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lara [203]
3 years ago
8

14. The flow water in a 10-in Schedule 40 pipe is to be metered. The temperature of the water is

Engineering
1 answer:
marta [7]3 years ago
4 0

Answer:

I don't know plead hrdffffdddffff

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A small pad subjected to a shearing force is deformed at the top of the pad 0.08 in. The height of the pad is 1.38 in. What is t
Aleksandr-060686 [28]

Answer:

The shear strain is 0.05797 rad.

Explanation:

Shear strain is the ratio of change in dimension along the shearing load direction to the height of the plate under application of shear load. Width of the plate remains same. Length of the plate slides under shear load.

Step1

Given:

Height of the pad is 1.38 in.

Deformation at the top of the pad is 0.08 in.

Calculation:

Step2

Shear strain is calculated as follows:

tan\phi=\frac{\bigtriangleup l}{h}

tan\phi=\frac{0.08}{1.38}

tan\phi= 0.05797

For small angle of \phi, tan\phi can take as\phi.

\phi = 0.05797 rad.

Thus, the shear strain is 0.05797 rad.

7 0
3 years ago
A horizontal curve of a two-lane undivided highway (12-foot lanes) has a radius of 678 feet to the center line of the roadway. A
OLEGan [10]

Answer:

maximum speed for safe vehicle operation = 55mph

Explanation:

Given data :

radius ( R ) = 678 ft

old building located ( m )= 30 ft

super elevation = 0.06

<u>Determine the maximum speed for safe vehicle operation </u>

firstly calculate the stopping sight distance

m = R ( 1 - cos \frac{28.655*S}{R} )  ----  ( 1 )

R = 678  

m ( horizontal sightline ) = 30 ft

back to equation 1

30 = 678 ( 1 - cos (28.655 *s / 678 ) )

( 1 - cos (28.655 *s / 678 ) )  = 30 / 678 = 0.044

cos \frac{28.65 *s }{678}  = 1.044

hence ; 28.65 * s = 678 * 0.2956

s = 6.99 ≈ 7 ft

next we will calculate the design speed ( u ) using the formula below

S = 1.47 ut  + \frac{u^2}{30(\frac{a}{3.2} )-G1}  ----  ( 2 )

t = reaction time,  a = vehicle acceleration, G1 = grade percentage

assuming ; t = 2.5 sec , a = 11.2 ft/sec^2, G1 = 0

back to equation 2

6.99 = 1.47 * u * 2.5 + \frac{u^2}{30[(11.2/32.2)-0 ]}

3.675 u  + 0.0958 u^2 - 6.99 = 0

u ( 3.675 + 0.0958 u ) = 6.99

5 0
3 years ago
Much of the workd went to bed hungry
Marysya12 [62]
The workers went to bed hungry probably because they are hard workers and so didn’t want to eat because they didn’t want to take break┌(; ̄◇ ̄)┘
7 0
3 years ago
When two or more simple machines are combined they form
Volgvan
Compound machine is the answer
8 0
3 years ago
A drilling operation is performed on a steel part using a 12.7 mm diameter twist drill with a point angle of 118 degrees. The ho
Masteriza [31]

Answer:

a. Rotational speed of the drill  = 375.96 rev/min

b. Feed rate  = 75 mm/min

c. Approach allowance  = 3.815 mm

d. Cutting time  = 0.67 minutes

e. Metal removal rate after the drill bit reaches full diameter. = 9525 mm³/min

Explanation:

Here we have

a. N = v/(πD) = 15/(0.0127·π) = 375.96 rev/min

b. Feed rate = fr = Nf = 375.96 × 0.2 = 75 mm/min

c. Approach allowance = tan 118/2 = (12.7/2)/tan 118/2 = 3.815 mm

d. Approach allowance T∞ =L/fr = 50/75 = 0.67 minutes

e. R = 0.25πD²fr = 9525 mm³/min.

7 0
3 years ago
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