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Nata [24]
3 years ago
11

Shear modulus is analogous to what material property that is determined in tensile testing? (a)- Percent reduction of area (b) Y

ield strength (c)- Elastic modulus (d)- Poisson's ratio
Engineering
1 answer:
yulyashka [42]3 years ago
8 0

Answer:

(c)- Elastic modulus

Explanation:

  We know that in tensile test we measure the properties of the material like yield strength,ultimate tensile strength ,Poisson ratio.

In tensile test

σ = ε E

Where σ is the stress

ε  is the strain.

E is the elastic modulus.

Now for shear tress

τ = Φ G

Where τ the shear stress

Φ  is the shear strain.

G  is the shear  modulus.

So we can say that Shear modulus is analogous to Elastic modulus.

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There are two types of cellular phones, handheld phones (H) that you carry and mobile phones (M) that are mounted in vehicles. P
nexus9112 [7]

Answer:

A) P(W) = 0.5

B) P(MF) = 0.3

C) P(H) = 0.6

Explanation:

We are told that there are two types of cellular phones which are handheld phones (H) that you carry and mobile phones (M) that are mounted in vehicles.

Also, Phone calls can be classified by the traveling speed of the user as fast (F) or slow (W).

Thus, the sample space is combination of types and classification we are given and it is written as;

S = {HF, HW, MF, MW}

A) Now, phones can either be fast(F) or slow(W). Thus, we can write;

P(F) + P(W) = 1

We are given P(F) = 0.5

Thus;

0.5 + P(W) = 1

P(W) = 1 - 0.5

P(W) = 0.5

B) Now, from the problem statement, a phone call can either be made with a handheld(H) or mobile(M). Thus the sample space partition is {H, M} and we can express as;

P(H ∩ F) + P(M ∩ F) = P(F)

We are given P[F] = 0.5 and P[HF] = 0.2.

P(H ∩ F) is same as P[HF]

Also, P(M ∩ F) is same as P(MF)

Thus;

0.2 + P(MF) = 0.5

P(MF) = 0.5 - 0.2

P(MF) = 0.3

C) Similarly, mobile Phone calls can either be fast or slow. It means the sample space partition is {F, W}

Thus;

P(M) = P(MW) + P(MF)

P(M) = 0.1 + 0.3

P(M) = 0.4

Now, since cellular phones can either be handheld(H) or Mobile(M), then we can say;

P(H) + P(M) = 1

P(H) + 0.4 = 1

P(H) = 1 - 0.4

P(H) = 0.6

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The bar DA is rigid and is originally held in the horizontal position when the weight W is supported from C. The wires are made
Rom4ik [11]

Answer:

W = 112 lb

Explanation:

Given:

- δb = 0.025 in

- E = 29000 ksi      (A-36)

- Area A_de = 0.002 in^2

Find:

Compute Weight W attached at C

Solution:

- Use proportion to determine δd:

                              δd/5 = δb/3

                              δd = (5/3) * 0.025

                              δd = 0.0417 in

- Compute εde i.e strain in DE:

                               εde = δd / Lde

                               εde = 0.0417 / 3*12

                               εde = 0.00116

- Compute stress in DE, σde:

                               σde = E*εde

                               σde = 29000*0.00116

                               σde = 33.56 ksi

- Compute the Force F_de:

                               F_de = σde *A_de

                               F_de = 33.56*0.002

                               F_de = 0.0672 kips

- Equilibrium conditions apply:

                               (M)_a = 0

                               3*W - 5*F_de = 0

                               W = (5/3)*F_de

                              W = (5/3)* 0.0672 = 112 lb

4 0
2 years ago
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