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Savatey [412]
2 years ago
15

Bending is defined as? A. the application of a load tending to distort a member in one direction. B. the application of opposing

load with a very narrow specific area. C. the application of pushing load data application of pulling loads. E.application of twisting a load.
Engineering
1 answer:
amm18122 years ago
4 0
Hi how are you today
You might be interested in
Find the Rectangular form of the following phasors?
almond37 [142]

Answer:

The angles are missing in the question.

The angles are :

45,     30,    60,     90,    -34,     -56,      20,     -42,  -65,    -15

P=10, P=5,  P=25, P=54, P=65, P=95, P=250, P=8, P=35, P=150

Explanation:

1. P = 10,   θ = 45°  rectangular coordinates

x = r cosθ  ,   y = r sinθ

So, rectangular form is x + iy

x = P cosθ = 10 cos 45°

  = 7.07

y =P sinθ = 10 sin 45°

  = 7.07

Therefore, rectangular form

x + iy = 7.07 + i (7.07)

2. P = 5 , θ = 30°

x = 5 cos  30° = 4.33

y = 5 sin  30° = 2.5

So, (x+iy) = 4.33 + i (2.5)

3. P = 25 , θ = 60°

x = 25 cos  60° = 12.5

y = 25 sin  60° = 21.65

So, (x+iy) = 12.5 + i (21.65)

4. P = 54 , θ = 90°

x = 54 cos  90° = 0

y = 54 sin  90° = 54

So, (x+iy) = 0+ i (54)

5. P = 65 , θ = -34°

x = 65 cos  (-34°) = 53.88

y = 65 sin  (-34°) = -36.34

So, (x+iy) = 53.88 - i (36.34)

6. P = 95 , θ = -56°

x = 95 cos  (-56)° = 53.12

y = 95 sin  (-56)° = -78.75

So, (x+iy) = 53.12 - i (78.75)

7. P = 250 , θ = 20°

x = 250 cos  20° = 234.92

y = 250 sin 20° = 85.5

So, (x+iy) = 234.92 + i (85.5)

8. P = 8 , θ = (-42)°

x = 8 cos  (-42)° = 5.94

y = 8 sin  (-42)° = -5.353

So, (x+iy) = 5.94 - i (5.353)

9. P = 35 , θ = (-65)°

x = 35 cos  (-65)° = 14.79

y = 35 sin  (-65)° = -31.72

So, (x+iy) = 14.79 - i (31.72)

10. P = 150 , θ = (-15)°

x = 150 cos  (-15)° = 144.88

y = 150 sin  (-15)° = -38.82

So, (x+iy) = 144.88 - i (38.82)

6 0
3 years ago
Calculate the wire pressure for a round copper bar with an original cross-sectional area of 12.56 mm2 to a 30% reduction of area
dybincka [34]

Answer:153.76 MPa

Explanation:

Initial Area\left ( A_0\right )=12.56 mm^2

Final Area\left ( A_f\right )=0.7\times 12.56 mm^2=8.792 mm^2

Die angle=30^{\circ}

\alpha =\frac{30}{2}=15^{\circ}

\mu =0.08

Yield stress\left ( \sigma _y \right )=350 MPa

B=\mu cot\left ( \aplha\right )=0.2985

\sigma _{pressure}=\sigma _y\left [\frac{1+B}{B}\right ]\left [ 1-\frac{A_f}{A_0}\right ]^B

\sigma _{pressure}=350\left [\frac{1+0.2985}{0.2985}\right ]\left [ 1-\frac{8.792}{12.56}\right ]^{0.2985}

\sigma _{pressure}=153.76 MPa

8 0
3 years ago
What is a ton of refrigeration?
AURORKA [14]

Explanation:

The unit refrigeration is generally is given in terms of tons.In refrigeration compressor consume some amount of work to produce the cooling effect  with the help of evaporator and condenser.

In the simple words ton is the cooling load of refrigeration system.

So  

1 ton = 3.5 KW

1 ton = 12,000 BTU/hr

 

6 0
3 years ago
After replacing a vacuum booster, the brakes lock up on a road test. Technician A says there is air trapped inside the brake lin
vitfil [10]

Answer:

Technician B

Explanation:

The brakes can lockup due to the following reasons

1) Overheating break systems

2) Use of wrong brake fluid

3) Broken or damaged drum brake backing plates, rotors, or calipers

4) A defective ABS part, or a defective parking mechanism or proportioning valve

5) Brake wheel cylinders, worn off

6) Misaligned power brake booster component

5 0
3 years ago
2 Blocks 1 and 2 rest on rough surfaces with coefficient of frictions ¢1 and ¢2 respectively. The blocks
Amiraneli [1.4K]

Answer:

  • 100N
  • 25N

Explanation:

a) On the verge of tipping over, reaction acts at the corner A

When slippage occurs,

Block moves w/ const. velocity  equilibrium

Three-force member: reaction at A must pass through B

tan b/2h, h b/ 2 θ µ = = ∴= k k ( µ )

b) When slippage occurs,

Block moves w/ const. velocity  equilibrium

Three-force member: reaction at C must pass through G

k tanθ µ =

tan x/ H/2 , x H/2

4 0
3 years ago
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