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3 years ago

What was the main drawback of Ford’s assembly line?

Engineering

2 answers:

nikklg [1K]3 years ago

7 0

Answer:

Explanation:

eimsori [14]3 years ago

3 0

Answer: B

Explanation:

Disadvantages of assembly line production are based on the worker's point of view. Because little training is generally required, wages may not be very competitive and required many hours of labor

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A large well-mixed tank of unknown volume, open to the atmosphere initially, contains pure water. The initial height of the solu

trasher [3.6K]

Answer:

The exact time when the sample was taken is = 0.4167337 hr

Explanation:

The diagram of a sketch of the tank is shown on the first uploaded image

Let A denote the first inlet

Let B denote the second inlet

Let C denote the single outflow from the tank

From the question we are given that the diameter of A is = 1 cm = 0.01 m

Area of A is = $\frac{\pi}{4}(0.01)^{2} m^{2}$

= $7.85 *10^{-5}m^{2}$

Velocity of liquid through A = 0.2 m/s

The rate at which the liquid would flow through the first inlet in terms of volume = $\frac{Volume of Inlet }{time}$ = $Velocity * Area$ i.e is $m^{2} * \frac{m}{s} = \frac{m^{3}}{s}$

= $0.2 *7.85*10^{-5} \frac{m^{3}}{s}$

The rate at which the liquid would flow through the first inlet in terms of mass of the liquid = mass of liquid × the rate of flow in terms of volume

= $1039.8 * 0.2 * 7.85 *10^{-5} Kg/s$

= 0.016324 $\frac{Kg}{s}$

From the question the diameter of B = 2 cm = 0.02 m

Area of B = $\frac{\pi}{4} * (0.02)^{2} m^{2} = 3.14 * 10^{-4}m^{2}$

Velocity of liquid through B = 0.01 m/s

The rate at which the liquid would flow through the first inlet in terms of volume = $\frac{Volume of Inlet }{time}$ = $Velocity * Area$ i.e is $m^{2} * \frac{m}{s} = \frac{m^{3}}{s}$

= $3.14*10^{-4} *0.01 \frac{m^{3}}{s}$

The rate at which the liquid would flow through the second inlet in terms of mass of the liquid = mass of liquid × the rate of flow in terms of volume

= $1053 * 3.14*10^{-6} \frac{Kg}{s}$

= 0.00330642 $\frac{Kg}{s}$

From the question The flow rate in term of volume of the outflow at the time of measurement is given as = 0.5 L/s

And also from the question the mass of potassium chloride at the time of measurement is given as 13 g/L

So The rate at which the liquid would flow through the outflow in terms of mass of the liquid = mass of liquid × the rate of flow in terms of volume

= $13\frac{g}{L} * 0.5 \frac{L}{s}$

= $\frac{6.5}{1000}\frac{Kg}{s}$ Note (1 Kg = 1000 g)

= 0.0065 kg/s

Considering potassium chloride

Let denote the rate at which liquid flows in terms of mass as as $\frac{dm}{dt}$ i.e change in mass with respect to time hence

Input(in terms of mass flow ) - output(in terms of mass flow ) = Accumulation in the Tank(in terms of mass flow )

(0.016324 + 0.00330642) - 0.0065 = $\frac{dm}{dt}$

$\int\limits {\frac{dm}{dt} } \, dx =\int\limits {0.01313122} \, dx$

=> 0.01313122 t = (m - $m_{o}$ )

From the question (m - $m_{o}$ ) is given as = 19.7 Kg

Hence the time when the sample was taken is given as

0.01313122 t = 19.7 Kg

=> t = 1500.2414 sec

t = .4167337 hours (1 hour = 3600 seconds)

5 0

3 years ago

For a bolted assembly with eight bolts, the stiffness of each bolt is kb = 1.0 MN/mm and the stiffness of the members is km = 2.

rjkz [21]

Answer:

a) 0.978

b) 0.9191

c) 1.056

d) 0.849

Explanation:

Given data :

Stiffness of each bolt = 1.0 MN/mm

Stiffness of the members = 2.6 MN/mm per bolt

Bolts are preloaded to 75% of proof strength

The bolts are M6 × 1 class 5.8 with rolled threads

Pmax =60 kN, Pmin = 20kN

a) Determine the yielding factor of safety

$n_{p} = \frac{S_{p}A_{t} }{CP_{max}+ F_{i} }$ ------ ( 1 )

Sp = 380 MPa, At = 20.1 mm^2, C = 0.277, Pmax = 7500 N, Fi = 5728.5 N

Input the given values into the equation above

equation 1 becomes ( np ) = $\frac{380*20.1}{0.277*7500*5728.5}$ = 0.978

note : values above are derived values whose solution are not basically part of the required solution hence they are not included

b) Determine the overload factor of safety

$n_{L} = \frac{S_{p}A_{t}-F_{i} }{C(P_{max} )}$ ------- ( 2 )

Sp = 380 MPa, At = 20.1 mm^2, C = 0.277, Pmax = 7500 N, Fi = 5728.5 N

input values into equation 2 above

hence : $n_{L}$ = 0.9191 $n_{L} = 0.9191$

C) Determine the factor of safety based on joint separation

$n_{0} = \frac{F_{i} }{P_{max}(1 - C ) }$

Fi = 5728.5 N, Pmax = 7500 N, C = 0.277,

input values into equation above

Hence $n_{0}$ = 1.056

D) Determine the fatigue factor of safety using the Goodman criterion.

nf = 0.849

attached below is the detailed solution .

4 0

2 years ago

Two Technicians are discussing shock absorbers. Technician A says most shock absorbers help support vehicle weight. Technician B

yKpoI14uk [10]

Answer:

Both Technician A and B

Explanation:

In a vehicle suspension system, a shock absorber has the functions such as; limiting the movement of the vehicle, stabilizing the ride, stabilizing the tires of the vehicle, minimizing wear and tear in the vehicle and decreasing the overall suspension tear. Some of the ways of maintaining shock absorbers is to inspect the strut for leaks.Shock oil sometimes leaks to cover the seals with a transparent liquid.

6 0

3 years ago

What’s another name for a service overcurrent device?

alexandr402 [8]

Overcurrent protective devices, or OCPDs

4 0

3 years ago

Is normally a large red cable connected to the battery

Igoryamba

Answer:

yes

Explanation:

its the ground cable

4 0

2 years ago

Read 2 more answers