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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

Tech A says that a cylinder leakage test is performed on a cylinder with low compression to determine the severity of the leak a

Karolina [17]

Tech A djjdjdndnndndbdbx

4 0

3 years ago

The process of generation of force by the high speed that pushes the jet engine forward is based on Newton’s _____ law of motion

jasenka [17]

Answer:

The process of generation of force by the high speed that pushes the jet engine forward is based on Newton’s 2 law of motion ?

Explanation:

1, Newton’s first law states that, if a body is at rest or moving at a constant speed in a straight line, it will remain at rest or keep moving in a straight line at constant speed unless it is acted upon by a force. This postulate is known as the law of inertia.

2,

Newton’s second law is a quantitative description of the changes that a force can produce on the motion of a body. It states that the time rate of change of the momentum of a body is equal in both magnitude and direction to the force imposed on it. The momentum of a body is equal to the product of its mass and its velocity. Momentum, like velocity, is a vector quantity, having both magnitude and direction. A force applied to a body can change the magnitude of the momentum, or its direction, or both.For a body whose mass m is constant, it can be written in the form F = ma, where F (force) and a (acceleration)

3, Newton’s third law states that when two bodies interact, they apply forces to one another that are equal in magnitude and opposite in direction.The third law is also known as the law of action and reaction. This law is important in analyzing problems of static equilibrium, where all forces are balanced, but it also applies to bodies in uniform or accelerated motion. The forces it describes are real ones, not mere bookkeeping devices. For example, a book resting on a table applies a downward force equal to its weight on the table. According to the third law, the table applies an equal and opposite force to the book.

5 0

3 years ago

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A pump is used to extract water from a reservoir and deliver it to another reservoir whose free surface elevation is 200 ft abov

babunello [35]

Answer:

a) the expected flow rate is 31.4 ft³/s

b) the required brake horsepower is 2808.4 bhp

c) the location of pump inlet to avoid cavitation is -8.4 ft

Explanation:

Given the data in the question;

free surface elevation = 200 ft

total length of pipe required = 1000 ft

diameter = 12 inch

Iron with relative roughness ( k/D ) = 0.0005

H $_{pump$ = 665-0.051Q² [Qinft ]

a) the expected flow rate

given that;

k/D = 0.0005

k/2R = 0.0005

R/k = 1000

now, we determine the friction factor;

1/√f = 2log₁₀( R/k ) + 1.74

we substitute

1/√f = 2log₁₀( 1000 ) + 1.74

1/√f = 6 + 1.74

1/√f = 7.74

√f = 1/7.74

√f = 0.1291989

f = (0.1291989)²

f = 0.01669

Now, Using Bernoulli theorem between two reservoirs;

(p/ρq)₁ + (v²/2g)₁ + z₁ + H $_p$ = (p/ρq)₂ + (v²/2g)₂ + z₂ + h $_L$

so

0 + 0 + 0 + 665-0.051Q² = 0 + 0 + 200 + flQ²/2gdA²

665-0.051Q² = 200 + flQ²/2gdA²

665-0.051Q² = 200 +[ ( 0.01669 × 1000 × Q² ) / (2 × 32.2 × (π/4)² × 1⁵ )

665 - 0.051Q² = 200 + [ 16.69Q² / 39.725 ]

665 - 200 - 0.051Q² = 0.420138Q²

665 - 200 = 0.420138Q² + 0.051Q²

465 = 0.471138Q²

Q² = 465 / 0.471138

Q² = 986.97196

Q = √986.97196

Q = 31.4 ft³/s

Therefore, the expected flow rate is 31.4 ft³/s

b) the brake horsepower required to drive the pump (assume an efficiency of 78%).

we know that;

P = ρgH $_p$ Q / η

where; H $_p$ = 665 - 0.051(986.97196) = 614.7

we substitute;

P = ( 62.42 × 614.7 × 31.4 ) / ( 0.78 × 550 )

P = 1204804.6236 / 429

P = 2808.4 bhp

Therefore, the required brake horsepower is 2808.4 bhp

c) the location of pump inlet to avoid cavitation (assume the required NPSH=25 ft).

NPSH = ( $P_{atom$ / ρg) - h $_s$ - ( P $_v$ / ρg )

we substitute

25 = ( 2116 / 62.42 ) - h $_s$ - ( 30 / 62.42 )

h $_s$ = 8.4 ft

Therefore, the location of pump inlet to avoid cavitation is -8.4 ft

6 0

2 years ago

Were you surprised by the “pie data”? Is it true for you, your family, and your friends? Why or why not?

Sergeu [11.5K]

Answer:

No because it is something to gain knowledge of peoples lives.

6 0

2 years ago

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