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Hoochie [10]
4 years ago
8

Technician A says that the first step in diagnosing engine condition is to perform a thorough visual inspection. Technician B sa

ys that oil leaks can be found by measuring straight up from where the oil drips on the floor. Who is​ right?
A. Technician A only

B. Technician B only

C. Both technicians A and B

D. Neither technician A nor B
Engineering
1 answer:
Mashutka [201]4 years ago
7 0
A- Technician A only, because A visual inspection can tell you a lot about the engine and previous maintenance routines
You might be interested in
The device whose operation closely matches the way the clamp-on ammeter works is
Ivanshal [37]

Answer:

The answer is

C. Split phase motor

Explanation:

Clamp meters rely on the principle of magnetic induction to make non contact AC current measurements. Electric current flowing through a wire produces a magnetic field.

Which is similar to basic mode of operation of electric motor and split phase motor is a type of electric motor.

What is a a clamp on meter?

Clamp meters are electrical testers which have wide jaws that are able to clamp around an electrical conductor. Originally designed as a single purpose tool for measuring AC current, clamp meters now include inputs for accepting test leads and other probes that support a wide range of electrical measurements, the jaws of a clamp meter permit work in tight spaces and permits current measurements on live conductors without circuit interruption.

6 0
3 years ago
Consider the following Moore’s law growth pattern (since 1980) for the number of transistors inside a particular commercial micr
Fantom [35]

Answer:

No. of transistors = $4.1524 \times 10^{10}$ transistors

Explanation:

Given that:

N = $1920 \times 10^{0.163(Y-1980)}$

Y = 2025

N = $1920 \times 10^{0.163(2025-1980)}$

N = $4.1524 \times 10^{10}$ transistors

Now at Y = 1980

Number of transistors N = 1920

Therefore,

$1000 = 10^{0.163(Y-1980)}$

$\log_{10} 1000=0.163(Y-1980)$

$\frac{3}{0.163}=Y-1980$

18 ≅ 18.4 = Y - 1980

Y = 1980 + 18

   = 1998

So, to increase multiples of 1000 transistors. it takes 18 years.

5 0
3 years ago
Refrigerant-134a enters a 28-cm-diameter pipe steadily at 200 kPa and 20°C with a velocity of 5 m/s. The refrigerant gains heat
Alexandra [31]

Answer:

V = 0.30787 m³/s

m = 2.6963 kg/s

v2 =  0.3705 m³/s

v2 = 6.017 m/s

Explanation:

given data

diameter = 28 cm

steadily =200 kPa

temperature = 20°C

velocity = 5 m/s

solution

we know mass flow rate is

m = ρ A v

floe rate V = Av

m = ρ V

flow rate = V = \frac{m}{\rho}

V = Av = \frac{\pi}{4} * d^2 * v1

V = \frac{\pi}{4} * 0.28^2 * 5

V = 0.30787 m³/s

and

mass flow rate of the refrigerant is

m = ρ A v

m = ρ V

m = \frac{V}{v} = \frac{0.30787}{0.11418}

m = 2.6963 kg/s

and

velocity and volume flow rate at exit

velocity = mass × v

v2 = 2.6963 × 0.13741 = 0.3705 m³/s

and

v2 = A2×v2

v2 = \frac{v2}{A2}

v2 = \frac{0.3705}{\frac{\pi}{4} * 0.28^2}

v2 = 6.017 m/s

7 0
4 years ago
Tadpoles raised in water with atrazine levels of 0.1 ppb should produce a higher percentage of male frogs with gonadal abnormali
Kobotan [32]

Answer:

correct option is c. a testable prediction leading to design of an experiment

Explanation:

The results of raising tadpoles were estimated to be water with an atrazine level of 0.1 ppb compared to those grown in pure water. So, this is not the question. If this assumption can now be tested, an experiment can be made in which a certain number of tadpoles can be raised in pure water and the same number of tadpoles can be raised in water with a 0.1ppb atrazine level can. The difference between the two populations can be estimated or compared.

8 0
3 years ago
Write an ALP to separate odd and even numbers from an array of N numbers; arrange odd
Marta_Voda [28]

Below is the program to separate odd and even numbers                                  

<u>Explanation</u>:

<u>L1:</u>

         mov ah,00

         mov al,[BX]

         mov dl,al

         div dh

         cmp ah,00

         je EVEN1

         mov [DI],dl

         add OddAdd,dl

         INC DI

         INC BX

         Loop L1

         jmp CAL

    <u>EVEN1:</u>

         mov [SI],dl

         add Even Add,dl

         INC SI

         INC BX

         Loop L1

    <u>CAL:   </u>  

         mov ax,0000

         mov bx,0000

         mov al,OddAdd

         mov bl,EvenAdd

         MOV  ax,4C00h

         int 21h

end

The above program separates odd and even numbers from the array using 8086 microprocessor. It has odd numbers in 2000h and even numbers in 3000h.

6 0
4 years ago
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