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AfilCa [17]
3 years ago
8

If datums are used as suggested below (A datum on shaft, B datum on face below the head, C datum on the hex flat) and order of p

recedence is in alphabetic order, name the datum that prevents: a. Translation along the X axis: _ b. Rotation about the X axis: _ c. Translation along the Z axis: _ d. Rotation about the Y axis: _
Engineering
1 answer:
allsm [11]3 years ago
6 0

Answer:

a. Use datum on shaft

b. Use datum on hex flat

c. Use datum on face below the head

d. Use datum on shaft

When these datum are used, they will prevent translation and rotation along axis which they act.

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Finger [1]

Answer:

specific heat

Explanation:

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3 years ago
Why is the drawdown cone for a well completed in a low permeability aquifer narrower and deeper than a drawdown cone for a well
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Answer:

c) It takes a greater hydraulic head to drive the groundwater laterally to the well casing in the lower permeability aquifer

Explanation:

The groundwater are contains under the rock and in the open spaces within the rocks and the unconsolidated sediments. Aquifer refers to the underground layers of the permeable sand or rocks that transmits the groundwater below water table which provides a sufficient supply of water to the well. Groundwater is present everywhere where there is porosity in the rocks and it depends on the permeability of the rocks to allow them flow.

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5 0
3 years ago
2. Similar to problem 1, assume your computer system has a 32-bit byte-addressable architecture where addresses and data are eac
andreev551 [17]

Question:

The question is not complete. The question to answer was not added. See below the possible question and the answer.

a. How many blocks are in the cache with this new arrangement?

b. Calculate the number of bits in each of the Tag, Index, and Offset fields of the memory address.

C. Using the values calculated in part b, what is the actual total size of the cache including data, tags, and valid bits?

Answer:

(a) Number of blocks =  512 blocks

(b) Tag is 18

(c)  Total size of the cache = 8388608 bytes

Explanation:

a .

block size = 32 bytes

cache size = 16384 bytes

No.of blocks = 16384 / 32

No.pf blocks = 512 blocks

b.

Total address size = 32 bits

Address bits = Tag + Line index +block offset

Block Size = 32 bytes.

So block size = 25 bytes.

Hence Offset is 5

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So , 32 =tag+9+5

tag = 32-(9+5)

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c.

Data bits = 32 bits

Tag=18 bits

Valid bit is 1 bit

so Total cache size = 25+218+20

                                  = 223

                                  =8388608 bytes

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