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Ipatiy [6.2K]
1 year ago
9

joe is an exceptional systems engineer who hopes to work at the techtonic group. he is handicapped and uses a wheelchair. techto

nic does not have wheelchair ramps and some doorways are not wide enough for a wheelchair. the hr department should
Engineering
1 answer:
ASHA 777 [7]1 year ago
5 0

The HR department should install wheelchair ramps and widen the doorways as they don't want to loose an exceptional systems engineer who hopes to work at the techtonic group.

What is a wheelchair ramps?

A wheelchair ramp is an inclined plane that can be installed in addition to or instead of stairs. Ramps make it easier for wheelchair users, as well as people pushing strollers, carts, or other wheeled objects, to enter a building.

The Americans with Disabilities Act requires wheelchair ramps (or other ways for wheelchair users to access a building, such as a wheelchair lift) in new construction for public accommodations in the United States.

A wheelchair ramp can be fixed, semi-fixed, or portable. Permanent ramps are intended to be bolted or otherwise attached to the ground. Semi-permanent ramps are commonly used in the short term and rest on top of the ground or concrete pad.

To know more about wheelchair ramps, visit: brainly.com/question/17395685

#SPJ4

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Rotating magnetic field inside a set of conducting wires is a simple description of a what
Advocard [28]

Answer:

hii there

It is called an electromagnet. The strength of the magnetic field produced is determined by the amount of current passing through the conductor. The rotating magnetic field is the rotor and the windings in which current is produced are in the fixed stator.

Explanation:

hope it helps

have a nice day : )

5 0
2 years ago
A strain gage is mounted at an angle of 30° with respect to the longitudinal axis of the cylindrical pressure. The pressure vess
GuDViN [60]

Answer:

1790 μrad.

Explanation:

Young's modulus, E is given as 10000 ksi,

μ is given as 0.33,

Inside diameter, d = 54 in,

Thickness, t = 1 in,

Pressure, p = 794 psi = 0.794 ksi

To determine shear strain, longitudinal strain and circumferential strain will be evaluated,

Longitudinal strain, eL = (pd/4tE)(1 - 2μ)

eL = (0.794 x 54)(1 - 0.66)/(4 x 1 x 10000)

eL = 3.64 x 10-⁴ radians

Circumferential strain , eH = (pd/4tE)(2-μ)

eH = (0.794 x 54)(2 - 0.33)/(4 x 1 x 10000)

eH = 1.79 x 10-³ radians

The maximum shear strain is 1790 μrad.

4 0
3 years ago
B)<br>State the essential difference between a plain carbon steel<br>and an alloy steel​
choli [55]

Answer:

Plain carbon steel has no or trace external elements while alloy steel has high amount of other elements.

Explanation:

Plain carbon steel has no or trace amount of other elements while alloy steel has high amount of other elements in their composition.

The presence of other elements in alloy steel improvise several physical properties of the steel while plain carbon steel has the basic properties.

6 0
3 years ago
A system consists of N very weakly interacting particles at a temperature T sufficiently high so that classical statistical mech
algol [13]

Answer:

the restoring force is = 3/4NKT

Explanation:

check the attached files for answer.

7 0
3 years ago
A large particle composite consisting of tungsten particles within a copper matrix is to be prepared. If the volume fractions of
OverLord2011 [107]

Answer:

Upper bounds 22.07 GPa

Lower bounds 17.59 GPa

Explanation:

Calculation to estimate the upper and lower bounds of the modulus of this composite.

First step is to calculate the maximum modulus for the combined material using this formula

Modulus of Elasticity for mixture

E= EcuVcu+EwVw

Let pug in the formula

E =( 110 x 0.40)+ (407 x 0.60)

E=44+244.2 GPa

E=288.2GPa

Second step is to calculate the combined specific gravity using this formula

p= pcuVcu+pwTw

Let plug in the formula

p = (19.3 x 0.40) + (8.9 x 0.60)

p=7.72+5.34

p=13.06

Now let calculate the UPPER BOUNDS and the LOWER BOUNDS of the Specific stiffness

UPPER BOUNDS

Using this formula

Upper bounds=E/p

Let plug in the formula

Upper bounds=288.2/13.06

Upper bounds=22.07 GPa

LOWER BOUNDS

Using this formula

Lower bounds=EcuVcu/pcu+EwVw/pw

Let plug in the formula

Lower bounds =( 110 x 0.40)/8.9+ (407 x 0.60)/19.3

Lower bounds=(44/8.9)+(244.2/19.3)

Lower bounds=4.94+12.65

Lower bounds=17.59 GPa

Therefore the Estimated upper and lower bounds of the modulus of this composite will be:

Upper bounds 22.07 GPa

Lower bounds 17.59 GPa

7 0
2 years ago
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