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timama [110]
3 years ago
7

A bolt is tightened, subjecting its shank to a tensile stress of 80 kpsi and a torsional shear stress of 50 kpsi at a critical p

oint. All of the other stresses are zero. The material is a brittle material with Sut =80 kpsi and Suc =150 kpsi. Find the factor of safety at the critical point by the more conservative failure theory and determine if the bolt will fracture.

Engineering
1 answer:
Stolb23 [73]3 years ago
5 0

Answer:

factor of safety is 0.77

fracture of bolt will occur since principle stress exceeding бmax on the bolt.

Explanation:

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torisob [31]

Answer:

The correct option is;

c. Leaving the chuck key in the drill chuck

Explanation:

A Common safety issues with a drill press leaving the chuck key in the drill chuck

It is required that, before turning the drill press power on, ensure that chuck key is removed from the chuck. A self ejecting chuck key reduces the likelihood of the chuck key being accidentally left in the chuck.

It is also required to ensure that the switch is in the OFF position before turning plugging in the power cable

Be sure that the chuck key is removed from the chuck before turning on the power. Using a self-ejecting chuck key is a good way of insuring that the key is not left in the chuck accidentally. Also to avoid accidental starting, make sure the switch is in the OFF position before plugging in the cord. Always disconnect the drill from the power source when making repairs.

5 0
3 years ago
Question # 3
tino4ka555 [31]

Answer:

I think it's False.

Apologies if I am wrong.

4 0
3 years ago
Read 2 more answers
Shortly after the introduction of a new​ coin, newspapers published articles claiming the coin is biased. The stories were based
garik1379 [7]

Answer:

(a) 0.12924

(b) Taking into consideration significance level of 0.05 yet the value of p is greater than 0.05, it suggests that the coin is fair hence the coin can be used at the beginning of any sport event.

Explanation:

(a)

n=200 for fair coin getting head, p= 0.5

Expectation = np =200*0.5=100

Variance = np(1 - p) = 100(1-0.5)=100*0.5=50

Standard deviation, s = \sqrt {variance}=\sqrt {50}= 7.071068

Z value for 108, z =\frac {108-100}{7.071068}= 1.131371

P( x ≥108) = P( z >1.13)= 0.12924

(b)

Taking into consideration significance level of 0.05 yet the value of p is greater than 0.05, it suggests that the coin is fair hence the coin can be used at the beginning of any sport event.

3 0
3 years ago
If we have silicon at 300K with 10 microns of p-type doping of 4.48*10^18/cc and 10 microns of n-type doping at 1000 times less
liq [111]

Answer:

The resistance is 24.9 Ω

Explanation:

The resistivity is equal to:

R=\frac{1}{N_{o}*u*V } =\frac{1}{4.48x10^{15}*1500*106x10^{-19}  } =0.93ohm*cm

The area is:

A = 60 * 60 = 3600 um² = 0.36x10⁻⁴cm²

w=\sqrt{\frac{2E(V_{o}-V) }{p}(\frac{1}{N_{A} }+\frac{1}{N_{D} })

If NA is greater, then, the term 1/NA can be neglected, thus the equation:

w=\sqrt{\frac{2E(V_{o}-V) }{p}(\frac{1}{N_{D} })

Where

V = 0.44 V

E = 11.68*8.85x10¹⁴ f/cm

V_{o} =\frac{KT}{p} ln(\frac{N_{A}*N_{D}}{n_{i}^{2}  } , if n_{i}=1.5x10^{10}cm^{-3}  \\V_{o}=0.02585ln(\frac{4.48x10^{18}*4.48x10^{15}  }{(1.5x10^{10})^{2}  } )=0.83V

w=\sqrt{\frac{2*11.68*8.85x10^{-14}*(0.83-0.44) }{1.6x10^{-19}*4.48x10^{15}  } } =3.35x10^{-5} cm=0.335um

The length is:

L = 10 - 0.335 = 9.665 um

The resistance is:

Re=\frac{pL}{A} =\frac{0.93*9.665x10^{-4} }{0.36x10^{-4} } =24.9ohm

7 0
3 years ago
(a)Compute the electrical conductivity of a cylindrical silicon specimen 7.0 mm (0.28 in.) diameter and 57 mm (2.25 in.) in leng
igor_vitrenko [27]

Answer:

a) \sigma = 12.2 (Ω-m)^{-1}

b) Resistance = 121.4 Ω

Explanation:

given data:

diameter is 7.0 mm

length 57 mm

current I = 0.25 A

voltage v = 24 v

distance between the probes is 45 mm

electrical conductivity is given as

\sigma = \frac{I l}{V \pi r^2}

\sigma  = \frac{0.25 \times 45\times 10^{-3}}{24 \pi [\frac{7 \times 10^{-3}}{2}]^2}

\sigma = 12.2(Ω-m)^{-1}[/tex]

b)

Resistance = \frac{l}{\sigma A}

                  = \frac{l}{ \sigma \pi r^2}

= \frac{57  \times 10^{-3}}{12.2 \times \pi [\frac{7 \times 10^{-3}}{2}]^2}

Resistance = 121.4 Ω

8 0
3 years ago
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