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natali 33 [55]
3 years ago
5

I. Draw the velocity diagram for the instant shown and determine the velocity of

Engineering
1 answer:
trapecia [35]3 years ago
3 0

Answer:

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A rod that was originally 100-cm-long experiences a strain of 82%. What is the new length of the rod?
Ierofanga [76]

Answer: (b)

Explanation:

Given

Original length of the rod is L=100\ cm

Strain experienced is \epsilon=82\%=0.82

Strain is the ratio of the change in length to the original length

\Rightarrow \epsilon =\dfrac{\Delta L}{L}\\\\\Rightarrow 0.82=\dfrac{\Delta L}{100}\\\\\Rightarrow \Delta L=82\ cm

Therefore, new length is given by (Considering the load is tensile in nature)

\Rightarrow L'=\Delta L+L\\\Rightarrow L'=82+100=182\ cm

Thus, option (b) is correct.

8 0
3 years ago
Mihuv8tr5qwertgyhjzxcvbnfr5y7nnbvcxzwertgyhujio vv solve the riddle
Inessa [10]

Answer:

v1QAZ3EDCRFV5TGB6YHNUJMIK,9OL0K9MIJNUHB7YGVTFCRDXESZWAq

Explanation:

qaAQzwsxedcnujmik,ol mkjuhtfcrxdZSWAQWSEDRFTGYHUJIKO,LP.; ,LMKJNUHTGDXESZWaEDRFTGHJKL,MNBVFDSWQAAWERTYUIOP;L,MNHGFDEWwertyuikolp;[l.,mnbvfre345678990098765434rtyhnbhju8765rtghjui875rfghji8765rfghju7654redfghu7643erfghji987yhjko987y

4 0
3 years ago
Read 2 more answers
The performance of a heat pump degrades (i.e., its COP decreases) as the temperature of the heat source decreases. This makes us
vfiekz [6]

Answer:

COP_max = 18.69

Explanation:

We are given;

Heated space temperature; T_H = 26°C = 273K + 26 = 299K

Temperature at which heat is extracted; T_L = 10°C = 273 + 10 = 283K

Now the Coefficient of performance (COP) of a heat pump will be a maximum when the heat pump operates in a reversible manner. The COP of a reversible heat pump depends on the temperature limits in the cycle only and is determined by the formula;

COP_max = 1/(1 - (T_L/T_H))

Thus,

COP_max = 1/(1 - (283/299))

COP_max = 1/(1 - 0.9465)

COP_max = 1/0.0535 = 18.69

7 0
3 years ago
. A normal-weight concrete has an average compressive strength of 20 MPa. What is the estimated flexure strength
bulgar [2K]

Answer:

2.77mpa

Explanation:

compressive strength = 20 MPa. We are to find the estimated flexure strength

We calculate the estimated flexural strength R as

R = 0.62√fc

Where fc is the compressive strength and it is in Mpa

When we substitute 20 for gc

Flexure strength is

0.62x√20

= 0.62x4.472

= 2.77Mpa

The estimated flexure strength is therefore 2.77Mpa

4 0
2 years ago
ASAE 1060 Steel wire (1 mm diameter) is coated with copper to form a composite with a diameter of 2mm. Use the following propert
k0ka [10]

Answer:

a) E_{m} = 133.75 Gpa

b) Fnet = 560 N

c) thermal expansion of the composite material = 14.31 10^{-6 } / °C

Explanation:

Solution:

a) Elastic Modulus of the composite:

Area of steel wire = \frac{\pi }{4} x (0.001^{2}) = 0.8 x 10^{-6} m^{2}

Area of Copper wire = \frac{\pi }{4} x (0.002^{2}) - 0.8 x 10^{-6} m^{2}

Area of Copper wire = 2.4 x 10^{-6} m^{2}

Young's Modulus of Composite mixture:

E_{m} = F_{st}E_{st} +  F_{Cu}E_{Cu}     Equation 1

here,

F_{st} = Stress in Steel

F_{Cu} = Stress in Copper.

We know that,

F = P/A

F is inversely proportional to Area, so if area is large, stress will less and vice versa. So, Take

Ratio for area of steel = \frac{0.8. 10^{-6} }{(0.8 + 2.4) .10^{-6} }

Ratio for area of steel = \frac{0.8}{3.2 } = 0.25

Similarly, for Copper,

Ratio for area of copper = \frac{2.4. 10^{-6} }{(0.8 + 2.4) .10^{-6} }

Ratio for area of copper = \frac{2.4 }{3.2} = 0.75

Put these values in equation 1:

E_{m} = F_{st}E_{st} +  F_{Cu}E_{Cu}    

E_{m} = (0.25) E_{st} +  (0.75)E_{Cu}

We are given that,

  E_{st} = 205 Gpa

E_{Cu}  = 110 Gpa

So,

E_{m} = (0.25) (205 Gpa) +  (0.75) (110 GPa)

E_{m} = 51.25GPa + 82.5 Gpa

Hence, the Elastic Modulus of the composite will be:

E_{m} = 133.75 Gpa

b) maximum force:

Fnet = Fst + Fcu

We know that F = (Yield Stress x Area)

F = fst x Ast + fcu x Acu

And we are given that,

Yield stress of Steel = 280 Mpa

Yield stress of Copper = 140 Mpa

And,

Ast = 0.8 x 10^{-6} m^{2}

Acu = 2.4 x 10^{-6} m^{2}

Just plugging in the values, we get:

F = (280 Mpa) (0.8 x 10^{-6} m^{2}) + (140 Mpa) (2.4 x 10^{-6} m^{2})

F = 224 + 336

Fnet = 560 N    ( because Mpa = 10^{6} N/m^{2})

So, it means the composite will carry the maximum force of 560N

c) Coefficient of Thermal Expansion:

Strain on both material is same upon loading so,

(ΔL/L)st = (ΔL/L)cu

by thermal expansion equation:

(\alpha .ΔT  + \frac{F}{A}. \frac{1}{Est}) = \alpha .ΔT  + \frac{F}{A}. \frac{1}{Ecu})

Where \alpha = Coefficient of Thermal expansion

Here, fst = -fcu = F

and ΔT = 1°

So,

Plugging in the values, we get.

( 10 x 10^{-6} x (1) + \frac{F}{0.8.10^{-6} } . \frac{1}{205 . 10^{9} } ) = ( 17 x 10^{-6} x (1) + \frac{-F}{2.4.10^{-6} } . \frac{1}{110 . 10^{9} } )

Solving for F, we get:

F = 0.71 N

Here,

fst = F = 0.71 N (Tension on Heating)

fcu = -F = 0.71 N ( Compression on Heating )

So, the combined thermal expansion of the composite material will be:

(ΔL/L)cu = ( 17 x 10^{-6} x (1°) + \frac{-0.71}{2.4.10^{-6} } . \frac{1}{110 . 10^{9} } )

(ΔL/L)cu = ( 17 x 10^{-6} x (1°) - 2.69 x 10^{-6}

combined thermal expansion of the composite material = 14.31 10^{-6 } / °C

4 0
3 years ago
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