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2 years ago

Wastewater flows into a once it is released into A floor drain

Engineering

1 answer:

Phoenix [80]2 years ago

8 0

Answer: C) Sump pit

Explanation:

Wastewater from the house is allowed to flow out of places such as the bathroom or the kitchen sink through a floor drain.

This floor drain is connected to a sump pit which is typically located at the lowest point in the house (lowest point in basement) so that water can flow into it easier.

When the wastewater is released from the floor drain, it will flow into the sump pit. This water is in turn removed from the pit by a sump pump.

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Air flows from a large reservoir in which the pressure and temperature are 1 MPa and 30°C, respectively, through a convergent–di

SSSSS [86.1K]

Answer:

The solution is attached in the attachment.

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2 years ago

Read 2 more answers

Compare the use of a low-strength, ductile material (1018 CD) in which the stress-concentration factor can be ignored to a high-

kicyunya [14]

Answer:

Step 1 of 3

Case A:

AISI 1018 CD steel,

Fillet radius at wall=0.1 in,

Diameter of bar

From table deterministic ASTM minimum tensile and yield strengths for some hot rolled and cold drawn steels for 1018 CD steel

Tensile strength

Yield strength

The cross section at A experiences maximum bending moment at wall and constant torsion throughout the length. Due to reasonably high length to diameter ratio transverse shear will be very small compared to bending and torsion.

At the critical stress elements on the top and bottom surfaces transverse shear is zero

Explanation:

See the next steps in the attached image

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2 years ago

Problem 4.079 SI A rigid tank whose volume is 3 m3, initially containing air at 1 bar, 295 K, is connected by a valve to a large

salantis [7]

Answer:

$Q_{cv} = -1007.86kJ$

Explanation:

Our values are,

State 1

$V=3m^3\\P_1=1bar\\T_1 = 295K$

We know moreover for the tables A-15 that

$u_1 = 210.49kJ/kg\\h_i = 295.17kJkg$

State 2

$P_2 =6bar\\T_2 = 296K\\T_f = 320K$

For tables we know at T=320K

$u_2 = 228.42kJ/kg$

We need to use the ideal gas equation to estimate the mass, so

$m_1 = \frac{p_1V}{RT_1}$

$m_1 = \frac{1bar*100kPa/1bar(3m^3)}{0.287kJ/kg.K(295k)}$

$m_1 = 3.54kg$

Using now for the final mass:

$m_2 = \frac{p_2V}{RT_2}$

$m_2 = \frac{1bar*100kPa/6bar(3m^3)}{0.287kJ/kg.K(320k)}$

$m_2 = 19.59kg$

We only need to apply a energy balance equation:

$Q_{cv}+m_ih_i = m_2u_2-m_1u_1$

$Q_{cv}=m_2u_2-m1_u_1-(m_2-m_1)h_i$

$Q_{cv} = (19.59)(228.42)-(3.54)(210.49)-(19.59-3.54)(295.17)$

$Q_{cv} = -1007.86kJ$

The negative value indidicates heat ransfer from the system

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2 years ago

what would my engine exhaust sound like if i made a custom muffler that was just smooth on the inside, would it have an echoing

Elena L [17]

Answer:

you know what sounds rlly nice? first have you seent he new corvettes? next u should here a nisan gtr. (like tanner foxes)

Explanation:

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2 years ago

What Member function places a new node at the end of the linked list?

Nadusha1986 [10]

<h2>˜”*°•.˜”*°• Question •°*”˜.•°*”˜ </h2>

<em>What Member function places a new node at the end of the linked list? </em>

<h2>˜”*°•.˜”*°• Answer •°*”˜.•°*”˜</h2>

appendNode

<h2>˜”*°•.˜”*°• Explanation •°*”˜.•°*”˜</h2>

The appendNode() member function places a new node at the end of the linked list. The appendNode() requires an integer representing the current data of the node.

<h2>˜”*°•.˜”*°• Details •°*”˜.•°*”˜</h2>

Subject: Coding (?)

Grade: College

Keywords: Function, linked list, appendNode, integer

Hope this helped. <3

3 0

2 years ago