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Dafna1 [17]
3 years ago
9

19. What part of the engine is shown at point "A" in the above figure?

Engineering
2 answers:
stepan [7]3 years ago
6 0

Answer:

C. Four bolt main

Explanation:

I think.. :)

erik [133]3 years ago
4 0
The answer is c I took the quiz my self
You might be interested in
A flat plate 1.5 m long and 1.0 m wide is towed in water at 20 o C in the direction of its length at a speed of 15 cm/s. Determi
beks73 [17]

Answer:

15.8

0.0944

Explanation:

L = 1.5

B = 1.0

Speed of water = 15cm

Temperature = 20⁰C

At 20⁰C

Specific weight = 9790

Kinematic viscosity v = 1.00x10^-4m²/s

Dynamic viscosity u = 1.00x10^-3

Density p = 998kg/m²

Reynolds number

= 0.15x1.5/1.00x10^-4

= 225000

S = 5

5x1.5/225000^1/2

= 0.0158

= 15.8mm

Resistance on one side of plate

F = 0.664x1x1.0x10^-3x0.15x225000^1/2

= 0.04724N

Total resistance

= 2N

= 2x0.04724

= 0.0944N

3 0
2 years ago
A large plate is fabricated from a steel alloy that has a plane strain fracture toughness of 55 MPa √m (50 ksi √in.). If, during
astra-53 [7]

Answer:

0.024 m = 24.07 mm

Explanation:

1) Notation

\sigma_c = tensile stress = 200 Mpa

K = plane strain fracture toughness= 55 Mpa\sqrt{m}

\lambda= length of a surface crack (Variable of interest)

2) Definition and Formulas

The Tensile strength is the ability of a material to withstand a pulling force. It is customarily measured in units (F/A), like the pressure. Is an important concept in engineering, especially in the fields of materials and structural engineering.

By definition we have the following formula for the tensile stress:

\sigma_c=\frac{K}{Y\sqrt{\pi\lambda}}   (1)

We are interested on the minimum length of a surface that will lead to a fracture, so we need to solve for \lambda

Multiplying both sides of equation (1) by Y\sqrt{\pi\lambda}

\sigma_c Y\sqrt{\pi\lambda}=K   (2)

Sequaring both sides of equation (2):

(\sigma_c Y\sqrt{\pi\lambda})^2=(K)^2  

\sigma^2_c Y^2 \pi\lambda=K^2   (3)

Dividing both sides by \sigma^2_c Y^2 \pi we got:

\lambda=\frac{1}{\pi}[\frac{K}{Y\sigma_c}]^2   (4)

Replacing the values into equation (4) we got:

\lambda=\frac{1}{\pi}[\frac{55 Mpa\sqrt{m}}{1.0(200Mpa)}]^2 =0.02407m

3) Final solution

So the minimum length of a surface crack that will lead to fracture, would be 24.07 mm or more.

7 0
3 years ago
*3–32. The rubber block is subjected to an elongation of 0.03 in. along the x axis, and its vertical faces are given a tilt so t
riadik2000 [5.3K]

Rubber block is not shown. I have attached an image of it.

Answer:

A) ε_x = 0.0075

B) ε_y = 0.00375

C) γ_xy = 0.0122 rad

Explanation:

We are given;

δ = 0.03 in

L = 4 in

ν_r = 0.5

θ = 89.3° = 89.3π/180 rad

Let's calculate ε_x in the direction of axis x

Thus, ε_x = δ/L = 0.03/4 = 0.0075

Let's calculate ε_y in the direction of axis y;

ε_y = v•ε_x = 0.5 x 0.0075 = 0.00375

Now, shear strain is angle between π/2 rad surfaces at that point.

Thus,

γ_xy =  π/2 - θ = π/2 - 89.3π/180

γ_xy = π(0.003889) = 0.0122 rad

3 0
2 years ago
(3) Calculate the heat flux through a sheet of brass 7.5 mm (0.30 in.) thick if the temperatures at the two faces are 150°Cand 5
bezimeni [28]

Answer:

a.) 1.453MW/m2,  b.)  2,477,933.33 BTU/hr  c.) 22,733.33 BTU/hr  d.) 1,238,966.67 BTU/hr

Explanation:

Heat flux is the rate at which thermal (heat) energy is transferred per unit surface area. It is measured in W/m2

Heat transfer(loss or gain) is unit of energy per unit time. It is measured in W or BTU/hr

1W = 3.41 BTU/hr

Given parameters:

thickness, t = 7.5mm = 7.5/1000 = 0.0075m

Temperatures 150 C = 150 + 273 = 423 K

                        50 C = 50 + 273 = 323 K

Temperature difference, T = 423 - 323 = 100 K

We are assuming steady heat flow;

a.) Heat flux, Q" = kT/t

K= thermal conductivity of the material

The thermal conductivity of brass, k = 109.0 W/m.K

Heat flux, Q" = \frac{109 * 100}{0.0075} = 1,453,333.33 W/m^{2} \\ Heat flux, Q" = 1.453MW/m^{2} \\

b.) Area of sheet, A = 0.5m2

Heat loss, Q = kAT/t

Heat loss, Q = \frac{109*0.5*100}{0.0075} = 726,666.667W

Heat loss, Q = 726,666.667 * 3.41 = 2,477,933.33 BTU/hr

c.) Material is now given as soda lime glass.

Thermal conductivity of soda lime glass, k is approximately 1W/m.K

Heat loss, Q=\frac{1*0.5*100}{0.0075} = 6,666.67W

Heat loss, Q = 6,666.67 * 3.41 = 22,733.33 BTU/hr

d.) Thickness, t is given as 15mm = 15/1000 = 0.015m

Heat loss, Q=\frac{109*0.5*100}{0.015} =363,333.33W

Heat loss, Q = 363,333.33 * 3.41 = 1,238,966.67 BTU/hr

5 0
2 years ago
1. When the pneumatic device was demonstrated, explain something that was different than what you expected.
Vera_Pavlovna [14]

Answer:

1. Effect of air pressure

2. air- powered wheel chair

3. Pneumatic valves

Explanation:

1. In any pneumatic device, the mipact of air pressure to produce the moving effect on an heavy object is unexpected.

2. pneumatice demultiplexer when air in comprressed tank is allowed released to cause movement of the  chair.

3. In industries, a pneumatic valve operates by force of air when actuated. A signal causes actuation of coil. When coil is energized, compressed high pressure air is allowe to enter in a small cylinder and cause operation of valve

5 0
3 years ago
Read 2 more answers
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