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2 years ago

12

An intake manifold gasket has been replaced due to a vacuum leak. Which of the following steps uses a scan tool to complete the

job? O A. Torquing the manifold bolts B. Idle relearn O C. Refilling the cooling system O D. Air cleaner check

1 answer:

matrenka [14]2 years ago

7 0

Answer: B.Idle relearn

Explanation:

You might be interested in

4.2 A vapor compression refrigeration machine uses 30kW of electric power to produce 50 tons of cooling. What is

stellarik [79]

Answer:

5.833

Explanation:

Coefficient of Perfomance (COP) is the ratio of refrigeration effect to power input.

$COP=\frac {RE}{P}$ where RE is refrigeration effect and P is power input

Here, the power input is given as 30 kW

We also know that 1 ton cooling is equivalent to 3.5 kW hence for 50 tons, RE=50*3.5=175 kW

Now the $COP=\frac {175}{30}=5.833$

6 0

3 years ago

Which of the following requirement statements is an example of a breakdown of the accuracy standard?

const2013 [10]

Answer:

<u>The automobile rental prices shall show all taxes (including a 6% state tax).</u>

Explanation:

Im pretty sure

4 0

3 years ago

Water flows through a nozzle at the end of a fire hose. If the nozzle exit velocity must be 20 m/s and the exit diameter is 40 m

lara31 [8.8K]

Answer:

minimum flow rate provided by pump is 0.02513 m^3/s

Explanation:

Given data:

Exit velocity of nozzle = 20m/s

Exit diameter = 40 mm

We know that flow rate Q is given as

$Q = A \times V$

where A is Area

$A =\frac{\pi}{4} \times (40\times 10^{-3})^2 = 1.256\times 10^{-3} m^2$

$Q = 1.256\times 10^{-3} \times 20 = 0.02513 m^3/s$

minimum flow rate provided by pump is 0.02513 m^3/s

5 0

3 years ago

Sawing stock to reduce its thickness is known as __________ .

rjkz [21]

Answer:

resawing

Explanation:

8 0

4 years ago

Problem 4.079 SI A rigid tank whose volume is 3 m3, initially containing air at 1 bar, 295 K, is connected by a valve to a large

salantis [7]

Answer:

$Q_{cv} = -1007.86kJ$

Explanation:

Our values are,

State 1

$V=3m^3\\P_1=1bar\\T_1 = 295K$

We know moreover for the tables A-15 that

$u_1 = 210.49kJ/kg\\h_i = 295.17kJkg$

State 2

$P_2 =6bar\\T_2 = 296K\\T_f = 320K$

For tables we know at T=320K

$u_2 = 228.42kJ/kg$

We need to use the ideal gas equation to estimate the mass, so

$m_1 = \frac{p_1V}{RT_1}$

$m_1 = \frac{1bar*100kPa/1bar(3m^3)}{0.287kJ/kg.K(295k)}$

$m_1 = 3.54kg$

Using now for the final mass:

$m_2 = \frac{p_2V}{RT_2}$

$m_2 = \frac{1bar*100kPa/6bar(3m^3)}{0.287kJ/kg.K(320k)}$

$m_2 = 19.59kg$

We only need to apply a energy balance equation:

$Q_{cv}+m_ih_i = m_2u_2-m_1u_1$

$Q_{cv}=m_2u_2-m1_u_1-(m_2-m_1)h_i$

$Q_{cv} = (19.59)(228.42)-(3.54)(210.49)-(19.59-3.54)(295.17)$

$Q_{cv} = -1007.86kJ$

The negative value indidicates heat ransfer from the system

7 0

3 years ago