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3 years ago

12

New ventures that are based on strategic value, such as valuable technology, are attractive while those with low or no strategic

value are
less attractive.
Select one:
O a. False
O b. True

2 answers:

Ksju [112]3 years ago

8 0

The answer is false I believe

Alecsey [184]3 years ago

7 0

Answer:

false

Explanation:

i hope answer is right

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A ____ is either in the pressure reducer or in the downstream side of the system to ensure that the control air pressure does no

Alika [10]

Answer:

A relief valve is either in the pressure reducer or in the downstream side of the system to ensure that the control air pressure does not exceed about 30 psig.

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2 years ago

What form of joining uses heat to create coalescence of the materials?

Allushta [10]

The answer is Soldering

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3 years ago

In a tensile test on a steel specimen, true strain is 0.171 at a stress of 263.8 MPa. When true stress is 346.2 MPa, true strain

swat32

Answer:n=0.973

Explanation:

Given

When True strain $\left ( \epsilon _T_1\right )=0.171$

at $\sigma _1=263.8 MPa$

When True stress $\left ( \sigma _2\right )$ =346.2 MPa

true strain $\left ( \epsilon _T_2\right )$ =0.226

We know

$\sigma =k\epsilon ^n$

where $\sigma$ =True stress

$\epsilon$ =true strain

n=strain hardening exponent

k=constant

Substituting value

$263.8=k\left ( 0.171\right )^n------1$

$346.2=k\left ( 0.226\right )^n-----2$

Divide 1 & 2 to get

$\frac{346.2}{263.8}=\left ( \frac{0.226}{0.171}\right )^n$

$1.312=\left ( 1.3216\right )^n$

Taking Log both side

$ln\left ( 1.312\right )=nln\left ( 1.3216\right )$

n=0.973

6 0

3 years ago

An aluminum bar 125 mm long with a square cross section 16 mm on an edge is pulled in tension with a load of 66,700 N and experi

AfilCa [17]

Answer: the modulus of elasticity of the aluminum is 75740.37 MPa

Explanation:

Given that;

Length of Aluminum bar L = 125 mm

square cross section s = 16 mm

so area of cross section of the aluminum bar is;

A = s² = 16² = 256 mm²

Tensile load acting the bar p = 66,700 N

elongation produced Δ = 0.43

so

Δ = PL / AE

we substitute

0.43 = (66,700 × 125) / (256 × E)

0.43(256 × E) = (66,700 × 125)

110.08E = 8337500

E = 8337500 / 110.08

E = 75740.37 MPa

Therefore, the modulus of elasticity of the aluminum is 75740.37 MPa

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2 years ago

Water is contained in a rigid vessel of 5 m3 at a quality of 0.8 and a pressure of 1 MPa. If the pressure is reduced to 270.3 kP

professor190 [17]

Answer:

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Explanation:

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3 years ago