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Elena-2011 [213]
3 years ago
7

What is the pressure at the bottom of a 25 ft volume of hydraulic fluid with a weight density of 55 lb/ft3 a. 114.6 psi b. 1375p

si c. 2.2 psf. d. 9.55 psi e.1395 psf
Engineering
1 answer:
Assoli18 [71]3 years ago
3 0

Answer:

d) 9.55 psi

Explanation:

pressure at the bottom is =ρgh

weight density is ρg=55 lb/ft³

h=25ft

pressure at the bottom is =55\times 25

                                  =1375psf

1 ft = 12 inch

pressure at bottom =\frac{1375}{12^2}

                                = 9.55 psi

so, answer will be option (d) which is 9.55 psi

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Cold forging makes metal more workable than hot forging.<br> True<br> False
Whitepunk [10]

Answer:

I don't really know but i have some info for you...

Explanation:

The cold forging manufacturing process increases the strength of a metal through strain hardening at a room temperature. On the contrary the hot forging manufacturing process keeps materials from strain hardening at high temperature, which results in optimum yield strength, low hardness and high ductility.

7 0
3 years ago
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A Transmission Control Protocol (TCP) connection is in working order and both sides can send each other data. What is the TCP so
kkurt [141]

Answer:

ESTABLISHED

Explanation:

What is TCP?

A Transmission Control Protocol (TCP) is a communication protocol which allows the exchange of data between computers in a network.  

When a Transmission Control Protocol connection is up and running meaning that both sides can send and receive data then the corresponding TCP socket states is known as "ESTABLISHED".

The most common socket states are:

LISTEN:

Before a TCP connection is made, there needs to be a server with a listener that will listen on incoming connection request.

ESTABLISHED:

When a TCP connection is up and running meaning that both sides can send and receive data.

CLOSED:

The CLOSED state means that there is no TCP connection.

There are a total of 11 TCP socket states:

1. LISTEN

2. SYN-SENT

3. SYN-RECEIVED

4. ESTABLISHED

5. FIN-WAIT-1

6. FIN-WAIT-2

7. CLOSE-WAIT

8. CLOSING

9. LAST-ACK

10. TIME-WAIT

11. CLOSED

3 0
3 years ago
An alloy has a yield strength of 818 MPa and an elastic modulus of 104 GPa. Calculate the modulus of resilience for this alloy [
crimeas [40]

Answer:

Modulus of resilience will be 3216942.308j/m^3

Explanation:

We have given yield strength \sigma _y=818MPa

Elastic modulus E = 104 GPa

We have to find the modulus

Modulus of resilience is given by

Modulus of resilience =\frac{\sigma _y^2}{2E}, here \sigma _y is yield strength and E is elastic modulus

Modulus of resilience =\frac{(818\times 10^6)^2}{2\times 104\times 10^9}=3216942.308j/m^3  

5 0
3 years ago
The wall of drying oven is constructed by sandwiching insulation material of thermal conductivity k = 0.05 W/m°K between thin me
masha68 [24]

Answer:

86 mm

Explanation:

From the attached thermal circuit diagram, equation for i-nodes will be

\frac {T_ \infty, i-T_{i}}{ R^{"}_{cv, i}} + \frac {T_{o}-T_{i}}{ R^{"}_{cd}} + q_{rad} = 0 Equation 1

Similarly, the equation for outer node “o” will be

\frac {T_{ i}-T_{o}}{ R^{"}_{cd}} + \frac {T_{\infty, o} -T_{o}}{ R^{"}_{cv, o}} = 0 Equation 2

The conventive thermal resistance in i-node will be

R^{"}_{cv, i}= \frac {1}{h_{i}}= \frac {1}{30}= 0.033 m^{2}K/w Equation 3

The conventive hermal resistance per unit area is

R^{"}_{cv, o}= \frac {1}{h_{o}}= \frac {1}{10}= 0.100 m^{2}K/w Equation 4

The conductive thermal resistance per unit area is

R^{"}_{cd}= \frac {L}{K}= \frac {L}{0.05} m^{2}K/w Equation 5

Since q_{rad}  is given as 100, T_{o}  is 40 T_ \infty  is 300 T_{\infty, o}  is 25  

Substituting the values in equations 3,4 and 5 into equations 1 and 2 we obtain

\frac {300-T_{i}}{0.033} +\frac {40-T_{i}}{L/0.05} +100=0  Equation 6

\frac {T_{ i}-40}{L/0.05}+ \frac {25-40}{0.100}=0

T_{i}-40= \frac {L}{0.05}*150

T_{i}-40=3000L

T_{i}=3000L+40 Equation 7

From equation 6 we can substitute wherever there’s T_{i} with 3000L+40 as seen in equation 7 hence we obtain

\frac {300- (3000L+40)}{0.033} + \frac {40- (3000L+40)}{L/0.05}+100=0

The above can be simplified to be

\frac {260-3000L}{0.033}+ \frac {(-3000L)}{L/0.05}+100=0

\frac {260-3000L}{0.033}=50

-3000L=1.665-260

L= \frac {-258.33}{-3000}=0.086*10^{-3}m= 86mm

Therefore, insulation thickness is 86mm

8 0
3 years ago
The difference between an initial condition and a boundary condition for conduction in a solid is:___________
leva [86]

Answer:

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Explanation:

Conduction refers to the transfer of thermal energy or electric charge as a result of the movement of particles. When the conduction relates to electric charge, it is known as electrical conduction while when it relates to thermal energy, it is known as heat conduction.

In the process of heat conduction, thermal energy is usually transferred from fast moving particles to slow moving particles during the collision of these particles. Also, thermal energy is typically transferred between objects that has different degrees of temperature and materials (particles) that are directly in contact with each other but differ in their ability to accept or give up electrons.

Any material or object that allow the conduction (transfer) of electric charge or thermal energy is generally referred to as a conductor. Conductors include metal, steel, aluminum, copper, frying pan, pot, spoon etc.

Hence, the difference between an initial condition and a boundary condition for conduction in a solid is that an initial condition specifies the temperature at the start of the problem and a boundary condition provides information about temperatures on the boundaries.

7 0
2 years ago
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