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velikii [3]
3 years ago
11

ferry boat steamed along at 8 mph through calm seas, passenger casey exercised by walking the perimeter of the rectangular deck,

at a steady 4 mph. Discuss the variations in casey's speed relative to the water.
Engineering
1 answer:
olga_2 [115]3 years ago
6 0
Tell me why i got this question got it right and now won’t remember but i’ll get back at you when i remember
You might be interested in
A steady tensile load of 5.00kN is applied to a square bar, 12mm on a side and having a length of 1.65m. compute the stress in t
Shtirlitz [24]

Answer:

The stress in the bar is 34.72 MPa.

The design factor (DF) for each case is:

A) DF=0.17

B) DF=0.09

C) DF=0.125

D) DF=0.12

E) DF=0.039

F) DF=1.26

G) DF=5.5

Explanation:

The design factor is the relation between design stress and failure stress. In the case of ductile materials like metals, the failure stress considered is the yield stress. In the case of plastics or ceramics, the failure stress considered is the breaking stress (ultimate stress). If the design factor is less than 1, the structure or bar will endure the applied stress. By the opposite side, when the DF is higher than 1, the structure will collapse or the bar will break.

we will calculate the design stress in this case:

\displaystyle \sigma_{dis}=\frac{T_l}{Sup}=\frac{5.00KN}{(12\cdot10^{-3}m)^2}=34.72MPa

The design factor for metals is:

DF=\displaystyle \frac{\sigma_{dis}}{\sigma_{f}}=\frac{\sigma_{dis}}{\sigma_{y}}

The design factor for plastic and ceramics is:

DF=\displaystyle \frac{\sigma_{dis}}{\sigma_{f}}=\frac{\sigma_{dis}}{\sigma_{u}}

We now need to know the yield stress or the ultimate stress for each material. We use the AISI and ASTM charts for steels, materials charts for non-ferrous materials and plastics safety charts for the plastic materials.

For these cases:

A) The yield stress of AISI 120 hot-rolled steel (actually is AISI 1020) is 205 MPa, therefore:

DF=\displaystyle\frac{34.72MPa}{205MPa}=0.17

B) The yield stress of AISI 8650 OQT 1000 steel is 385 MPa, therefore:

DF=\displaystyle\frac{34.72MPa}{385MPa}=0.09

C) The yield stress of ductile iron A536-84 (60-40-18) is 40Kpsi, this is 275.8 MPa, therefore:

DF=\displaystyle\frac{34.72MPa}{275.8MPa}=0.125

D) The yield stress of aluminum allot 6061-T6 is 290 MPa, therefore:

DF=\displaystyle\frac{34.72MPa}{290MPa}=0.12

E) The yield stress of titanium alloy Ti-6Al-4V annealed (certified by manufacturers) is 880 MPa, therefore:

DF=\displaystyle\frac{34.72MPa}{880MPa}=0.039

F) The ultimate stress of rigid PVC plastic (certified by PVC Pipe Association) is 4Kpsi or 27.58 MPa, therefore:

DF=\displaystyle\frac{34.72MPa}{27.58 MPa}=1.26

In this case, the bar will break.

F) You have to consider that phenolic plastics are used as matrix in composite materials and seldom are used alone with no reinforcement. In this question is not explained if this material is reinforced or not, therefore I will use the ultimate stress of most pure phenolic plastics, in this case, 6.31 MPa:

DF=\displaystyle\frac{34.72MPa}{6.31 MPa}=5.5

This material will break.

3 0
3 years ago
Which of the following is not caused by alcohol?
Brrunno [24]

Answer:

Inhibited comprehension

4 0
3 years ago
Read 2 more answers
2. Given that a quality-control inspection can ensure that a structural ceramic part will have no flaws greater than 100 µm (100
MA_775_DIABLO [31]

Answer:

SiC=169.26 Mpa

Partially stabilized zirconia=507.77 Mpa

Explanation:

<u>SiC </u>

Stress intensity, K is given by

K=\sigma Y\sqrt{\pi a} hence making \sigma the subject where \sigma is applied stress, Y is shape factor, a is crack length

\sigma=\frac {K}{Y\sqrt{\pi a}} substituting the given figures and assuming shape factor, Y of 1

\sigma=\frac {3}{1\sqrt{\pi 100*10^{-6}}}= 169.2569\approx 169.26 Mpa

<u>Stabilized zirconia </u>

Stress intensity, K is given by

K=\sigma Y\sqrt{\pi a} hence making \sigma the subject where \sigma is applied stress, Y is shape factor, a is crack length

\sigma=\frac {K}{Y\sqrt{\pi a}} substituting the given figures and assuming shape factor, Y of 1

\sigma=\frac {9}{1\sqrt{\pi 100*10^{-6}}}= 507.7706\approx 507.77 Mpa

7 0
3 years ago
Drilling is an example of what type of manufacturing process?
k0ka [10]

Answer:

  Machining

Explanation:

Drilling is a machining operation. It involves using a cutting tool to alter the shape.

5 0
3 years ago
A cartridge electrical heater is shaped as a cylinder of length L=200 mm and outer diameter D=20 mm. Under normal operating cond
Setler79 [48]

Answer:

When water is surrounding T_s = 34.17 degree C

When air surrounding T_S = 1434.7 degree C

from above calculation we can conclude that air is less effective than water  as heat transfer agent

Explanation:

Given data:

length  = 300 mm

Outer diameter  = 30 mm

Dissipated energy = 2 kw = 2000 w

Heat transfer coefficient IN WATER = 5000 W/m^2 K

Heat transfer coefficient in air  = 50 W/m^2 K

we know that q_{convection} =  P

From newton law of coding we have

q_{convection} =  hA(T_s -  T_{\infity})

T_s is surface temp.

T - temperature at surrounding

P = hA(T_s -  T_{\infity})[tex]\frac{P}{\pi hDL} =  (T_s -  T_{\infity})

solving for[/tex] T_s [/tex] w have

T_s = T_{\infty} + \frac{P}{\pi hDL}

T_s = 20 + \frac{2000}{\pi 5000\times 0.03\times 0.3}

T_s = 34.17 degree C

When air is surrounding we have

T_s = T_{\infty} + \frac{P}{\pi hDL}

T_s = 20 + \frac{2000}{\pi 2000\times 0.03\times 0.3}

T_s = 1434.7 degree C

from above calculation we can conclude that air is less effective than water  as heat transfer agent

5 0
4 years ago
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