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Mashcka [7]
3 years ago
13

1. When the pneumatic device was demonstrated, explain something that was different than what you expected.

Engineering
2 answers:
Vera_Pavlovna [14]3 years ago
5 0

Answer:

1. Effect of air pressure

2. air- powered wheel chair

3. Pneumatic valves

Explanation:

1. In any pneumatic device, the mipact of air pressure to produce the moving effect on an heavy object is unexpected.

2. pneumatice demultiplexer when air in comprressed tank is allowed released to cause movement of the  chair.

3. In industries, a pneumatic valve operates by force of air when actuated. A signal causes actuation of coil. When coil is energized, compressed high pressure air is allowe to enter in a small cylinder and cause operation of valve

Alex Ar [27]3 years ago
4 0

Answer (1)

<em>The difference in the pneumatic device to what i was expecting  </em>was<em> </em>that i was expecting a device with a very complicated working system but was amazed at how simple but yet a little complex the pneumatic device was.

Answer (2)

<em>A typical device used help a disabled person </em>the pneumatic wheel chair. It uses compressed air to lift and lower the seat of the wheel chair. The compressed air is stored in a compressor below the seat and connected to the seat through piston cylinders. In order to lift, actuators activate valves that open and close at certain place to force air upward lifting the chair. Coming down is usually done by slowly leaking out the air and letting the chair descend under the weight of the user.

Answer:

<em>A pneumatic device used an automated piece of machinery </em>is the pneumatic press. The pneumatic press is a compressed air operated device in most manufacturing setting, frequently used in the automotive parts manufacturing industries. It uses compressed air to force down a ram which is usually used for baling scraps or for shaping metals.

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What are the Basic requirements of drinking Water ?
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Explanation:

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2 years ago
The angle of twist can be computed using the material’s shear modulus if and only if: (a)- The shear stress is still in the elas
ollegr [7]

Answer:

The angle of twist can be computed using the material’s shear modulus if and only if the shear stress is still in the elastic region

Explanation:

The shear modulus (G) is the ratio of shear stress to shear strain. Like the modulus of elasticity, the shear modulus is governed by Hooke’s Law: the relationship between shear stress and shear strain is proportional up to the proportional limit of the material. The angle of twist can be computed using the material’s shear modulus if and only if the shear stress is still in the elastic region.

3 0
3 years ago
The fracture strength of glass may be increased by etching away a thin surface layer. It is believed that the etching may alter
Korvikt [17]

Answer:

the ratio of the etched to the original crack tip radius is 30.24

Explanation:

Given the data in the question;

we determine the initial fracture stress using the following expression;

(σf)₁ = 2(σ₀)₁ [ α₁/(p_t)₁ ]^{1/2 ----- let this be equation 1

where; (σ₀)₁ is the initial fracture strength

(p_t)₁ is the original crack tip radius

α₁ is the original crack length.

first, we determine the final crack length;

α₂ = α₁ - 16% of α₁

α₂ = α₁ - ( 0.16 × α₁)

α₂ = α₁ - 0.16α₁

α₂ = 0.84α₁

next, we calculate the final fracture stress;

the fracture strength is increased by a factor of 6;

(σ₀)₂ = 6( σ₀ )₁

Now, expression for the final fracture stress

(σf)₂ = 2(σ₀)₂ [ α₂/(p_t)₂ ]^{1/2 ------- let this be equation 2

where (p_t)₂ is the etched crack tip radius

value of fracture stress of glass is constant

Now, we substitute 2(σ₀)₁ [ α₁/(p_t)₁ ]^{1/2 from equation for (σf)₂  in equation 2.

0.84α₁ for α₂.

6( σ₀ )₁ for (σ₀)₂.

∴

2(σ₀)₁ [ α₁/(p_t)₁ ]^{1/2  = 2(6( σ₀ )₁) [ 0.84α₁/(p_t)₂ ]^{1/2  

divide both sides by 2(σ₀)₁

[ α₁/(p_t)₁ ]^{1/2  =  6 [ 0.84α₁/(p_t)₂ ]^{1/2

[ 1/(p_t)₁ ]^{1/2  =  6 [ 0.84/(p_t)₂ ]^{1/2

[ 1/(p_t)₁ ]  =  36 [ 0.84/(p_t)₂ ]

1 / (p_t)₁ = 30.24 / (p_t)₂

(p_t)₂ = 30.24(p_t)₁

(p_t)₂/(p_t)₁ = 30.24

Therefore, the ratio of the etched to the original crack tip radius is 30.24

6 0
3 years ago
The human eye, as well as the light-sensitive chemicals on color photographic film, respond differently to light sources with di
jeka57 [31]

Answer:

a) at T = 5800 k  

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at T = 2900 k

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Explanation:

To Calculate the band emission fractions we will apply the Wien's displacement Law

The ban emission fraction in spectral range λ1 to λ2 at a blackbody temperature T can be expressed as

F ( λ1 - λ2, T ) = F( 0 ----> λ2,T) - F( 0 ----> λ1,T )

<em>Values are gotten from the table named: blackbody radiati</em>on functions

<u>a) Calculate the band emission fractions for the visible region</u>

at T = 5800 k  

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at T = 2900 k

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attached below is a detailed solution to the problem

<u>b)calculate wavelength corresponding to the maximum spectral intensity</u>

For daylight ( d ) = 2898 μm *k / 5800 k  = 0.50 μm

For Incandescent ( i ) = 2898 μm *k / 2900 k = 1 μm

3 0
3 years ago
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