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SashulF [63]
3 years ago
7

The Stefan-Boltzmann law can be employed to estimate the rate of radiation of energy H from a surface, as in

Engineering
1 answer:
Mazyrski [523]3 years ago
5 0

Explanation:

A.

H = Aeσ^4

Using the stefan Boltzmann law

When we differentiate

dH/dT = 4AeσT³

dH/dT = 4(0.15)(0.9)(5.67)(10^-8)(650)³

= 8.4085

Exact error = 8.4085x20

= 168.17

H(650) = 0.15(0.9)(5.67)(10^-8)(650)⁴

= 1366.376watts

B.

Verifying values

H(T+ΔT) = 0.15(0.9)(5.67)(10)^-8(670)⁴

= 1542.468

H(T+ΔT) = 0.15(0.9)(5.67)(10^-8)(630)⁴

= 1205.8104

Error = 1542.468-1205.8104/2

= 168.329

ΔT = 40

H(T+ΔT) = 0.15(0.9)(5.67)(10)^-8(690)⁴

= 1735.05

H(T-ΔT) = 0.15(0.9)(5.67)(10^-8)(610)⁴

= 1735.05-1059.83/2

= 675.22/2

= 337.61

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Your class is using engineering principles to improve the design of football helmets to prevent brain injury. your teacher divid
nika2105 [10]

Answer:

a. students participate and apply engineering principles

Explanation:

We can rule out option B because the question states students are working in groups.

We can rule out option C because following directions, while appropriate, is not a benefit from STEM education specifically.

We can rule out option D because students are building off of a current design, not creating a new product.

5 0
1 year ago
Mention verious medium level and higher level human resources related to engineering​
aliya0001 [1]

Answer:

Engineering aims to allow technical parts, structures and/or systems, such as those of a machine, to fulfill their function. To this end, occurring and/or desired processes (the operation thereof) are investigated and elaborated. Appropriate parts are designed individually or in a team, occurring stresses are calculated and the parts to be produced are modeled and/or specified on technical drawings. In some fields, maintenance is a standard part of the work to be performed. For a large part of engineering, standards and agreements have been drawn up with a view to the desired safety and practical applicability.

8 0
2 years ago
For the system in problem 4, suppose a main memory access requires 30ns, the page fault rate is .01%, it costs 12ms to access a
raketka [301]

Answer:

a. 7.75

b. 24.4

Explanation:

The Operating system uses virtual memory and page tables maps these virtual address to physical address. TLB works as a cache for such mapping.

program >>> TLB >>> cache >>> Ram

A program search for a page in TLB, if it doesn't find that page it's a TLB miss and then further looks for the page in cache.

If the page is not in cache then it's a cache miss and further looks for the page in RAM.

If the page is not in RAM, then it's a page fault and program look for the data in secondary storage.

So, typical flow would be

Page Requested >> TLB miss >> cache miss >>main memory>> page fault >> looks in secondary memory.

Here,

Main memory access time= 30 ns

Page fault rate=.01%

page fault service time= 12ns

TLB access time=7 ns

TLB hit rate= .95%

TLB miss rate =1-.95=.05%

cache access time = 15 ns

cache miss rate= .3%

cache hit rate = 1-.3=.97%

So,

a) TLB hit time= TLB access time = 7 ns

cache hit time = TLB hit rate * TLB access time + TLB miss rate * ( TLB access time + cache hit time)

= .95 * 7 + .05 * (7+15)

= 7.75 ns

b) EAT for TLB hit= 7ns

Total EAT = TLB hit rate *( TLB access time + Cache hit rate * cache access time + cache miss rate * (cache + main memory access time))+ TLB miss rate ( TLB access time + main memory access time + cache hit rate * cache access time + cache miss rate ( cache + main memory access time))

= .95 *( 7 + (.97*15) + .03(15+30))+ .05*(7+30+(.97*15) + .03 ( 15 + 30))=24.4 ns

8 0
2 years ago
For a bolted assembly with eight bolts, the stiffness of each bolt is kb = 1.0 MN/mm and the stiffness of the members is km = 2.
rjkz [21]

Answer:

a) 0.978

b) 0.9191

c) 1.056

d) 0.849

Explanation:

Given data :

Stiffness of each bolt = 1.0 MN/mm

Stiffness of the members = 2.6 MN/mm per bolt

Bolts are preloaded to 75% of proof strength

The bolts are M6 × 1 class 5.8 with rolled threads

Pmax =60 kN,  Pmin = 20kN

<u>a) Determine the yielding factor of safety</u>

n_{p} = \frac{S_{p}A_{t}  }{CP_{max}+ F_{i}  }  ------ ( 1 )

Sp = 380 MPa,   At = 20.1 mm^2,   C = 0.277,  Pmax = 7500 N,  Fi = 5728.5 N

Input the given values into the equation above

equation 1 becomes ( np ) = \frac{380*20.1}{0.277*7500*5728.5} = 0.978

note : values above are derived values whose solution are not basically part of the required solution hence they are not included

<u>b) Determine the overload factor of safety</u>

n_{L} =  \frac{S_{p}A_{t}-F_{i}   }{C(P_{max} )}  ------- ( 2 )

Sp =  380 MPa,   At =  20.1 mm^2, C = 0.277,  Pmax = 7500 N,  Fi = 5728.5 N

input values into equation 2 above

hence : n_{L} = 0.9191n_{L}  = 0.9191

<u>C)  Determine the factor of safety based on joint separation</u>

n_{0} = \frac{F_{i} }{P_{max}(1 - C ) }

Fi =  5728.5 N,  Pmax = 7500 N,  C = 0.277,

input values into equation above

Hence n_{0} = 1.056

<u>D)  Determine the fatigue factor of safety using the Goodman criterion.</u>

nf = 0.849

attached below is the detailed solution .

4 0
2 years ago
A commuter train traveling at 50 mi/h is 3 mi from a station. The train then decelerates so that its speed is 15 mi/h when it is
jonny [76]

Answer:

a) t = 277.477\,s\,(4.625\min), b) v_{f} = 0\,\frac{mi}{h}, c) a = -0.128\,\frac{ft}{s^{2}}

Explanation:

a) The deceleration experimented by the commuter train in the first 2.5 miles is:

a=\frac{[(15\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,mi} )\cdot (\frac{1\,h}{3600\,s} )]^{2}-[(50\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,mi} )\cdot (\frac{1\,h}{3600\,s} )]^{2}}{2\cdot (2.5\,mi)\cdot (\frac{5280\,ft}{1\,mi} )}

a = -0.185\,\frac{ft}{s^{2}}

The time required to travel is:

t = \frac{(15\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,fi} )\cdot(\frac{1\,h}{3600\,s} )-(50\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,fi} )\cdot(\frac{1\,h}{3600\,s} )}{-0.185\,\frac{ft}{s^{2}} }

t = 277.477\,s\,(4.625\min)

b) The commuter train must stop when it reaches the station to receive passengers. Hence, speed of train must be v_{f} = 0\,\frac{mi}{h}.

c) The final constant deceleration is:

a = \frac{(0\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,mi} )\cdot(\frac{1\,h}{3600\,s} )-(15\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,mi} )\cdot(\frac{1\,h}{3600\,s} )}{(2.875\,min)\cdot (\frac{60\,s}{1\,min} )}

a = -0.128\,\frac{ft}{s^{2}}

7 0
3 years ago
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