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Gnoma [55]
3 years ago
13

In a chassis, the path along which air from a cool air source is conducted, past equipment to cool it, and then out of the rack.

Typically, air moves form front to back. Clutter in the rack should be minimized to prevent airflow blockages.
______________________
Computers and Technology
1 answer:
Lerok [7]3 years ago
5 0

Answer: AIRFLOW

Explanation:Chassis is the part of a physical structure which helps to carry its load and gives some support to the physical structure or equipment. Vehicles generally have chassis,most are numbered, it numbering is to ensure easy identification. The AIRFLOW helps to guarantee efficiency by controlling the movement of air into the working parts of equipment which could be the ENGINE or in a GENERATOR . Airflow system has also been adopted in cooling equipments like REFRIGERATOR or an AIR-CONDITIONING. Typically,air moves from Front to back.

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Which statement is false? Classes are reusable software components. A class is to an object as a blueprint is to a house. Perfor
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Answer:

The last option i.e "A class is an instance of its object" is the correct answer of this question.

Explanation:

A class is a collection of variable and method . The class is the blueprint of the object that means class is physical space entity and object is a runtime space entity.It is the the reusability feature of the programming component.

Following are the  syntax to declare any class.

                     Class classname

<u>For Example</u>: class test1

With the help of class we can achieve the inheritance, encapsulation and abstraction etc.

All the other options except the last one are correct option. Because option(1), option(2) and option(3) follow the property of class and object .  that's why last option is False.

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Translate the following MIPS code to C. Assume that the variables f, g, h, i, and j are assigned to registers $s0, $s1, $s2, $s3
Romashka [77]

Answer:

f = 2 * (&A[0])

See explaination for the details.

Explanation:

The registers $s0, $s1, $s2, $s3, and $s4 have values of the variables f, g, h, i, and j respectively. The register $s6 stores the base address of the array A and the register $s7 stores the base address of the array B. The given MIPS code can be converted into the C code as follows:

The first instruction addi $t0, $s6, 4 adding 4 to the base address of the array A and stores it into the register $t0.

Explanation:

If 4 is added to the base address of the array A, then it becomes the address of the second element of the array A i.e., &A[1] and address of A[1] is stored into the register $t0.

C statement:

$t0 = $s6 + 4

$t0 = &A[1]

The second instruction add $t1, $s6, $0 adding the value of the register $0 i.e., 32 0’s to the base address of the array A and stores the result into the register $t1.

Explanation:

Adding 32 0’s into the base address of the array A does not change the base address. The base address of the array i.e., &A[0] is stored into the register $t1.

C statement:

$t1 = $s6 + $0

$t1 = $s6

$t1 = &A[0]

The third instruction sw $t1, 0($t0) stores the value of the register $t1 into the memory address (0 + $t0).

Explanation:

The register $t0 has the address of the second element of the array A (A[1]) and adding 0 to this address will make it to point to the second element of the array i.e., A[1].

C statement:

($t0 + 0) = A[1]

A[1] = $t1

A[1] = &A[0]

The fourth instruction lw $t0, 0($t0) load the value at the address ($t0 + 0) into the register $t0.

Explanation:

The memory address ($t0 + 0) has the value stored at the address of the second element of the array i.e., A[1] and it is loaded into the register $t0.

C statement:

$t0 = ($t0 + 0)

$t0 = A[1]

$t0 = &A[0]

The fifth instruction add $s0, $t1, $t0 adds the value of the registers $t1 and $t0 and stores the result into the register $s0.

Explanation:

The register $s0 has the value of the variable f. The addition of the values stored in the regsters $t0 and $t1 will be assigned to the variable f.

C statement:

$s0 = $t1 + $t0

$s0 = &A[0] + &A[0]

f = 2 * (&A[0])

The final C code corresponding to the MIPS code will be f = 2 * (&A[0]) or f = 2 * A where A is the base address of the array.

3 0
2 years ago
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