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Gala2k [10]
2 years ago
10

15- Vipsana's Gyros House sells gyros. The cost of ingredients (pita, meat, spices, etc.) to make a gyro is $2.00. Vipsana pays

her employee $60 per day. She also incurs a fixed cost of $120 per day. Calculate Vipsana's variable cost per day when she produces 50 gyros using two workers?
A) $100
B) $124.40
C) $220
D) $240
Engineering
1 answer:
sineoko [7]2 years ago
4 0
D is a great answer
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Can crushers help us recycle in a space efficient way which is good for saving the earth and for giving you more room in your ap
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Explanation:

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Which factor has the most significant impact in design?
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D) cost


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3 0
2 years ago
A hollow shaft of diameter ratio 3/8 (internal dia to outer dia) is to transmit 375 kW power at 100 rpm. The maximum torque bein
Schach [20]

Answer:

External diameter = 158.15 mm mm

Internal diameter = 59.31 mm

Explanation:

We are given;

Diameter ratio; d_i = ⅜d_o

Where d_i is internal diameter and d_o is external diameter

Power;P = 375 KW = 375000 W

Rotational speed;N = 100 rpm

Max torque is 20% greater than mean torque; T_max = 1.2T_avg

Shear stress;τ = 60 N/mm²

Length; L = 4m = 4000 mm

Angle of twist; θ = 2° = 2π/180 radians

Modulus of rigidity;G = 0.85 X 10^(5) N/mm²

Formula for the power transmitted by the shaft is;

P = 2πNT_avg/60

Plugging in the relevant values, we have ;

375000 = 2π × 100T_avg/60

T_avg = (375000 × 60)/(2π × 100) = 35809.862 N.m = 35809862 N.mm

Since T_max = 1.20T_avg

Thus, T_max = 1.20(35809862) = 42971834.4 N.mm

Checking for strength, we'll use;

τ = Tr/J_p

Or since r = d/2

It can be written as;

τ = T(d_o)/2J_p - - - (1)

Where T is T_max

But Polar moment of inertia of hollow shaft is;

J_p = [π(d_o)⁴ - π(d_i)⁴]/32

Now, we are told that d_i = ⅜d_o

Thus;

J_p = [π(d_o)⁴ - π(⅜d_o)⁴]/32

J_p = (π/32) × d_o⁴(1 - 3⁴/8⁴)

J_p = 0.0926 d_o⁴

Plugging this for J_p in eq 1,we have;

τ = T(d_o)/2(0.0926d_o⁴)

Making d_o the subject gives;

d_o³ = T/(2 × 0.0926τ)

Plugging in the relevant values to give;

d_o³ = 42971834.4/(2 × 0.0926 × 60)

d_o³ = 3867155.7235421166

d_o = ∛3867155.7235421166

d_o = 156.96 mm

Thus, d_i = ⅜ × 156.96 = 58.86 mm

Checking for stiffness, we'll use;

T/J_p = Gθ/L

Again T is T_max

Plugging in the relevant values, we have;

42971834.4/0.0926 d_o⁴ = (0.85 × 10^(5) × 2π/180)/4000

464058686.825054/d_o⁴ = 0.7417649321

d_o⁴ = 464058686.825054/0.7417649321

d_o⁴ = 625614216.5028806

d_o = ∜625614216.5028806

d_o = 158.15 mm

d_i = ⅜ × 158.15 = 59.31 mm

So we will pick the highest values.

Thus;

d_o = 158.15 mm

d_i = 59.31 mm

3 0
3 years ago
A single crystal of zinc is oriented for a tensile test such that its slip plane normal makes an angle of 65 degree with the ten
Crank

Answer:

a. 30°

b. 0.9MPa

Explanation:

The slip will occur along that direction for which the Schmid factor is maximum. The three possible slip directions are mentioned as 30°, 48°, 78°

The cosines for the possible λ values are given as

For 30°, cos 30 = 0.867

For 48°, cos 48 = 0.67

For 78°, cos 78 = 0.21

Among the three-calculated cosine values, the largest cos(λ) gives the favored slip direction

The maximum value of Schmid factor is 0.87. Thus, the most favored slip direction is 30° with the tensile axis.

The plastic deformation begins at a tensile stress of 2.5MPa. Also, the value of the angle between the slip plane normal and the tensile axis is mentioned as 65°

Thus, calculate the value of critical resolved shear stress for zinc:

From the expression for Schmid’s law:

τ = σ*cos(Φ)*cos(λ)

Substituting 2.5MPa for σ, 30° for λ and 65° for Φ

We obtain The critical resolved shear stress for zinc, τ = 0.9 MPa

4 0
2 years ago
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