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3 years ago

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Movimiento

1 answer:

goblinko [34]3 years ago

5 0

wth u should clear ur question

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When using fall arrest, free fall must be kept at or below how many feet

FromTheMoon [43]

6 feet
Explanation: A few extra feet of free fall can significantly increase the arresting force on the employee, possibly to the point of causing injury. Because of this, the free fall distance should be kept at a minimum, and, as required by the standard, in no case greater than 6 feet (1.8 m).

5 0

3 years ago

It is the tool used to measure the amount of electric current

sergeinik [125]

Answer:

Ammeter

Explanation:

Instrument for measuring either direct or alternating electric current, in amperes. Ammeters vary in their operating principles and accuracies

7 0

3 years ago

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A 5-mm-thick stainless steel strip (k = 21 W/m•K, rho = 8000 kg/m3, and cp = 570 J/kg•K) is being heat treated as it moves throu

Drupady [299]

Answer:

The temperature of the strip as it exits the furnace is 819.15 °C

Explanation:

The characteristic length of the strip is given by;

$L_c = \frac{V}{A} = \frac{LA}{2A} = \frac{5*10^{-3}}{2} = 0.0025 \ m$

The Biot number is given as;

$B_i = \frac{h L_c}{k}\\\\B_i = \frac{80*0.0025}{21} \\\\B_i = 0.00952$

$B_i$ < 0.1, thus apply lumped system approximation to determine the constant time for the process;

$\tau = \frac{\rho C_p V}{hA_s} = \frac{\rho C_p L_c}{h}\\\\\tau = \frac{8000* 570* 0.0025}{80}\\\\\tau = 142.5 s$

The time for the heating process is given as;

$t = \frac{d}{V} \\\\t = \frac{3 \ m}{0.01 \ m/s} = 300 s$

Apply the lumped system approximation relation to determine the temperature of the strip as it exits the furnace;

$T(t) = T_{ \infty} + (T_i -T_{\infty})e^{-t/ \tau}\\\\T(t) = 930 + (20 -930)e^{-300/ 142.5}\\\\T(t) = 930 + (-110.85)\\\\T_{(t)} = 819.15 \ ^0 C$

Therefore, the temperature of the strip as it exits the furnace is 819.15 °C

5 0

3 years ago

Two routes connect an origin and a destination. Routes 1 and 2 have performance functions t1 = 2 + X1 and t2 = 1 + X2, where the

Musya8 [376]

Solution :

Given

$t_1=2+x_1$

$t_2=1+x_2$

Now,

$$P(h$

$0.4=1-P(h \geq5)$

$0.6=P(h \geq5)$

$0.6= e^{\frac{-x_1 5}{3600}}$

Therefore, $x_1=368 \ veh/h$

$=\frac{368}{1000} = 0.368$

Given, $t_1=2+x_1$

= 2 + 0.368

= 2.368 min

At user equilibrium, $t_2=t_1$

∴ $t_2$ = 2.368 min

$t_2=1+x_2$

$2.368=1+x_2$

$x_2 = 1.368$

$x_2 = 1.368 \times 1000$

= 1368 veh/h

7 0

3 years ago

The loneliest people are to kindest

valkas [14]

Answer:

The most damaged people are the wisest is a fact

Explanation:

5 0

3 years ago

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