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Ne4ueva [31]
1 year ago
10

the left rear brake drum is scored, but the right rear drum looks as good as new. technician a says the left-side drum should be

resurfaced and the right-side drum should be reinstalled without resurfacing. technician b says both sides should be remachined to a similar diameter. who is correct?
Engineering
1 answer:
Romashka-Z-Leto [24]1 year ago
7 0

Technician b is correct because the left rear brake drum is scored, but the right rear drum looks as good as new then both sides should be remachined to a similar diameter.

A brake drum is a revolving cylinder-shaped component that is friction-activated by a set of shoes or pads pressing outward against it. Drum brakes are brakes where shoes are pressed against the inner surface of the drum. Drum brakes are frequently used at the rear. Compared to disc brakes, these brakes have more parts and are more difficult to maintain, but they cost less to make and are simple to add an emergency braking mechanism. The front and rear brakes are standard equipment on all automobiles. Your car may have a "disc brake" system on all four wheels, a "drum brake" system on all four wheels, or a combination of the two. Typically, the front brakes of your car have a disc system, and the rear brakes have a drum system.

Learn more about brake drum here:

brainly.com/question/14937026

#SPJ4

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An 1800-W toaster, a 1400-W electric frying pan, and a 75-W lamp are plugged into the same outlet in a 15-A, 120-V circuit. The
Mila [183]

Answer:

a) Current drawn by the toaster = 15A

Current drawn by the electric frying pan = 11.67A

Current drawn by the lamp = 0.625A

b) This combination will blow the 15A fuse as the total current requirement for this setup exceeds the 15A rating of the fuse.

Explanation:

a) For parallel connection, there exists, the same voltage and different currents across all the devices.

Voltage cross each of the 3 devices = outlet voltage of 120V

From their respective power rating, current drawn by each device will be calculated.

P = IV

For the toaster, P = 1800 W, V = 120V

I = 1800/120 = 15A

For the electric frying pan, P = 1400 W, V = 120 V

I = 1400/120 = 11.67 A

For the lamp, P = 75 W, V = 120V

I = 75/120 = 0.625 A

b) Total current in a parallel connection setup = Sum total of all the currents.

Total current drawn by all 3 devices = 15 + 11.67 + 0.625 = 27.295A = 27.3 A

This total current requirement surpasses the 15A current rating of the fuse, therefore, this combination will blow the fuse.

Hope this Helps!!!

6 0
3 years ago
An incandescent light bulb can be regarded as a resistor. If its power output is 100W, calculate the resistance of the light bul
stira [4]

Answer:121\ \Omega

0.909\ A

Explanation:

Given

Power P=100\ W

Voltage applied V=110\ V

Resistance of the bulb is given by

P=\frac{V^2}{R}

100=\frac{110^2}{R}

R=\frac{12100}{100}

R=121\ \Omega

Current drawn by the Power source is given by

P=V\cdot I

I=\frac{P}{V}

I=\frac{100}{110}

I=0.909\ A

8 0
3 years ago
Acertain foundation will experience a bearing capacity failurewhen it is subjected to a downward load of 2200 kN. Using ASD with
ehidna [41]

Answer:

Um...

Explanation:

This is what I like to see teachers giving out.

7 0
3 years ago
A liquid stream containing 52.0 mole% benzene and the balance toluene at 20.0°C is fed to a continuous single-stage evaporator a
OLga [1]

Answer:

Operating Pressure P = 793.716 mmHg

Y_Benzene y1 = 0.541

Explanation:

Given that;

liquid phase leaving the evaporator = 32.5 mole%

Equi Temp T = 99.0°C = 99 + 273.15 = 372.15 K

Now let 1 and 2 represent Benzene and Toluene respectively.

Antoine's Constant for these components are;

COMPONENETS        A                B                    C

Benzene 1             4.72583     1660.652        -1.461

Toluene  2            4.07827     1343.943         -53.773

Antoine's equation is expressed as;

Ps = 10^(A - (B/(T+C)))

Ps is in Bar and T is in Kelvin

so

P1s = 10^( 4.72583 - (1660.652/(372.15 - (-1.461)))) = 1.7617 Bar

P2s = 10^( 4.07827 - (1343.943/(372.15 - (-53.773)))) = 0.7195 Bar

now here, liquid leaving and vapor are both in equilibrium

composition of liquid leaving are;

X1 = 32.5%    = 0.325

X2 = 1 - X1 = 1 - 0.325 = 0.675

Now

Raoult's Law is expressed as;

p × y1=x1 × pis     for all components

So for Benzene ; p × y1=x1 × p1s   ------let this be equation 1

for Toluene ; p × y2=x2 × p2s   ------let this be equation 2

lets add equ 1 and 2

p × y1=x1 × p1s + p × y2=x2 × p2s

p(y1 + y2) = x1 × p1s + x2 × p2s

buy y1 + y2 = 1

therefore we substitute

p(1) = 0.325 × 1.7617 + 0.675 × 0.7195 = 1.0582 Bar

we know that 1 Bar = 750.062 mmHg

so p = 1.0582 × 750.062

p = 793.716 mmHg

Also from equation 1

p × y1=x1 × p1s

y1 = (x1 × p1s) / p

y1 = (0.325 × 1.7617) / 1.0582

y1 = 0.541

Therefore;

Operating Pressure P = 793.716 mmHg

Y_Benzene y1 = 0.541

5 0
2 years ago
CODING HELPPP Erin’s classmate said that the program that they were working on is not running properly because the classmate acc
mr Goodwill [35]
Syntax

Compiling the code, punctuation and characters must be corrected.

4 0
3 years ago
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