# In this type of projection, the angles between the three axes are different:- A) Isometric B) Axonometric C) Trimetric D) Dimetnic

The correct answer is C) Trimetric

Explanation:

The most suitable answer is a trimetric projection because, in this type of projection, we see that the projection of the three angles between the axes are not equal. Therefore, to generate a trimetric projection of an object, it is necessary to have three separate scales.

## Related Questions

A cylindrical specimen of a hypothetical metal alloy is stressed in compression. If its original and final diameters are 20.000 and 20.025 mm, respectively, and its final length is 74.96 mm, compute its original length if the deformation is totally elastic. The elastic and shear moduli for this alloy are 105 GPa and 39.7 GPa, respectively.

L = 75.25 mm

Explanation:

First we need to find the lateral strain:

Lateral Strain = Change in Diameter/Original Diameter

Lateral Strain = (20.025 mm - 20 mm)/20 mm

Lateral Strain = 1.25 x 10⁻³

Now, we will find the Poisson's Ratio:

Poisson's Ratio = (E/2G) - 1

where,

E = Elastic Modulus = 105 GPa

G = Shear Modulus = 39.7 GPa

Therefore,

Poisson's Ratio = [(105 GPa)/(2)(39.7 GPa)] - 1

Poisson's Ratio = 0.322

Now, we find longitudinal strain by following formula:

Poisson's Ratio = - Lateral Strain/Longitudinal Strain

Longitudinal Strain = - Lateral Strain/Poisson's Ratio

Longitudinal Strain = - (1.25 x 10⁻³)/0.322

Longitudinal Strain = - 3.87 x 10⁻³

Now, we can fin the original length:

Longitudinal Strain = Change in Length/L

where,

L = Original Length = ?

Therefore,

- 3.87 x 10⁻³ = (74.96 mm - L)/L

(- 3.87 x 10⁻³)(L) + L = 74.96 mm

0.99612 L = 74.96 mm

L = 74.96 mm/0.99612

L = 75.25 mm

The process in which the system pressure remain constant is called a)-isobaric b)-isochoric c)-isolated d)-isothermal

Isobaric process

Explanation:

The process in which the system pressure remain constant is called is called isobaric process. The word "iso"means same and baric means pressure.

At constant pressure, the work done is given by :

Where

W is the work done by the system

p is the constant pressure

is the change in volume

So, the correct option is (c) " isobaric process ".

Q/For the circuit showm bellow:a) find the mathematical expression for the transient behavior of ve and ic after closing the switeh,
b) sketch vc and ic.

hello your question is incomplete attached is the complete question

A) Vc =  15 ( 1 - ) ,   ic =

B) attached is the relevant sketch

Explanation:

applying Thevenin's theorem to find the mathematical expression for the transient behavior of Vc and ic after closing the switch

= 8 k ohms || 24 k ohms = 6 k ohms

=  =  15 v

t = RC = (10 k ohms( 15 uF) = 0.15 s

Also; Vc =

hence Vc = 15 ( 1 - )

ic = = =

attached

A 5000-ft long X-65 pipeline is laid down on seabed with two PLETS (One at each end). The pipe OD=7-in with 0.5-in wall thickness. The pipeline was laid at environmental temperature of 40 °F (As- laid temperature). When pipeline is put into operation, the oil flow was produced at 140 °F. If the thermal expansion coefficient of the pipe material is 6.5*10-/°F and its modulus of elasticity is 30,000 ksi, determine the compressive load applied by the pipeline on a PLET due to its thermal expansion. Assume no temperature change and no seabed friction along the pipeline span.

Explanation:

Given that

Outer diameter is Do = 7 in

Inner diameter Di = ( Do - ( 2×0.5)) = 6 in

Length = 5000 ft = 60000 in

Now change in length of the pipe due to temperature difference

SL = L∝ΔT

= 60000 × 6.5×10^-6(140-40)

SL = 39 in

Also

sL = PL/AE

A = cross sectional area of pipe = π/4(Do^2 - Di^2)

so

P = SL×A×E / L

= (39 × π/4(7^2 - 6^2)×30000) / 60000

= 199.1 kip

compressive load applied by the pipeline on a PLET due to its thermal expansion is 199.1 kip

In an adiabatic process, the temperature of the system remains constant. a)- True b)- False