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3 years ago

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Technician A says that a voltage drop of 0.8 volts on the starter ground circuit is within specifications. Technician B says tha

t high starter draw current could be caused by a spun main bearing in the engine. Who is correct?

1 answer:

Romashka-Z-Leto [24]3 years ago

8 0

Answer:

Technician A is wrong

Technician B is right

Explanation:

voltage drop of 0.8 volts on the starter ground circuit is not within specifications. Voltage drop should be within the range of 0.2 V to 0.6 V but not more than that.

A spun bearing can seize itself around the crankshaft journal causing it not to move. As the car ignition system is turned on, the stater may draw high current in order to counter this seizure.

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Assume a program requires the execution of 50 x 106 FP instructions, 110 x 106 INT instructions, 80 x 106 L/S instructions, and

Pavlova-9 [17]

Answer:

Part A:

$1.3568*10^{-5}=\frac{5300* New\ CPI_1+11660*1+8480*4+1696*2}{2*10^9\ Hz} \\ New\ CPI_1=-4.12$

CPI cannot be negative so it is not possible to for program to run two times faster.

Part B:

$1.3568*10^{-5}=\frac{5300*1+11660*1+8480*New\ CPI_3+1696*2}{2*10^9\ Hz} \\ New\ CPI_3=0.8$

CPI reduced by $1-\frac{0.8}{4} = 0.80$ =80%

Part C:

New Execution Time= $\frac{5300*0.6+11660*0.6+8480*2.8+1696*1.4}{2*10^9\ Hz}=1.81472*10^{-5}\ s$

Increase in speed= $1-\frac{1.81472*10^{-5}}{2.7136*10^{-5}} =0.33125= 33.125\%$

Explanation:

FP Instructions=50*106=5300

INT Instructions=110*106=11660

L/S Instructions=80*106=8480

Branch Instructions=16*106=1696

Calculating Execution Time:

Execution Time= $\frac{\sum^4_{i=1} Number\ of\ Instruction*\ CPI_{i}}{Clock\ Rate}$

Execution Time= $\frac{5300*1+11660*1+8480*4+1696*2}{2*10^9\ Hz}$

Execution Time= $2.7136*10^{-5}\ s$

Part A:

For Program to run two times faster,Execution Time (Calculated above) is reduced to half.

New Execution Time= $\frac{2.7136*10^{-5}}{2}=1.3568*10^{-5}\ s$

$1.3568*10^{-5}=\frac{5300* New\ CPI_1+11660*1+8480*4+1696*2}{2*10^9\ Hz} \\ New\ CPI_1=-4.12$

CPI cannot be negative so it is not possible to for program to run two times faster.

Part B:

For Program to run two times faster,Execution Time (Calculated above) is reduced to half.

New Execution Time= $\frac{2.7136*10^{-5}}{2}=1.3568*10^{-5}\ s$

$1.3568*10^{-5}=\frac{5300*1+11660*1+8480*New\ CPI_3+1696*2}{2*10^9\ Hz} \\ New\ CPI_3=0.8$

CPI reduced by $1-\frac{0.8}{4} = 0.80$ =80%

Part C:

$New\ CPI_1=0.6*Old\ CPI_1=0.6*1=0.6\\New\ CPI_2=0.6*Old\ CPI_2=0.6*1=0.6\\New\ CPI_3=0.7*Old\ CPI_3=0.7*4=2.8\\New\ CPI_4=0.7*Old\ CPI_4=0.7*2=1.4$

New Execution Time= $\frac{\sum^4_{i=1} Number\ of\ Instruction*\ CPI_{i}}{Clock\ Rate}$

New Execution Time= $\frac{5300*0.6+11660*0.6+8480*2.8+1696*1.4}{2*10^9\ Hz}=1.81472*10^{-5}\ s$

Increase in speed= $1-\frac{1.81472*10^{-5}}{2.7136*10^{-5}} =0.33125= 33.125\%$

8 0

3 years ago

A pump with a power of 5 kW (pump power, and not useful pump power) and an efficiency of 72 percent is used to pump water from a

almond37 [142]

Answer:

a) The mass flow rate of water is 14.683 kilograms per second.

b) The pressure difference across the pump is 245.175 kilopascals.

Explanation:

a) Let suppose that pump works at steady state. The mass flow rate of the water ( $\dot m$ ), in kilograms per second, is determined by following formula:

$\dot m = \frac{\eta \cdot \dot W}{g\cdot H}$ (1)

Where:

$\dot W$ - Pump power, in watts.

$\eta$ - Efficiency, no unit.

$g$ - Gravitational acceleration, in meters per square second.

$H$ - Hydrostatic column, in meters.

If we know that $\eta = 0.72$ , $\dot W = 5000\,W$ , $g = 9.807\,\frac{m}{s^{2}}$ and $H = 25\,m$ , then the mass flow rate of water is:

$\dot m = 14.683\,\frac{kg}{s}$

The mass flow rate of water is 14.683 kilograms per second.

b) The pressure difference across the pump ( $\Delta P$ ), in pascals, is determined by this equation:

$\Delta P = \rho\cdot g\cdot H$ (2)

Where $\rho$ is the density of water, in kilograms per cubic meter.

If we know that $\rho = 1000\,\frac{kg}{m^{3}}$ , $g = 9.807\,\frac{m}{s^{2}}$ and $H = 25\,m$ , then the pressure difference is:

$\Delta P = 245175\,Pa$

The pressure difference across the pump is 245.175 kilopascals.

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3 years ago

What are the advantages of studying a scheduled course over a more freeform one?

anyanavicka [17]

Answer:

Taking responsibility for your own learning makes it easier to identify your strengths and weaknesses. Once these have been identified you can work on a learning plan that focuses on the areas that you need most help with, increasing the speed of your learning, and build the skills you have been trying to perfect.

Explanation:

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2 years ago

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A 0.2-m^3 rigid tank equipped with a pressure regulator contains steam at 2MPa and 320C. The steam in the tank is now heated. Th

prohojiy [21]

Answer:

Q=486.49 KJ/kg

Explanation:

Given that

V= 0.2 m³

At initial condition

P= 2 MPa

T=320 °C

Final condition

P= 2 MPa

T=540°C

From steam table

At P= 2 MPa and T=320 °C

h₁=3070.15 KJ/kg

At P= 2 MPa and T=540°C

h₂=3556.64 KJ/kg

So the heat transfer ,Q=h₂ - h₁

Q= 3556.64 - 3070.15 KJ/kg

Q=486.49 KJ/kg

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3 years ago

Analyzing Manufacturing Production Process Development Tasks

artcher [175]

Answer:

4, 5, 6

Explanation:

i just did it

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3 years ago

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