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3 years ago

4.

Engineering

2 answers:

sergiy2304 [10]3 years ago

7 0

D because there’s telephone poles everywhere

erica [24]3 years ago

3 0

I would say C because GPS can’t spot construction sites

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In a mechanical assembly operation, the first work unit required 7.83 min to complete and the learning rate for mechanical assem

Marianna [84]

Answer:

(a) 6.5774 minutes

(b) 4.3879 minutes

(d) 245.902 minutes

Explanation:

The learning curve model follows the following equation:

$Y=aX^b$

where Y are the units time to produce X units, and a is the time for assembly the first unit. Additionally, b is calculated as:

$b=\frac{ln(r)}{ln(2)}$

Where r is the learning rate for mechanical assembly. So, b is equal to:

$b=\frac{ln(0.84)}{ln(2)}=-0.2515$

Then, the equation is:

$Y=7.83X^{-0.2515}$

Finally, the unit times to produce the second unit are:

$Y=7.83(2)^{-0.2515}=6.5774$

The 10th unit:

$Y=7.83(10)^{-0.2515}=4.3879$

The 100th unit:

$Y=7.83(100)^{-0.2515}=2.459$

Then, the total cumulative time T to produce 100 units in the Crawford model is calculated as:

$T=7.83\frac{(100.5)^{1-0.2515}-(0.5)^{1-0.2515})}{1-0.2515} \\T=323.5383$

7 0

3 years ago

A woodcutter wishes to cause the tree trunk to fall uphill, even though the trunk is leaning downhill. With the aid of the winch

Sedbober [7]

Answer:

The correct answer will be "400.4 N". The further explanation is given below.

Explanation:

The given values are:

Mass of truck,

m = 600 kg

g = 9.8 m/s²

On equating torques at the point O,

⇒ $T\times Cos(10+5)\times (1.3+4)=mg\times Sin(5)\times 4$

So that,

On putting the values, we get

⇒ $T\times Cos(15^{\circ})\times 5.3=600\times 9.8\times Sin(5^{\circ})\times 4$

⇒ $T=400.4 \ N$

8 0

4 years ago

Hello it's my new id<br>I am numu

Sonja [21]

Answer:

i am felix

Exp lanation:

nice to meet you

6 0

3 years ago

A car crusher has one hydraulic input piston with diameter 2 ft and 6 crushing pistons with diameter 10 feet. If this car needs

IrinaK [193]

Not sure of the answers

4 0

3 years ago

Four race cars are traveling on a 2.5-mile tri-oval track. The four cars are traveling at constant speeds of 195 mi/h, 190 mi/h,

Snezhnost [94]

Answer:

Explanation:

1) The number of times, the car with the speed of 195 mph will cross the given point is equal to 30 minutes divided by the time taken by car to cross the 2.5 miles.

0 .5*195/2.5 = 39

Likewise, the car with the speed of 190 mph crosses the point 38 times; the car with the speed of 185 mph crosses the point 37 times

and car with the speed of 180 mph crosses it 36 times

here, the time-mean speed, vt is given below,

vt = (39*195 +38*190+37*185+36*180)/(39+38+37+38)

= 186.433 mph

and space mean speed is given by,

= (39+38+37+36)/(39/195+38/190+37/1850+36/180)

1) The number of times, the car with the speed of 195 mph will cross the given point is equal to 30 minutes divided by the time taken by car to cross the 2.5 miles.

0 .5*195/2.5 = 39

Likewise, the car with the speed of 190 mph crosses the point 38 times; the car with the speed of 185 mph crosses the point 37 times

and car with the speed of 180 mph crosses it 36 times

here, the time-mean speed, vt is given below,

vt = (39*195 +38*190+37*185+36*180)/(39+38+37+38)

= 186.433 mph

and space mean speed is given by,

= (39+38+37+36)/(39/195+38/190+37/1850+36/180)

=187.5 mph

2) There would be only four number of observations when the aerial photo is given, therefore time mean speed, vt in that condition will be calculated as

Vt = 195+190+185+180/4

= 187.5

Vs= 4/(1/195+1/190+1/185+1/180)

= 188.36 mph

2) There would be only four number of observations when the aerial photo is given, therefore time mean speed, vt, in that condition will be calculated as

Vt = 195+190+185+180/4

= 187.5

Vs= 4/(1/195+1/190+1/185+1/180)

= 188.36 mph

4 0

4 years ago