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d1i1m1o1n [39]
1 year ago
12

I need help with this question

Engineering
1 answer:
Roman55 [17]1 year ago
6 0

Answer:

last one 8;2per leg

Explanation:

sorry if I got it wrong

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i believe the correct answer is c but i’m sorry if i’m not correct

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The ________ is the part of the drill press that holds and rotates the cutting tool.
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Spindle

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Relay contacts that are defined as being normally open (n.o.) have contacts that are:_____.
Reptile [31]

Relay contacts that are defined as being normally open (n.o.) have contacts that are open only if  the relay coil is known to have de-energized.

<h3>What is meant by normally open contacts?</h3>

Normally open (NO) are  known to be open if there is no measure of current that is flowing through a given coil but it often close as soon as the coil is said to be energized.

Note that  Normally closed (NO) contacts are said to be closed only if the coil is said to be de-energized and open only if the coil is said to carry current or is known to have energized.

The role of relay contact is wide. The Relays are tools that are often used in the work of  switching of control circuits and it is one that  a person cannot used for power switching that has relatively bigger ampacity.

Therefore, Relay contacts that are defined as being normally open (n.o.) have contacts that are open only if  the relay coil is known to have de-energized.

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1 year ago
An airline ticket counter forecasts that 220 people per hour will need to check in. It takes an average of 2 minutes to service
Nadusha1986 [10]

A) Number of agents required to achieve a wait time of 10 minutes or less = 8 agents

B) The number of agents required on duty to reduce cost = 9 agents

<u>Given data : </u>

Arrival rate of customers ( β ) = 220 per hour

Service rate ( mu ) = 60 minutes / 2 minutes = 30 customer per hour

utilization ( rho ) = 220 / 30 ≈ 7

at least 8 server personnel are required for stability of the queue

A<u>) Determine the number of agents required to achieve a wait time of 10 minutes or less per customer</u>

waiting time = 10 - 2 = 8 minutes

number of customers waiting ( ∝ ) = 7 and required server = 8

assuming   Lq = 5.2266

Hence the waiting time in line = Lq / arrival rate

                                                  = 5.2266 / 220 = 0.0238 hour

                                                  = 0.0238 * 60 = 1.428 minutes

Since the waiting time ( 1.428 minutes ) is less than the original waiting time ( 2 minutes ) the number of agents that will achieve a wait time of 10 minutes or less is = 8 agents

<u>B) Determine the number of</u><u> ticket agents</u><u> that should be on duty to minimize cost </u>

salary of ticket agent = £12 per hour

cost of customer waiting in queue = £5 per hour per customer

<em> </em><u>i) When 8 agents are used </u>

waiting time of customers = 0.0238 * 220 = 5.236

waiting cost for customers = 5.236 * 5 = £26.18

employee cost = 8 * 12 = £96

∴ Total cost = 96 + 26.18

                    = £ 122.18

<u>ii) When 9 agents are used </u>

waiting time for customers = 0.0074 * 220 = 1.628

Wq = 1.6367 / 220 = 0.0074

waiting cost for customers = 1.6367 * 5 = £ 8.1835

assuming Lq = 1.6367

employee cost = 9 * 12 = £ 108

∴ Total cost = 108 + 8.1835 = £ 116.18

From the calculations in ( i ) and ( ii ) the Ideal number of ticket agents that should be on duty to minimize cost should be 9 agents.

Hence we can conclude that A) Number of agents required to achieve a wait time of 10 minutes or less = 8 agents and The number of agents required on duty to reduce cost = 9 agents.

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In the engineering design process, testing is the least important step.<br> True<br> False
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