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Hoochie [10]
3 years ago
7

Select the correct answer. Ruby wants to create a cube puzzle game. For that she has to create a cube. Which drafting tool do yo

u think she should use to create the cube? ОА. a French curve and a ruler OB a compass and a protractor Oc. a ruler and a protractor OD. a triangle and a T-Square O E. a ruler and a compass​
Engineering
1 answer:
Alexxx [7]3 years ago
3 0

Answer:

D. a triangle and a T-Square

Explanation:

A T-Square is the best drawing tool to create squares. You would need a squares to create cubes.

You might be interested in
If you have a hole diameter of 0.250 with a tolerance of ±0.005, what are the limits of the hole size?
sleet_krkn [62]

Answer:

The limits of the hole size are;

The maximum limit of the hole diameter 0.255

The minimum limit of the hole diameter = 0.245

Explanation:

Tolerance is a standardized form of language that can be used to define the intended 'tightness' or 'clearance' degree between mating parts in a mechanical assembly process and in metal joining processes such as welding and brazing processes

In tolerancing, the size used in the description of a part is known as the nominal size while allowable variation of the nominal size that will still allow the part to function properly is known as the tolerance

A tolerance given in the form ±P is known as bilateral tolerancing, with the value being added to or subtracted from the nominal size to get the maximum and minimum allowable limits of the dimensions of the nominal size

Therefore;

The given nominal dimension of the hole diameter = 0.250

The bilateral tolerance of the dimension, = ±0.005

Therefore;

The maximum limit of the diameter of the hole = 0.250 + 0.005 = 0.255

The minimum limit of the diameter of the hole = 0.250 - 0.005 = 0.245

7 0
3 years ago
Consider a vortex filament of strength in the shape of a closed circular loop of radius R. Obtain an expression for the velocity
zysi [14]

Answer:

<em>v</em><em> </em>= T/(2R)

Explanation:

Given

R = radius

T = strength

From Biot - Savart Law

d<em>v</em> = (T/4π)* (d<em>l</em> x <em>r</em>)/r³

Velocity induced at center

<em>v </em>= ∫  (T/4π)* (d<em>l</em> x <em>r</em>)/r³

⇒   <em>v </em>= ∫  (T/4π)* (d<em>l</em> x <em>R</em>)/R³  (<em>k</em>)        <em>k</em><em>:</em> unit vector perpendicular to plane of loop

⇒   <em>v </em>= (T/4π)(1/R²) ∫ dl

If l ∈  (0, 2πR)

⇒   <em>v </em>= (T/4π)(1/R²)(2πR)  (<em>k</em>)    ⇒   <em>v </em>= T/(2R)  (<em>k</em>)  

3 0
3 years ago
Exercises
Feliz [49]

Answer:

Rocket

Gas

Explanation:

5 0
3 years ago
An Ideal gas is being heated in a circular duct as while flowing over an electric heater of 130 kW. The diameter of duct is 500
Assoli18 [71]

Answer: The exit temperature of the gas in deg C is 32^{o}C.

Explanation:

The given data is as follows.

C_{p} = 1000 J/kg K,   R = 500 J/kg K = 0.5 kJ/kg K (as 1 kJ = 1000 J)

P_{1} = 100 kPa,     V_{1} = 15 m^{3}/s

T_{1} = 27^{o}C = (27 + 273) K = 300 K

We know that for an ideal gas the mass flow rate will be calculated as follows.

     P_{1}V_{1} = mRT_{1}

or,         m = \frac{P_{1}V_{1}}{RT_{1}}

                = \frac{100 \times 15}{0.5 \times 300}  

                = 10 kg/s

Now, according to the steady flow energy equation:

mh_{1} + Q = mh_{2} + W

h_{1} + \frac{Q}{m} = h_{2} + \frac{W}{m}

C_{p}T_{1} - \frac{80}{10} = C_{p}T_{2} - \frac{130}{10}

(T_{2} - T_{1})C_{p} = \frac{130 - 80}{10}

(T_{2} - T_{1}) = 5 K

T_{2} = 5 K + 300 K

T_{2} = 305 K

           = (305 K - 273 K)

           = 32^{o}C

Therefore, we can conclude that the exit temperature of the gas in deg C is 32^{o}C.

8 0
3 years ago
A three-point bending test was performed on an aluminum oxide specimen having a circular cross section of radius 5.6 mm; the spe
ankoles [38]

Answer:

F =  8849 N

Explanation:

Given:

Load at a given point = F =  4250 N

Support span = L = 44 mm

Radius = R = 5.6 mm

length thickness of tested material = 12 mm

First compute the flexural strength for circular cross section using the formula below:

σ_{fs} = F_{f} L / \pi  R^{3}

σ = FL / π R³

Putting the given values in the above formula:

σ = 4250 ( 44 x 10⁻³ ) / π  ( 5.6 x 10⁻³ ) ³

  = 4250 ( 44 x 10⁻³ )  / 3.141593 ( 5.6 x 10⁻³ ) ³

  = 4250 (44 x 1 /1000 )) / 3.141593 ( 5.6 x 10⁻³ ) ³

  = 4250 ( 11 / 250  ) / 3.141593 ( 5.6 x 10⁻³ ) ³

  = 187 / 3.141593 ( 5.6 x 1 / 1000 ) ³

  = 187 / 3.141593 (0.0056)³

  = 338943767.745358

  = 338.943768 x 10⁶

σ = 338 x 10⁶ N/m²

Now we compute the load i.e. F from the following formula:

F_{f} = 2 σ_{fs} d³/3 L

F = 2σd³/3L

  = 2(338 x 10⁶)(12 x 10⁻³)³ / 3(44 x 10⁻³)

  = 2 ( 338 x 1000000 ) ( 12 x 10⁻³)³ / 3 ( 44 x 10⁻³)

  = 2 ( 338000000 ) ( 12 x 10⁻³)³ / 3 ( 44 x 10⁻³)

  = 676000000 ( 12 x 10⁻³)³ / 3 ( 44 x 10⁻³)

  = 676000000 ( 12  x  1/1000  )³ / 3 ( 44 x 10⁻³)

  = 676000000 (  3  / 250  )³ / 3 ( 44 x 10⁻³)

  = 676000000 (  27  / 15625000 )  / 3 ( 44 x 10⁻³)

  = 146016  / 125 / 3 ( 44 x 1 / 1000  )

  = ( 146016  / 125 ) /  (3 ( 11 /  250 ))

  =  97344  / 11

F =  8849 N

4 0
3 years ago
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