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USPshnik [31]
3 years ago
10

1)A wheel is used to turn a valve stem on a water valve. If the wheel radius is 1 foot and the stem, (axle), radius is .5 inches

, what is the mechanical advantage of the wheel and axle? 2)How much resistance force can ideally be overcome when an effort of 80 lbs is applied to the wheel of the water valve in problem 1? 3)What is the linear distance traveled when a 2.5' diameter wheel makes one revolution? 4)A compressor motor has a 1.00 inch output shaft (axle) that drives a 4.00inch pulley (wheel). If the motor output shaft supplies 24 lbs of force, ideally how much force does the pulley apply to the drive belt wrapped around it? 5) If the belt pulley's force is measured to be 4 lbs, what is the AMA wheel and axle here?6) What is the efficiency of the compressor motor – belt pulley wheel and axle?
Engineering
1 answer:
Novay_Z [31]3 years ago
3 0

Answer:

1.) 2.4

2.) 112 lbs

3.) 7.85 inches

4.) 6 lbs

5.) 2 lbs

6.) 67%

Explanation:

Given that

Radius of the wheel R = 1 foot

1 foot = 12 inches.

Radius of the axle r = 5 inches

1.) The mechanical advantage MA is :

MA = R/r = 12/5 = 2.4

2.) How much resistance force can ideally be overcome when an effort of 80 lbs is applied to the wheel of the water valve in problem 1?

MA = Load / effort

Where effort = 80 lbs

Substitute MA and effort into the formula

2.4 = Load / 80

Cross multiply

Load = 2.4 × 80 = 192 lbs

The resistance force to be overcome will be

Force = load - effort

Resistance force = 192 - 80 = 112 lbs

3) What is the linear distance traveled when a 2.5' diameter wheel makes one revolution

One revolution = 2π

Radius = 2.5 /2 = 1.25 inches

Linear distance S = angular distance Ø × radius r

S = Ør

S = 2π × 1.25

S = 7.85 inches

4. ) given that

Wheel radius R = 4

Axle radius r = 1

MA = 4/1 = 4

MA = Load / effort

4 = 24/ effort

Effort = 24/4 = 6 lbs

5.) 6 - 4 = 2lb

6.) Efficiency = MA / VR × 100

Efficiency = 4 / 6 × 100

Efficiency = 67%

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5.21e-2mm

Explanation:

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8 0
3 years ago
Who can use NIST resources?
sukhopar [10]

Answer:

Federal agencies

Explanation:

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7 0
3 years ago
Q4. (20 points) For a bronze alloy, the stress at which plastic deformation begins is 271 MPa and the modulus of elasticity is 1
babunello [35]

Answer:

a) P = 86720 N

b) L = 131.2983 mm

Explanation:

σ = 271 MPa = 271*10⁶ Pa

E = 119 GPa = 119*10⁹ Pa

A = 320 mm² = (320 mm²)(1 m² / 10⁶ mm²) = 3.2*10⁻⁴ m²

a) P = ?

We can apply the equation

σ = P / A     ⇒    P = σ*A = (271*10⁶ Pa)(3.2*10⁻⁴ m²) = 86720 N

b) L₀ = 131 mm = 0.131 m

We can get ΔL applying the following formula (Hooke's Law):

ΔL = (P*L₀) / (A*E)    ⇒  ΔL = (86720 N*0.131 m) / (3.2*10⁻⁴ m²*119*10⁹ Pa)

⇒  ΔL = 2.9832*10⁻⁴ m = 0.2983 mm

Finally we obtain

L = L₀ + ΔL = 131 mm + 0.2983 mm = 131.2983 mm

3 0
3 years ago
An air-standard Diesel cycle engine operates as follows: The temperatures at the beginning and end of the compression stroke are
Vika [28.1K]

This question is incomplete, the complete question is;

An air-standard Diesel cycle engine operates as follows: The temperatures at the beginning and end of the compression stroke are 30 °C and 700 °C, respectively. The net work per cycle is 590.1 kJ/kg, and the heat transfer input per cycle is 925 kJ/kg. Determine the a) compression ratio, b) maximum temperature of the cycle, and c) the cutoff ratio, v3/v2.

Use the cold air standard assumptions.

Answer:

a) The compression ratio is 18.48

b) The maximum temperature of the cycle is 1893.4 K

c) The cutoff ratio, v₃/v₂ is 1.946

Explanation:

Given the data in the question;

Temperature at the start of a compression T₁ = 30°C = (30 + 273) = 303 K

Temperature at the end of a compression T₂ = 700°C = (700 + 273) = 973 K

Net work per cycle W_{net = 590.1 kJ/kg

Heat transfer input per cycle Qs = 925 kJ/kg

a) compression ratio;

As illustrated in the diagram below, 1 - 2 is adiabatic compression;

so,

Tγ^{Y-1 = constant { For Air, γ = 1.4 }

hence;

⇒ V₁ / V₂ = ( T₂ / T₁ )^{\frac{1}{Y-1}

so we substitute

⇒ V₁ / V₂ = (  973 K / 303 K  )^{\frac{1}{1.4-1}

= (  3.21122  )^{\frac{1}{0.4}

= 18.4788 ≈ 18.48

Therefore, The compression ratio is 18.48

b) maximum temperature of the cycle

We know that for Air, Cp = 1.005 kJ/kgK

Now,

Heat transfer input per cycle Qs = Cp( T₃ - T₂ )

we substitute

925 = 1.005( T₃ - 700 )

( T₃ - 700 ) = 925 / 1.005

( T₃ - 700 ) = 920.398

T₃ = 920.398 + 700

T₃ = 1620.398 °C

T₃ = ( 1620.398 + 273 ) K

T₃ = 1893.396 K ≈ 1893.4 K

Therefore, The maximum temperature of the cycle is 1893.4 K

c)  the cutoff ratio, v₃/v₂;

Since pressure is constant, V ∝ T

So,

cutoff ratio S = v₃ / v₂  = T₃ / T₂

we substitute

cutoff ratio S = 1893.396 K / 973 K

cutoff ratio S = 1.9459 ≈ 1.946

Therefore, the cutoff ratio, v₃/v₂ is 1.946

8 0
3 years ago
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Eduardwww [97]
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2.Proteins

3.DNA is made up of molecules called nucleotides. Each nucleotide contains a phosphate group, a sugar group and a nitrogen base. The four types of nitrogen bases are adenine (A), thymine (T), guanine (G) and cytosine (C). The order of these bases is what determines DNA's instructions, or genetic code.

4. any of a group of basic proteins found in chromatin.

5.46 chromosomes

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