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elena55 [62]
2 years ago
11

3.8 LAB - Select lesson schedule with multiple joins

Engineering
1 answer:
dem82 [27]2 years ago
4 0

Answer:

The database has three tables for tracking horse-riding lessons: Horse with columns: ID - primary key; RegisteredName; Breed; Height; BirthDate.

Explanation:

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Water at 15°C is to be discharged from a reservoir at a rate of 18 L/s using two horizontal cast iron pipes connected in series
love history [14]

Answer:

The required pumping head is 1344.55 m and the pumping power is 236.96 kW

Explanation:

The energy equation is equal to:

\frac{P_{1} }{\gamma } +\frac{V_{1}^{2}  }{2g} +z_{1} =\frac{P_{2} }{\gamma } +\frac{V_{2}^{2}  }{2g} +z_{2}+h_{i} -h_{pump} , if V_{1} =0,z_{2} =0\\h_{pump} =\frac{V_{2}^{2}}{2} +h_{i}-z_{1}

For the pipe 1, the flow velocity is:

V_{1} =\frac{Q}{\frac{\pi D^{2} }{4} }

Q = 18 L/s = 0.018 m³/s

D = 6 cm = 0.06 m

V_{1} =\frac{0.018}{\frac{\pi *0.06^{2} }{4} } =6.366m/s

The Reynold´s number is:

Re=\frac{\rho *V*D}{u} =\frac{999.1*6.366*0.06}{1.138x10^{-3} } =335339.4

\frac{\epsilon }{D} =\frac{0.00026}{0.06} =0.0043

Using the graph of Moody, I will select the f value at 0.0043 and 335339.4, as 0.02941

The head of pipe 1 is:

h_{1} =\frac{V_{1}^{2}  }{2g} (k_{L}+\frac{fL}{D}  )=\frac{6.366^{2} }{2*9.8} *(0.5+\frac{0.0294*20}{0.06} )=21.3m

For the pipe 2, the flow velocity is:

V_{2} =\frac{0.018}{\frac{\pi *0.03^{2} }{4} } =25.46m/s

The Reynold´s number is:

Re=\frac{\rho *V*D}{u} =\frac{999.1*25.46*0.03}{1.138x10^{-3} } =670573.4

\frac{\epsilon }{D} =\frac{0.00026}{0.03} =0.0087

The head of pipe 1 is:

h_{2} =\frac{V_{2}^{2}  }{2g} (k_{L}+\frac{fL}{D}  )=\frac{25.46^{2} }{2*9.8} *(0.5+\frac{0.033*36}{0.03} )=1326.18m

The total head is:

hi = 1326.18 + 21.3 = 1347.48 m

The required pump head is:

h_{pump} =\frac{25.46^{2} }{2*9.8} +1347.48-36=1344.55m

The required pumping power is:

P=Q\rho *g*h_{pump}  =0.018*999.1*9.8*1344.55=236965.16W=236.96kW

8 0
3 years ago
The Ethernet (CSMA/CD) alternates between contention intervals and successful transmissions. Assume a 100 Mbps Ethernet over 1 k
Vesnalui [34]
<h3><u>CSMA/CD Protocol: </u></h3>

Carrier sensing can transmit the data at anytime only the condition is before sending the data sense carrier if the carrier is free then send the data.

But the problem is the standing at one end of channel, we can’t send the entire carrier. Because of this 2 stations can transmit the data (use the channel) at the same time resulting in collisions.

There are no acknowledgement to detect collisions, It's stations responsibility to detect whether its data is falling into collisions or not.

<u>Example: </u>

T_{P}=1 H r, at time t = 10.00 AM, A starts, 10:59:59 AM B starts at time 11:00 AM collision starts.

12:00 AM A will see collisions

Pocket Size to detect the collision.

\begin{aligned}&T_{t} \geq 2 T_{P}\\&\frac{L}{B} \geq 2 T_{P}\\&L \geq 2 \times T_{P} \times B\end{aligned}

CSMA/CD is widely used in Ethernet.

<u>Efficiency of CSMA/CD:</u>

  • In the previous example we have seen that in worst case 2 T_{P} time require to detect a collision.
  • There could be many collisions may happen before a successful completion of transmission of a packet.

We are given number of collisions (contentions slots)=4.

\text { Propagation day }=\frac{\text {distance}}{\text {speed}}

Distance = 1km = 1000m

\begin{aligned}&\text { Speed }=2 \times 10^{8} \mathrm{m} / \mathrm{sec}\\ &T_{P}=\frac{1000}{2 \times 10^{8}}=(0.5) \times 10^{-5}=5 \times 10^{-6}\\ &T_{t}=5 \mu \mathrm{sec}\end{aligned}

7 0
3 years ago
Select the correct answer.
olya-2409 [2.1K]

Answer:

screw is the answer of the question

6 0
3 years ago
Read 2 more answers
Ben İngiliz oldum düzelte bilirmiyim
Lelu [443]

Answer:

What laguange is that?

Explanation:

7 0
2 years ago
A 60-cm-high, 40-cm-diameter cylindrical water tank is being transported on a level road. The highest acceleration anticipated i
dlinn [17]

Answer:

h_{max} = 51.8 cm

Explanation:

given data:

height of tank = 60cm

diameter of tank =40cm

accelration = 4 m/s2

suppose x- axis - direction of motion

z -axis - vertical direction

\theta = water surface angle with horizontal surface

a_x =accelration in x direction

a_z =accelration in z direction

slope in xz plane is

tan\theta = \frac{a_x}{g +a_z}

tan\theta = \frac{4}{9.81+0}

tan\theta =0.4077

the maximum height of water surface at mid of inclination is

\Delta h = \frac{d}{2} tan\theta

            =\frac{0.4}{2}0.4077

\Delta h  0.082 cm

the maximu height of wwater to avoid spilling is

h_{max} = h_{tank} -\Delta h

            = 60 - 8.2

h_{max} = 51.8 cm

the height requird if no spill water is h_{max} = 51.8 cm

3 0
3 years ago
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