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VikaD [51]
3 years ago
15

A plastic molding machine produces a product whose annual demand is in the millions.

Engineering
1 answer:
BabaBlast [244]3 years ago
8 0

Answer:

a. 78.4 pieces/hr

b. $0.1806/min

c. $1.34/piece

Explanation:

(a) With a cycle time Tc = 45 sec = 0.75 min.

Production rate, Rp = 60/0.75 = 80 pieces/hr.

Let us factor in the 98% proportion uptime, so Production rate, Rp = 0.98*(80) = 78.4 pieces/hr

Quantity of product annualy = 6000 *(78.4) = 470,400 pieces/yr

(b) Equipment cost rate, Ceq = 500,000(1.30)/(60 x 10 x 6000) = $0.1806/min.

(c) Cost per piece of mould, Ct = 100,000/1,000,000 = $0.10/piece

Cost rate of labour ,CL = 18.00(0.20) = $3.60/hr = $0.06/min

Conclusively, cost per piece, Cpc = 1.20(0.88) + (0.06 + 0.1806)(0.75) + 0.10 = $1.34/piece

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true

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Here are some other words as well.

potential, voltage, potential drop, potential difference.

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3 years ago
Suppose there are 93 packets entering a queue at the same time. Each packet is of size 4 MiB. The link transmission rate is 1.4
Ghella [55]

Answer:

0.19s

Explanation:

Queueing delay is the time a job waits in a queue before it can be executed. it is the difference in time betwen when the packet data reaches it destination and the time when it was executed.

Queueing delay =(N-1) L /2R

where N = no of packet =93

L = size of packet = 4MB

R = bandwidth = 1.4Gbps = 1×10⁹ bps

4 MB = 4194304 Bytes

(93 - 1)4194304 / 2× 10⁹

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5 0
3 years ago
Consider a half-wave rectifier circuit with a triangular-wave input of LaTeX: 6V6 V peak-to-peak amplitude and zero average valu
Varvara68 [4.7K]

The question is not complete. We are supposed to find the average value of v_o.

Answer:

v_o,avg = 0.441V

Explanation:

Let t1 and t2 be the start and stop times of the output waveforms. Thus, from the diagram i attached, using similar triangles, we have;

3/(T/4) = 0.7/t1

So, 12/T = 0.7/t1

So, t1 = 0.7T/12

t1 = 0.0583 T

Also, from symmetry of triangles,

t2 = T/2 - t1

So, t2 = T/2 - 0.0583 T

t2 = 0.4417T

Average of voltage output is;

v_o,avg = (1/T) x Area under small triangle

v_o,avg = (1/T) x (3 - 0.7) x (T/4 - t1)

v_o,avg = (1/T) x (2.3) x (T/4 - 0.0583 T)

v_o,avg = (1/T) x 2.3 x 0.1917T

T will cancel out to give;

v_o,avg = 0.441V

8 0
3 years ago
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