Engineering College

2.) Technician A says that milky colored ATF could indicate a leaking transmission cooler in the radiator.
Technician B says that milky colored ATF could indicate the presence of leak detection dye.
Who is right?
OA. A only
OB. B only
OC. Both A and B
OD. Neither A nor B

Answers

Answer 1
I think a is the correct answer

Related Questions

A brass alloy rod having a cross sectional area of 100 mm2 and a modulus of 110 GPa is subjected to a tensile load. Plastic deformation was observed to begin at a load of 39872 N. a. Determine the maximum stress that can be applied without plastic deformation. b. If the maximum length to which a specimen may be stretched without causing plastic deformation is 67.21 mm, what is the original specimen length

Answers

Answer:

a) the maximum stress that can be applied without plastic deformation is 398.72 N/mm²  

b) length of the specimen is 66.97 mm

Explanation:

Given the data in the question;

a) Determine the maximum stress that can be applied without plastic deformation

when know that; maximum stress σ[tex]_{max}[/tex]  = F / A

where F is the force in the rod ( 39872 N )

A is the cross-sectional area of the rod ( 100 mm² )

so we substitute;

σ[tex]_{max}[/tex]  = 39872 N / 100 mm²

σ[tex]_{max}[/tex]  = 398.72 N/mm²

Therefore, the maximum stress that can be applied without plastic deformation is 398.72 N/mm²  

b)  

strain in the members can be calculated using the expression

ε = σ / E

where σ is the stress in the rod

E is the module of elasticity (  110 GPa = 110000 N/mm² )

(Sl-L) / L = σ/E

where Sl-L is the change in length of the member

L is the original length of the specimen

so we substitute

(67.21 - L) / L = 398.72 / 110000

110000( 67.21 - L) = 398.72L

7393100 - 110000L = 398.72L

7393100 = 398.72L+ 110000L

7393100 = 110398.72L

L = 7393100 / 110398.72

L = 66.97 mm

Therefore; length of the specimen is 66.97 mm

 

Perform the following unit conversions. Please do not use an on-line unit converter since this problem is given to you as practice in preparation for what you need to be proficient in:

a. 180 in^3 to L
b. 750 ft-lbf to kJ
c. 75.0 hp to kW
d. 2500.0 lb/h to kg/s
e. 120 psia to kPa
f. 120 psig to kPa
g. 300 ft/min to m/s
h. 125 km/h to miles/h
i. 6000 N to Ibf
j. 6000 N to ton

Answers

Answer:

The answers are below

Explanation:

a. 180 in^3 to L

1 in³ = 0.0164L

180 in³ = [tex]180\ in^3*\frac{0.0164\ L}{1\ in^3}= 2.95\ L[/tex]

b. 750 ft-lbf to kJ

1 ft-lbf = 0.00136 kJ

750 ft-lbf  = [tex]750\ ft-lbf *\frac{0.00136\ kJ}{1\ ft-lbf} =1.02\ kJ[/tex]

c. 75.0 hp to kW

1 hp = 0.746 kW

75 hp = [tex]75\ hp*\frac{0.746\ kW}{1\ hp}=55.95\ kW[/tex]

d. 2500.0 lb/h to kg/s

1 lb/h = 0.000126 kg/s

2500.0 lb/h = [tex]2500.0\ lb/h*\frac{0.000126\ kg/s}{1\ lb/h} =0.315\ kg/s[/tex]

e. 120 psia to kPa

1 psia = 6.89 kPa

120 psia = [tex]120\ psia*\frac{6.89\ kPa}{1\ psia} =826.8\ kPa[/tex]

f. 120 psig to kPa

1 psig = 6.89 kPa

120 psig = [tex]120\ psia*\frac{6.89\ kPa}{1\ psig} =826.8\ kPa[/tex]

g. 300 ft/min to m/s

1 ft/min = 0.005 m/s

300 ft/min = [tex]300\ ft/min*\frac{0.005\ m/s}{1\ ft/min} = 1.5\ m/s\\[/tex]

h. 125 km/h to miles/h

1 km/h = 0.62 mph

125 km/h = [tex]125\ km/h*\frac{0.62\ mph}{1\ km/h} =77.5\ mph[/tex]

i) 6000 N to Ibf

1 N = 0.2248 lbf

6000 N = [tex]6000\ N*\frac{ 0.2248\ lbf}{1\ N}=1348.8\ N[/tex]

j. 6000 N to ton

1 N =  0.000102 Ton-force

6000 N = [tex]6000\ N*\frac{ 0.000102\ Ton-force}{1\ N}=0.612\ N[/tex]

Determine the voltage drop from the top terminal to the bottom terminal, vab, in the right hand branch and, vcd, in the left hand branch of the circuit. Determine each voltage drop based on the elements in the corresponding branch.

Answers

Answer:

Hello your question is incomplete attached below is the missing part of the question

answer ;

voltage drop in the Vcd branch = 30 V

Voltage drop in the middle branch = 40v - 30v = 10 volts

voltage drop in AB = 60 + ( -600 * 0.05 ) = 60 - 30 = 30 volts

Explanation:

Determine voltage drop from top terminal to bottom terminal ( Vab ) in the right hand branch and Vcd in left hand branch

40v and 50mA are in series hence;  Ix = 50mA

also Vcd = 30V

CD is parallel to AB hence; Vcd = Vab = 30 V

Vab = ∝*Ix + 60 v

 30v  = ∝ ( 50mA ) + 60

therefore ∝ = -600

voltage drop in the Vcd branch = 30 V

Voltage drop in the middle branch = 40v - 30v = 10 volts

voltage drop in AB = 60 + ( -600 * 0.05 ) = 60 - 30 = 30 volts

A river has an average rate of water flow of 59.6 M3/s. This river has three tributaries, tributary A, B and C, which account for 36%, 47% and 17% of water flow respectively. How much water is discharged in 30 minutes from tributary B?

Answers

Answer:

50421.6 m³

Explanation:

The river has an average rate of water flow of 59.6 m³/s.

Tributary B accounts for 47% of the rate of water flow. Therefore the rate of water flow through tributary B is:

Flow rate of water through tributary B = 47% of 59.6 m³/s = 0.47 * 59.6 m³/s = 28.012 m³/s

The volume of water that has been discharged through tributary B = Flow rate of water through tributary B * time taken

time = 30 minutes = 30 minutes * 60 seconds / minute = 1800 seconds

The volume of water that has been discharged through tributary B in 30 seconds = 28.012 m³/s * 1800 seconds = 50421.6 m³

3.) Technician A says that a scan tool can be used to verify engine operating temperature,
Technician B says that a refractometer can be used to verify engine operating temperature.
Who is right?
OA. A only
OB. B only
OC. Both A and B
OD. Neither A nor B

Answers

I think it is Both A and B

PWM input and output signals are often converted to analog voltage signals using low-pass filters. Design and simulate the following: a PWM signal source with 1 kHz base frequency and adjustable pulse-width modulation(PWM source), an analog filter with time constant of 0.01 s. Simulate the input and output of the low pass filter for a PWM duty cycle of 25, 50, and 100%

Answers

Answer:

Attached below

Explanation:

PWM signal source has 1 KHz base frequency

Analog filter : with time constant = 0.01 s

low pass transfer function = [tex]\frac{1}{0.01s + 1 }[/tex]

PWM duty cycle is a constant block

Attached below is the design and simulation into Simulink at 25% , 50% and 100% respectively

Engineer drawing:
How can i draw this? Any simple way?

Answers

Make 4 triangles left right up down and they must be connected with no gaps then make more triangles into the triangle about three times for each of them then add rectangles or lines to the drawing
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