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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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You want a potof water to boil at 105 celcius. How heavy a
ankoles [38]

Answer:

35.7 kg lid we put

Explanation:

given data

temperature = 105 celcius

diameter = 15 cm

Patm = 101 kPa

to find out

How heavy a  lid should you put

solution

we know Psaturated from table for temperature is 105 celcius is

Psat = 120.8 kPa

so

area will be here

area = \frac{\pi }{4} d^2    ..................1

here d is diameter

put the value in equation 1

area = \frac{\pi }{4} 0.15^2

area = 0.01767 m²

so net force is

Fnet = ( Psat - Patm ) × area

Fnet = ( 120.8 - 101 ) × 0.01767

Fnet = 0.3498 KN = 350 N

we know

Fnet = mg

mass = \frac{Fnet}{g}

mass  = \frac{350}{9.8}

mass = 35.7 kg

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3 0
3 years ago
A loss in value caused by an undesirable or hazardous influence offsite is which type of depreciation?
Lubov Fominskaja [6]

External depreciation may be defined as a loss in value caused by an undesirable or hazardous influence offsite.

<h3>What is depreciation?</h3>

Depreciation may be defined as a situation when the financial value of an acquisition declines over time due to exploitation, fray, and incision, or obsolescence.

External depreciation may also be referred to as "economic obsolescence". It causes a negative influence on the financial value gradually.

Therefore, it is well described above.

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5 0
2 years ago
Consider an 8-car caravan, where the propagation speed is 100 km/hour, each car takes 1 minute to pass a toll both. The caravan
melamori03 [73]

Answer:

A. 36 minutes

B. 120 minutes

C.

i. 144 minutes

ii. 984 minutes

D. Car 1 is 1.67km ahead of Cat 2 when Car 2 passed the toll B.

E. 98.33km

Explanation

A.

Given

dAb = 10km

dBc = 10km

Propagation Speed = 100km/hr

Delay time = 1 minute

Numbers of cars = 8

Number of tolls = 3

Total End to End delay = Propagation delay + Transition delay

Calculating Propagation Delay

Propagation delay = Total Distance/Propagation speed

Total distance = 10km + 10km = 20km

So, Propagation delay = 20km/100km/hr

Propagation delay = 0.2 hour

                               

Translation delay = delay time* numbers of tolls * numbers of cars

Transitional delay = 1 * 3 * 8

Transitional delay = 24 minutes

Total End delay = 24 minutes + 0.2 hours

= 24 minutes + 0.2 * 60 minutes

= 24 minutes + 12 minutes

= 36 minutes

B.

Total End to End delay = Propagation delay + Transition delay

Calculating Propagation Delay

Propagation delay = Total Distance/Propagation speed

Total distance = 10km + 10km = 20km

So, Propagation delay = 20km/100km/hr

Propagation delay = 0.2 hour

                               

Translation delay = delay time* numbers of tolls ------ Cars traveling separately

Transitional delay = 1 * 3

Transitional delay = 3 minutes

Total End delay for one car = 3 minutes + 0.2 hours

= 3 minutes + 0.2 * 60 minutes

= 3 minutes + 12 minutes

= 15 minutes

Total End delay for 8 cars = 8 * 15 = 120 minutes

C.

Given

dAb = 100km

dBc = 100km

Propagation Speed = 100km/hr

Delay time = 1 minute

Numbers of cars = 8

Number of tolls = 3

i. Cars travelling together

Total End to End delay = Propagation delay + Transition delay

Calculating Propagation Delay

Propagation delay = Total Distance/Propagation speed

Total distance = 100km + 100km = 200km

So, Propagation delay = 200km/100km/hr

Propagation delay = 2 hours

                               

Translation delay = delay time* numbers of tolls * numbers of cars

Transitional delay = 1 * 3 * 8

Transitional delay = 24 minutes

Total End delay = 24 minutes + 2 hours

= 24 minutes + 2 * 60 minutes

= 24 minutes + 120 minutes

= 144 minutes

ii. Cars travelling separately

Total End to End delay = Propagation delay + Transition delay

Calculating Propagation Delay

Propagation delay = Total Distance/Propagation speed

Total distance = 100km + 100km = 200km

So, Propagation delay = 200km/100km/hr

Propagation delay = 2 hours

                               

Translation delay = delay time* numbers of tolls ------ Cars traveling separately

Transitional delay = 1 * 3

Transitional delay = 3 minutes

Total End delay for one car = 3 minutes + 2 hours

= 3 minutes + 2 * 60 minutes

= 3 minutes + 120 minutes

= 123 minutes

Total End delay for 8 cars = 8 * 123 = 984 minutes

D.

Distance = 100km

Time = 1 min/car

Car 1 is 1 minute ahead of car 2 --- at toll A and B

If car 1 leaves toll B after 10 minutes then cat 2 leaves after 11 minutes

Time delay = 11 - 10 = 1 minute

Distance = time * speed

= 1 minute * 100km/hr

= 1 hr/60 * 100 km/hr

= 100/60

= 1.67km

E.

Given

Distance = 100km

Distance behind = 1.67

Maximum value of dBc = 100km - 1.67km = 98.33km

The maximum distance that can be reached is 98.33km

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