

Strength of hollow shaft = $\tau '_{max}=\frac{T\times R}{I_{hollow}}$

Weight of hollow shaft = $\rho \times \pi (R^2-0.7R^2)\times L$

Strength per unit weight of hollow shaft = $\frac{\tau _{max}}{W}=\frac{T\times R}{I_{hollow}}\times \frac{1}{\rho \times \pi (R^2-0.7^2)\times L}=\frac{5.16T}{\rho \pi ^2R^5L}$

Thus $\frac{Strength/Weight _{hollow}}{Strength/Weight _{Solid}}=5.16$

3 0

4 years ago

a(n) ? is an intentionally constructed, low-impedance electrically conductive path designed and intended to carry current during

tester [92]

A effective ground-fault current path is an intentionally constructed, low-impedance electrically conductive path designed and intended to carry current during ground-fault conditions from the point of grounding on a wiring system to the electrical supply source.

<h3>Is earth an effective ground fault current path?</h3>

Sticking the wire in the ground is not sufficient since the earth is not thought to be a reliable ground-fault current channel.
The electrical system of a building or other structure is based on grounding.
To give a fault current a secure path to travel, grounding is used.
When installing switches, light fixtures, appliances, and receptacles, a complete ground route must be kept.
The undesired current flow trips circuit breakers or blows fuses in a system that is correctly grounded.
Through the use of a grounding bank, effective grounding maintains voltages within predetermined limits during a line-to-ground fault (short-circuit condition).

To learn more about ground-fault current channel refer,

brainly.com/question/28498355

#SPJ4

5 0

1 year ago