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ExtremeBDS [4]
2 years ago
5

PLS HELP!!!

Engineering
2 answers:
Oksana_A [137]2 years ago
4 0
Answer: C

Synthesizing is basically mixed or together, sometimes creating something new
Advocard [28]2 years ago
3 0

Answer:

I would say C

Explanation:

let me know if im right

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What resources did Margaret Hutchinson Rousseau use ?
blagie [28]

Answer:

Explanation:

During the Second World War, she oversaw the design of production plants for the strategically important materials of penicillin and synthetic rubber.Her development of deep-tank fermentation of penicillium mold enabled large-scale production of penicillin.She worked on the development of high-octane gasoline for aviation fuel.Her later work included improved distillation column design and plants for the production of ethylene glycol and glacial acetic acid.

3 0
3 years ago
4. Fill in the blanks with the words from the Word Bank below.
lyudmila [28]

Answer:

space race

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mission control

failure

success

5 0
3 years ago
. What are the benefits of synthetic motor oil compared to conventional oil?
madam [21]

Explanation:

the answet is E. all of the above

4 0
3 years ago
What is the velocity of flow in an asphalt channel that has a hydraulic radius of 3.404 m, length of 200 m and bed slope of 0.00
yKpoI14uk [10]

Answer:

The velocity of flow is 10.0 m/s.

Explanation:

We shall use Manning's equation to calculate the velocity of flow

Velocity of flow by manning's equation is given by

V=\frac{1}{n}R^{2/3}S^{1/2}

where

n = manning's roughness coefficient

R = hydraulic radius

S = bed slope of the channel

We know that for an asphalt channel value of manning's roughness coefficient = 0.016

Applying values in the above equation we obtain velocity of flow as

V=\frac{1}{0.016}\times 3.404^{2/3}\times 0.005^{1/2}\\\\\therefore V=10.000m/s

7 0
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

queueing delay =192937984 ×10⁻⁹

=0.19s

5 0
3 years ago
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