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3 years ago

What are the disadvantages of using 3D ink jet printing ??

Engineering

1 answer:

viva [34]3 years ago

4 0

Answer:

The disadvantage of 3D ink jet printers are as follows

1.These printers are slow printers.

2.These printers are more expensive.

3.These are not eco friendly printers.These printers produces harmful effect.

4.These printers consume more power.

5.Thee printers are too much reliance on the plastic.

6.These printers increase the risk of asthma.

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A cannon ball is fired with an arching trajectory such that at the highest point of the trajectory the cannon ball is traveling

ELEN [110]

Answer:

The radius of curvature is 979 meter

Explanation:

We have given velocity of the canon ball v = 98 m/sec

Acceleration due to gravity $g=9.81m/sec^2$

We know that at highest point of trajectory angular acceleration is equal to acceleration due to gravity

Acceleration due to gravity is given by $a_c=\frac{v^2}{r}$ , here v is velocity and r is radius of curvature

So $\frac{98^2}{r}=9.81$

r = 979 meter

So the radius of curvature is 979 meter

8 0

4 years ago

Assume you are an observer standing at a point along a three-lane roadway. All vehicles in lane 1 are traveling at 30 mi/h, all

garik1379 [7]

Answer:

Explanation:

Given data;

In line 1, v1 = 30mi/hr

in line 2, v2 = 45mi/hr

in line 3, v3 = 60mi/hr

therefore time mean speed = v1 + v2 + v3 /n

= VT = 45mi/hr

space mean speed ; Vs

harmonic mean = 1/V = 1/v1 + 1/v2 + 1/v3

V = 13.85mi/hr

Hence Vs = V x n = 3 x 13.85 = 41.55mi/hr

5 0

3 years ago

when removing the pistons and rods assemblies from a cylinder block technician a positions the throw of the crankshaft at the to

defon

Answer: Technician A

Reason: Piston and rod are connected with crank shaft and connecting rod. Smaller end of connecting rod is connected with piston and bigger end is connected with crankshaft.

<u>Explanation: </u>

When removing the piston and rod assemblies from a cylinder block: The technician with correct approach. So to remove piston from cylinder technician must throw crankshaft first and the remove connecting rod by losing nuts and caps...So technician A is in right way. Technician B using hammer to remove piston from the rod not possible because connecting rod and piston connected by nuts and caps can’t be separate by hammer.

Technician A positions the throw of the crankshaft at the top of its stroke and removes the connecting rod nuts and cap.

Technician B covers the rod bolts with hammers and pushes the piston and rod assembly out with the wooden hammer handle or wooden drift and supports the piston as it comes out of the cylinder.

4 0

3 years ago

Shear plane angle and shear strain: In an orthogonal cutting operation, the tool has a rake angle = 16°. The chip thickness befo

Oduvanchick [21]

Answer:

shear plane angle Ф = 26.28°

shear strain 2.20

Explanation:

given data

angle = 16°

chip thickness t1 = 0.32 mm

cut yields chip thickness t2 = 0.72 mm

solution

we get here first chip thickness ratio that is

chip thickness ratio = $\frac{t1}{t2}$ ................. 1

put here value

chip thickness ratio = $\frac{0.32}{0.72}$

chip thickness ratio r = 0.45

so here shear angle will be Ф

tan Ф = $\frac{r*cos\alpha }{1-rsin\alpha}$ ............2

tan Ф = $\frac{0.45*cos16 }{1-rsin16}$

tan Ф = 0.4938

Ф = 26.28°

and

now we get shear strain that is

shear strain r = cot Ф + tan (Ф - α ) ................3

shear strain r = cot(26.28) + tan (26.28 - 16 )

shear strain r = 2.20

6 0

3 years ago

A cylindrical insulation for a steam pipe has an inside radius rt = 6 cm, outside radius r0 = 8 cm, and a thermal conductivity k

goldfiish [28.3K]

Answer:

heat loss per 1-m length of this insulation is 4368.145 W

Explanation:

given data

inside radius r1 = 6 cm

outside radius r2 = 8 cm

thermal conductivity k = 0.5 W/m°C

inside temperature t1 = 430°C

outside temperature t2 = 30°C

to find out

Determine the heat loss per 1-m length of this insulation

solution

we know thermal resistance formula for cylinder that is express as

Rth = $\frac{ln\frac{r2}{r1}}{2 \pi *k * L}$ .................1

here r1 is inside radius and r2 is outside radius L is length and k is thermal conductivity

heat loss is change in temperature divide thermal resistance

Q = $\frac{t1- t2}{\frac{ln\frac{r2}{r1}}{2 \pi *k * L}}$

Q = $\frac{(430-30)*(2 \pi * 0.5 * 1}{ln\frac{8}{6} }$

Q = 4368.145 W

so heat loss per 1-m length of this insulation is 4368.145 W

4 0

4 years ago