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antiseptic1488 [7]
2 years ago
10

1. A copper block of volume 1 L is heat treated at 500ºC and now cooled in a 200-L oil bath initially at 20◦C. Assuming no heat

transfer with the surroundings, what is the final temperature?
Engineering
1 answer:
MaRussiya [10]2 years ago
7 0

Answer:

final temperature T = 24.84ºC

Explanation:

given data

copper volume = 1 L

temperature t1 = 500ºC

oil volume = 200 L

temperature t2 = 20ºC

solution

Density of copper \rho cu = 8940 Kg/m³

Density of light oil  \rho oil = 889 Kg/m³

Specific heat capacity of copper Cv = 0.384  KJ/Kg.K

Specific heat capacity of light oil Cv = 1.880 KJ/kg.K

so fist we get here mass of oil and copper that is

mass = density × volume   ................1

mass of copper = 8940 × 1 ×  10^{-3}  = 8.94 kg  

mass of oil = 889 × 200 × 10^{-3}  =  177.8 kg  

so we apply here now energy balance equation that is

M(cu)\times Cv \times (T-T1)_{cu} + M(oil) \times Cv\times (T-T2)_{oil}  = 0

put here value and we get T2

8.94\times 0.384 \times (T-500) + 177.8 \times 1.890\times (T-20)  = 0

solve it we get

T = 24.84ºC

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The fracture toughness of a stainless steel is 137 MPa*m12. What is the tensile impact load sustainable before fracture that a r
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Answer:

7.7 kN

Explanation:

The capacity of a material having a crack to withstand fracture is referred to as fracture toughness.

It can be expressed by using the formula:

K = \sigma Y \sqrt{\pi a}

where;

fracture toughness K = 137 MPam^{1/2}

geometry factor Y = 1

applied stress \sigma = ???

crack length a = 2mm = 0.002

∴

137 =\sigma \times 1  \sqrt{ \pi \times 0.002 }

137 =\sigma \times 0.07926

\dfrac{137}{0.07926} =\sigma

\sigma = 1728.489 MPa

Now, the tensile impact obtained is:

\sigma = \dfrac{P}{A}

P = A × σ

P = 1728.289 × 4.5

P = 7777.30 N

P = 7.7 kN

7 0
3 years ago
 what can be done to prevent bridges from collapsing? ( give at least two examples)
ryzh [129]

Explanation:

the owner of the bridge and some workers

4 0
2 years ago
Reception of signals from a radio facility, located off the airway being flown, may be inadequate at the designated mea to ident
lakkis [162]

The altitude ensures acceptable navigational signal coverage only within 22 NM of a VOR.

<h3>What is altitude?</h3>

Altitude or height exists as distance measurement, usually in the vertical or "up" approach, between a reference datum and a point or object. The exact meaning and reference datum change according to the context.

The MOCA exists in the lower published altitude in effect between fixes on VOR airways, off-airway routes, or route segments that satisfy obstacle support conditions for the whole route segment. This altitude also ensures acceptable navigational signal coverage only within 22 NM of a VOR.

The altitude ensures acceptable navigational signal coverage only within 22 NM of a VOR.

Therefore, the correct answer is 22 NM of a VOR.

To learn more about altitudes refer to:

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3 0
1 year ago
Assume a program requires the execution of 50 x 106 FP instructions, 110 x 106 INT instructions, 80 x 106 L/S instructions, and
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Answer:

Part A:

1.3568*10^{-5}=\frac{5300* New\  CPI_1+11660*1+8480*4+1696*2}{2*10^9\ Hz} \\ New\ CPI_1=-4.12

CPI cannot be negative so it is not possible to for program to run two times faster.

Part B:

1.3568*10^{-5}=\frac{5300*1+11660*1+8480*New\ CPI_3+1696*2}{2*10^9\ Hz} \\ New\ CPI_3=0.8

CPI reduced by 1-\frac{0.8}{4} = 0.80=80%

Part C:

New Execution Time=\frac{5300*0.6+11660*0.6+8480*2.8+1696*1.4}{2*10^9\ Hz}=1.81472*10^{-5}\ s

Increase in speed=1-\frac{1.81472*10^{-5}}{2.7136*10^{-5}} =0.33125= 33.125\%

Explanation:

FP Instructions=50*106=5300

INT  Instructions=110*106=11660

L/S  Instructions=80*106=8480

Branch  Instructions=16*106=1696

Calculating Execution Time:

Execution Time=\frac{\sum^4_{i=1} Number\ of\ Instruction*\ CPI_{i}}{Clock\ Rate}

Execution Time=\frac{5300*1+11660*1+8480*4+1696*2}{2*10^9\ Hz}

Execution Time=2.7136*10^{-5}\ s

Part A:

For Program to run two times faster,Execution Time (Calculated above) is reduced to half.

New Execution Time=\frac{2.7136*10^{-5}}{2}=1.3568*10^{-5}\ s

1.3568*10^{-5}=\frac{5300* New\  CPI_1+11660*1+8480*4+1696*2}{2*10^9\ Hz} \\ New\ CPI_1=-4.12

CPI cannot be negative so it is not possible to for program to run two times faster.

Part B:

For Program to run two times faster,Execution Time (Calculated above) is reduced to half.

New Execution Time=\frac{2.7136*10^{-5}}{2}=1.3568*10^{-5}\ s

1.3568*10^{-5}=\frac{5300*1+11660*1+8480*New\ CPI_3+1696*2}{2*10^9\ Hz} \\ New\ CPI_3=0.8

CPI reduced by 1-\frac{0.8}{4} = 0.80=80%

Part C:

New\ CPI_1=0.6*Old\ CPI_1=0.6*1=0.6\\New\ CPI_2=0.6*Old\ CPI_2=0.6*1=0.6\\New\ CPI_3=0.7*Old\ CPI_3=0.7*4=2.8\\New\ CPI_4=0.7*Old\ CPI_4=0.7*2=1.4

New Execution Time=\frac{\sum^4_{i=1} Number\ of\ Instruction*\ CPI_{i}}{Clock\ Rate}

New Execution Time=\frac{5300*0.6+11660*0.6+8480*2.8+1696*1.4}{2*10^9\ Hz}=1.81472*10^{-5}\ s

Increase in speed=1-\frac{1.81472*10^{-5}}{2.7136*10^{-5}} =0.33125= 33.125\%

8 0
3 years ago
Without motorcycle riders are at risk of severe injury in a crash ?
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Answer:

The correct option is;

A. proper protection

Explanation:

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Injuries sustained from motorcycle accidents are several times more severe than injuries sustained by occupants of a car that is fully protected by the metallic panel in the same and even more serious accident scenarios

Hence, motorcycle riders require adequate protection by putting on available motorcyclist safety gear

Therefore, to reduce the risk of severe injury n a crash, motorcycle riders require proper protection.

5 0
3 years ago
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