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3 years ago

Kjchgyew More pts for my peeps

Engineering

2 answers:

liq [111]3 years ago

8 0

Ty and have a nice day and a nice holidays

aivan3 [116]3 years ago

7 0

Thank you very much :).

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A three-point bending test is performed on a glass specimen having a rectangular cross section of height d = 5.4 mm (0.21 in.) a

Fudgin [204]

Answer:

5.21e-2mm

Explanation:

Please see attachment

8 0

4 years ago

4. Which 2D shape on the left would be used to make the 3D shape on the right? (1 pt.)

dexar [7]

it would be a bc its a sqare?well 3d is like you can say a cube 2d is like flat

Explanation:

7 0

2 years ago

When calculating gear ratios, which formula is different from the others?

BartSMP [9]

Answer:

The number of threads in a worm is the number of teeth in a worm. The speed transmission ratio of a worm and worm gear set is obtained by dividing the number of teeth of the worm gear by the number of threads of the worm.

6 0

3 years ago

A cylindrical specimen of some metal alloy having an elastic modulus of 108 GPa and an original cross-sectional diameter of 4.5

IgorC [24]

Answer:

The maximum length of the specimen before the deformation was 358 mm or 0.358 m.

Explanation:

The specific deformation ε for the material is:

$\epsilon = \deltaL /L$ (1)

Where δL and L represent the elongation and initial length respectively. From the HOOK's law we also now that for a linear deformation, the deformation and the normal stress applied relation can be written as:

$\sigma = E/ \epsilon$ (2)

Where E represents the elasticity modulus. By combining equations (1) and (2) in the following form:

$L= \delta L/ \epsilon$

$\epsilon =\sigma /E$

So by calculating ε then will be possible to find L. The normal stress σ is computing with the applied force F and the cross-sectional area A:

$\sigma=F/A$

$\sigma=\frac{F} {\pi*d^2/4}$

$\sigma=\frac{2170 N}{\pi*4.5 mm^2/4}$

$\sigma= 136000000 Pa= 136 Mpa$

Then de specific defotmation:

$\epsilon =136 MPa / 108 GPa = 1.26*10^{-3}$

Finally the maximum specimen lenght for a elongation 0f 0.45 mm is:

$L= 0.45 mm/ 1.26*10^{-3} = 358 mm = 0.358 m$

4 0

3 years ago

A man can swim at 4 ft/s in still water. He wishes to cross tje 40-ft wide river to point B, 30 ft downstream. If the river flow

iVinArrow [24]

Answer:

$v \approx 4.472\,\frac{ft}{s}$ , $t = 10\,s$

Explanation:

Since man and river report constant speeds and velocities are mutually perpendicullar, the absolute speed of the man is calculated by the Pythagorean Theorem:

$v = \sqrt{(4\,\frac{ft}{s} )^{2}+(2\,\frac{ft}{s} )^{2}}$

$v \approx 4.472\,\frac{ft}{s}$

The required time to make the crossing is:

$t = \frac{40\,ft}{4\,\frac{ft}{s} }$

$t = 10\,s$

6 0

4 years ago