2.) Technician A says that milky colored ATF could indicate a leaking transmission cooler in the radiator.

The maximum gage pressure is known to be 10 MPa in a spherical steel pressure vessel having a 200 mm outer diameter and a(n) 2 m

VikaD [51]

Answer:

factor of safety = 0.8

Explanation:

given data

maximum gauge pressure in steel p = 10 MPa

outer diameter do = 200 mm

wall thickness t = 2 mm

ultimate stress of the steel σU = 400 MPa

solution

We check the type of shell that is

$\frac{t}{do} =\frac{6}{200}$

so we can see that that is

so that it mean vessel is a thin shell now we can we get maximum pressure by the maximum tensile stress is

σ = $\frac{p\times do}{2t}$ .......................1

put here value and we get

σ = $\frac{10\times 200}{2\times 2}$

σ = 500 MPa

so factor of safety will be express as

factor of safety = $\frac{\sigma U}{\sigma}$ ..........2

factor of safety = $\frac{400}{500}$

factor of safety = 0.8

6 0

4 years ago

Chemical engineers determine how to transport chemicals. O True False

Anna35 [415]

Answer:

Chemical Engineers determine how to transport chemicals:

TRUE

6 0

2 years ago

Read 2 more answers

Steep safety ramps are built beside mountain highways to enable vehicles with defective brakes to stop safely. A truck enters a

Virty [35]

Answer:

a) The additional time required for the truck to stop is 8.5 seconds

b) The additional distance traveled by the truck is 230.05 ft

Explanation:

Since the acceleration is constant, the average speed is:

(final speed - initial speed) / 2 = 0.75 v0

Since travelling at this speed for 8.5 seconds causes the vehicle to travel 690 ft, we can solve for v0:

0.75v0 * 8.5 = 690

v0 = 108.24 ft/s

The speed after 8.5 seconds is: 108.24 / 2 = 54.12 ft/s

We can now use the following equation to solve for acceleration:

$v^2 - u^2 = 2*a*s$

$54.12^2 - 108.24^2 = 2*a*690$

a = -6.367 m/s^2

Additional time taken to decelerate: 54.12/6.367 = 8.5 seconds

Total distance traveled:

$v^2 - u^2 = 2*a*s$

0 - 108.24^2 = 2 * (-6.367) * s

solving for s we get total distance traveled = 920.05 ft

Additional Distance Traveled: 920.05 - 690 = 230.05 ft

5 0

4 years ago

A 10-ft-long simply supported laminated wood beam consists of eight 1.5-in. by 6-in. planks glued together to form a section 6 i

ruslelena [56]

Answer:

point B where $Q_B = 101.25 \ in^3$ has the largest Q value at section a–a

Explanation:

The missing diagram that is suppose to be attached to this question can be found in the attached file below.

So from the given information ;we are to determine the point that has the largest Q value at section a–a

In order to do that; we will work hand in hand with the image attached below.

From the image attached ; we will realize that there are 8 blocks aligned on top on another in the R.H.S of the image with the total of 12 in; meaning that each block contains 1.5 in each.

We also have block partitioned into different point segments . i,e A,B,C, D

For point A ;

Let Q be the moment of the Area A;

SO ; $Q_A = Area \times y_1$

where ;

$y_1 = (6 - \dfrac{1.5}{2})$

$y_1 = (6- 0.75)$

$y_1 = 5.25 \ in$

$Q_A =(L \times B) \times y_1$

$Q_A =(6 \times 1.5) \times 5.25$

$Q_A =47.25 \ in^3$

For point B ;

Let Q be the moment of the Area B;

SO ; $Q_B = Area \times y_2$

where ;

$y_2 = (6 - \dfrac{1.5 \times 3}{2})$

$y_2= (6 - \dfrac{4.5}{2}})$

$y_2 = (6 -2.25})$

$y_2 = 3.75 \ in$

$Q_B =(L \times B) \times y_1$

$Q_B=(6 \times 4.5) \times 3.75$

$Q_B = 101.25 \ in^3$

For point C ;

Let Q be the moment of the Area C;

SO ; $Q_C = Area \times y_3$

where ;

$y_3 = (6 - \dfrac{1.5 \times 2}{2})$

$y_3 = (6 - 1.5})$

$y_3= 4.5 \ in$

$Q_C =(L \times B) \times y_1$

$Q_C =(6 \times 3) \times 4.5$

$Q_C=81 \ in^3$

For point D ;

Let Q be the moment of the Area D;

SO ; $Q_D = Area \times y_4$

since there is no area about point D

Area = 0

$Q_D =0 \times y_4$

$Q_D = 0$

Thus; from the foregoing ; point B where $Q_B = 101.25 \ in^3$ has the largest Q value at section a–a

3 0

4 years ago

Water leaves a penstock (the flow path through a hydroelectric dam) at a velocity of 100 ft/s. How deep is the water behind the

Marysya12 [62]

Answer:

155fts

Explanation:

We apply the bernoulli's equation to get the depth of water.

We have the following information

P1 = pressure at top water surface = 0

V1 = velocity at too water surface = 0

X1 = height of water surface = h

Hf = friction loss = 0

P2 = pressure at exit = 0

V2 = velocity at exit if penstock = 100ft/s

X2 = height of penstock = 0

g = acceleration due to gravity = 32.2ft/s²

Applying these values to the equation

0 + 0 + h = 0 + v2²/2g +0 + 0

= h = 100²/2x32.2

= 10000/64.4

= 155.28ft

= 155

8 0

3 years ago