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lukranit [14]
3 years ago
10

Using the tables for water, determine the specified property data at the indicated states. In each case, locate the state on ske

tches of the p–υ and T–υ diagrams.a. At p = 3 bar, υ = 0.5 m3 /kg, find T in °C and u in kJ/kg.b. At T = 320°C, υ = 0.03 m3 /kg, find p in MPa and u in kJ/kg.c. At p = 28 MPa, T = 520°C, find υ in m3 /kg and h in kJ/kg.d. At T = 10°C, υ = 100 m3 /kg, find p in kPa and h in kJ/kg.e. At p = 4 MPa, T = 160°C, find υ in m3 /kg and u in kJ/kg.
Engineering
1 answer:
Anarel [89]3 years ago
5 0

heres a link! ;)

http://www.econsulat.ro/

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List two reasons why machined parts often require a high degree of precision.
Gekata [30.6K]

Explanation:

Precision machining is a subtractive process used in cases where material needs to be removed from a raw product to create the finished product. Precision machining can be used to create a wide variety of products, items, and parts for any number of different objects and materials. These parts usually require tight tolerances variation from nominal dimensions and from part to part, which means that there is not much room for error in the production of the piece. Repeatability and well-controlled tolerances are hallmarks of precision machining. Components, parts and finished durable products that are designed to maintain extremely tight tolerance margins and a high degree of durability are essential and common drivers for utilization of precision machining. For example, parts that need to work together as part of a machine may need to always align within a certain margin of 0.01mm to 0.05mm. Precision engineering and machining help to ensure these parts can not only be made precisely but can be produced with this level of accuracy over and over again.

3 0
3 years ago
Construction support involves mostly what kind of work?
Paraphin [41]

Answer:

Outdoors

Explanation:

Construction workers perform outdoors.

6 0
2 years ago
Technologies that allow for instant worldwide communication include
hram777 [196]

Answer:

mobile phones and internet access

Explanation:

I got it right on my quiz

7 0
3 years ago
Read 2 more answers
4. The instant the ignition switch is turned to the start position,
geniusboy [140]

Answer:

D. Both pull-in and hold-in windings are energized.

Explanation:

The instant the ignition switch is turned to the start position, "Both pull-in and hold-in windings are energized." This is because the moment the ignition switch is turned to the start position, voltage passes through to the S terminal of the solenoid.

The hold-in winding is attached to the case of the solenoid. Similarly, the pull-in winding is also attached to the starter motor. Thereby, the current will move across both windings by getting energized to generate a strong magnetic field.

4 0
3 years ago
Two identical billiard balls can move freely on a horizontal table. Ball a has a velocity V0 and hits balls B, which is at rest,
Lyrx [107]

Answer:

Velocity of ball B after impact is 0.6364v_0 and ball A is 0.711v_0

Explanation:

v_0 = Initial velocity of ball A

v_A=v_0\cos45^{\circ}

v_B = Initial velocity of ball B = 0

(v_A)_n' = Final velocity of ball A

v_B' = Final velocity of ball B

e = Coefficient of restitution = 0.8

From the conservation of momentum along the normal we have

mv_A+mv_B=m(v_A)_n'+mv_B'\\\Rightarrow v_0\cos45^{\circ}+0=(v_A)_n'+v_B'\\\Rightarrow (v_A)_n'+v_B'=\dfrac{1}{\sqrt{2}}v_0

Coefficient of restitution is given by

e=\dfrac{v_B'-(v_A)_n'}{v_A-v_B}\\\Rightarrow 0.8=\dfrac{v_B'-(v_A)_n'}{v_0\cos45^{\circ}}\\\Rightarrow v_B'-(v_A)_n'=\dfrac{0.8}{\sqrt{2}}v_0

(v_A)_n'+v_B'=\dfrac{1}{\sqrt{2}}v_0

v_B'-(v_A)_n'=\dfrac{0.8}{\sqrt{2}}v_0

Adding the above two equations we get

2v_B'=\dfrac{1.8}{\sqrt{2}}v_0\\\Rightarrow v_B'=\dfrac{0.9}{\sqrt{2}}v_0

\boldsymbol{\therefore v_B'=0.6364v_0}

(v_A)_n'=\dfrac{1}{\sqrt{2}}v_0-0.6364v_0\\\Rightarrow (v_A)_n'=0.07071v_0

From the conservation of momentum along the plane of contact we have

(v_A)_t'=(v_A)_t=v_0\sin45^{\circ}\\\Rightarrow (v_A)_t'=\dfrac{v_0}{\sqrt{2}}

v_A'=\sqrt{(v_A)_t'^2+(v_A)_n'^2}\\\Rightarrow v_A'=\sqrt{(\dfrac{v_0}{\sqrt{2}})^2+(0.07071v_0)^2}\\\Rightarrow \boldsymbol{v_A'=0.711v_0}

Velocity of ball B after impact is 0.6364v_0 and ball A is 0.711v_0.

5 0
3 years ago
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