All categories

2 years ago

The first step to merging is entering the ramp and _____.

Engineering

1 answer:

Yuri [45]2 years ago

7 0

Answer:

D.telling your passengers where you are going

You might be interested in

Evan notices a small fire in his workplace. Since the fire is small and the atmosphere is not smoky he decides to fight the fire

Norma-Jean [14]

Answer:

not calling the firemean

Explanation:

7 0

2 years ago

The 10-kg block slides down 2 m on the rough surface with kinetic friction coefficient μk = 0.2. What is the work done by the fr

Rashid [163]

Answer:

153.2 J

Explanation:

Let's first list our given parameters;

mass (m) of the block = 10 kg

which slides down ( i.e displacement) = 2 m

kinetic coefficient of friction (μk) = 0.2

In the diagram shown below; if we take an integral look at the component of force in the direction of the displacement; we have

$F_x=$ Fcos 40°

$F_x=$ 100 (cos 40°)

$F_x=$ 76.60 N

Workdone by the friction force can now be determined as:

W = $F_x$ × displacement

W = 76.60 × 2

W = 153.2 J

∴ the work done by the friction force = 153.2 J

7 0

2 years ago

An annealed copper strip of 228 mm wide and 25 mm thick being rolled to a thickness of 20 mm, in one pass. The roll radius is 30

sdas [7]

Answer:

The roll force is 1.59 MN

The power required in this operation is 644.96 kW

Explanation:

Given;

width of the annealed copper, w = 228 m

thickness of the copper, h₀ = 25 mm

final thickness, hf = 20 mm

roll radius, R = 300 mm

The roll force is given by;

$F = LwY_{avg}$

where;

w is the width of the annealed copper

$Y_{avg}$ is average true stress of the strip in the roll gap

L is length of arc in contact, and for frictionless situation it is given as;

$L = \sqrt{R(h_o-h_f)} \\\\L = \sqrt{300(25-20)}\\\\L = 38.73 \ mm$

Now, determine the average true stress, $Y_{avg}$ , for the annealed copper;

The absolute value of the true strain, ε = ln(25/20)

ε = 0.223

from true stress vs true strain graph; at true strain of 0.223, the true stress is 280 MPa.

Then, the average true stress = ¹/₂(280 MPa.) = 180 MPa

Finally determine the roll force;

$F = LwY_{avg}$

$F = (\frac{38.73 }{1000})(\frac{228}{1000})*180 \ MPa\\\\F = 1.59 \ MN$

The power required in this operation is given by;

$P = \frac{2\pi FLN}{60}\\\\P = \frac{2\pi (1.59*10^6)(0.03873)(100)}{60}\\\\P = 644955.2 \ W\\\\P = 644.96 \ kW$

5 0

2 years ago

The extruder head in a fused- deposition modeling setup has a diameter of 1.25 mm (0.05 in) and produces layers that are 0.25mm

Angelina_Jolie [31]

Answer:

The time taken will be "1 hour 51 min". The further explanation is given below.

Explanation:

The given values are:

Number of required layers:

= $\frac{38}{0.25}$

= $152 \ layers$

Diameter (d):

= 1.25 mm

Velocity (v):

= 40 mm/s

Now,

The area of one layer will be:

= $38\times 38 \ mm^2$

= $1444 \ mm^2$

The area covered every \second will be:

= $d\times v$

= $1.25\times 40$

= $50 \ mm^2$

The time required to deposit one layer will be:

= $\frac{1444}{50}$

= $28.88 \ sec$

The time required for one layer will be:

= $15 \ sec$

∴ Total times required for one layer will be:

= $15+28.88$

= $43.88 \ sec$

So,

Number of layers = 152

Therefore,

Total time will be:

= $152\times 43.88$

= $6669.76 \ sec$

= $1 \ hour \ 51 \ min$

6 0

2 years ago

Superheated steam at an average temperature 200 C is transported through a steel pipe (k=50 W/mK, D_0=8.0 cm,D_i=6.0 cm,and L=20

Zarrin [17]

The total amount of daily heat transfer is 1382.38 M w.

The temperature on the outside surface of the gypsum plaster insulation is 17.96 ° C.

<u>Explanation:</u>

Given data,

$T_{\infty}$ = 10° C

$h_{0}$ = 250 w/ $m^{2}$ k

Pipe length = 20 m

Inner diameter $d_{1}$ = 6 cm, $r_{1}$ = 3 cm

Outer diameter $d_{2}$ = 8 cm, $r_{2}$ = 4 cm

The thickness of insulation is 4 cm.

$r_{3}$ = $r_{2}$ + 4

= 4+4

$r_{3}$ = 8 cm

$h_{0}$ is the heat transfer coefficient of convection inside, $h_{i}$ is the heat transfer coefficient of convection outside.

The heat transfer rate between ambient and steam is

$q=\frac{T_{S}-T_{\infty}}{\frac{1}{h_{i}\left(2 \pi r_{1} L\right)}+\frac{\ln \left(r_{2} / r_{1}\right)}{2 \pi K_{1} L}+\frac{\ln \left(r_{3}/ r_{2}\right)}{2 \pi K_{2} L}+\frac{1}{h_{0}\left(2 \pi r_{3} L\right)}}$ watt

= $\begin{aligned}&\frac{1}{800(2 \pi x \cdot 03 \times 20)}\++\frac{\ln (4 / 3)}{2 \pi \times 50 \times 20}+\frac{\ln (8 / 4)}{2 \pi \times 0.5 \times 20}+\frac{1}{200(2 \pi x \cdot 08 \times 20)}\end{aligned}$ watt