What is the minimum hose diameter of an ideal vacuum cleaner that could lift a 14 kg dog off the floor?


Answer 1
Answer: The diameter would be 267km speed suction i think

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A 120-volt fluorescent ballast has an input current of 0.34 ampere and an input power rating of 22 watts. The power factor of the ballast is ____.

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It is called an Cause-and-effect diagram.

A 5000-ft long X-65 pipeline is laid down on seabed with two PLETS (One at each end). The pipe OD=7-in with 0.5-in wall thickness. The pipeline was laid at environmental temperature of 40 °F (As- laid temperature). When pipeline is put into operation, the oil flow was produced at 140 °F. If the thermal expansion coefficient of the pipe material is 6.5*10-/°F and its modulus of elasticity is 30,000 ksi, determine the compressive load applied by the pipeline on a PLET due to its thermal expansion. Assume no temperature change and no seabed friction along the pipeline span.


Answer: 199.1 kip


Given that

Outer diameter is Do = 7 in

Inner diameter Di = ( Do - ( 2×0.5)) = 6 in

Length = 5000 ft = 60000 in

Now change in length of the pipe due to temperature difference


= 60000 × 6.5×10^-6(140-40)

SL = 39 in


sL = PL/AE

A = cross sectional area of pipe = π/4(Do^2 - Di^2)


P = SL×A×E / L

= (39 × π/4(7^2 - 6^2)×30000) / 60000

= 199.1 kip

compressive load applied by the pipeline on a PLET due to its thermal expansion is 199.1 kip

Although the viscoelastic response of a polymer can be very complex (time-dependent stress cycling for instance), two special loading scenarios are fairly simple to describe mathematically. _____________refers to scenarios for which the stress applied to a polymer must decay over time in order to maintain a constant strain. Otherwise, over time, the polymer chains will slip and slide past one another in response to a constant applied load and the strain will increase (in magnitude).
_____________refers to scenarios for which a polymer will permanently flow over time in response a constant applied stress.



Viscoelastic stress relaxation

Viscoelastic creep


  • Viscoelastic stress refers to the reduction of tensile stress with the relaxation time that occurs when the body is kept at a certain length under tensile stress. The purpose of this study was to demonstrate viscoelastic stress relaxation
  • Therefore, when the viscous tension stress occurs, the pressure decreases due to the steady-state stress of the phases.
  • When under constant pressure, the viscoelastic material experiences a time-dependent increase in pressure. This phenomenon is called viscoelastic creep.
  • Therefore the phase constant pressure decreases when there is relaxation in the viscoelastic stress

Q1. A truck traveling at 40 mph is approaching a stop sign. At time ????0 and at a distance of 80ft, the truck begins to slow down by decelerating rate of 12 ft/sec2 . Will the truck be able to stop in time?



The truck will not stop in time. The truck passes the stop sign by about 63.41 ft before it stops.


The distance that the truck starts slowing down = 80 ft from the stop sign

Using equations of motion, we can calculate the distance it will take the truck to stop, then check of it is less than or more than 80 ft.

u = initial velocity of the truck = 40 mph = 58.667 ft/s

v = final velocity of the truck = 0 ft/s (since it comes to a stop eventually)

x = horizontal distance covered during the deceleration

a = Deceleration = -12 ft/s² (it'll have a negative sign, since it is negative acceleration

v² = u² + 2ax

0² = 58.667² + 2(-12)(x)

24x = 3441.816889

x = 143.41 ft

143.41 ft > 80 ft; hence, the truck will not stop in time. The truck passes the stop sign by about 63.41 ft before it stops.

Corrected Question:

A truck traveling at 40 mph is approaching a stop sign. At time t₀ and at a distance of 80 ft, the truck begins to slow down by decelerating at 12 ft/s2, will the truck be able to stop in time?


The truck will not be able to stop in time.


==> First lets convert all variables to SI units

1 mph = 0.45m/s

40mph = 40 miles per hour = 40 x 0.45 m/s

40mph = 18m/s

1 ft = 0.3048m

80 ft = 80 x 0.3048m

80 ft = 24.38m


12ft/s² = 12 x 0.3048m/s²

12ft/s² = 3.66m/s²

==> Now, consider one of the equations of motion as follows;

v² = u² + 2as               -----------------(i)


v = final velocity of motion

u = initial velocity of motion

a= acceleration/deceleration of motion

s = distance covered during motion

Using this equation, lets calculate the distance, s, covered during the acceleration;

We know that;

v = 0               [since the truck comes to a stop]

u = 40mph = 18m/s

a = -12ft/s² = -3.66m/s²    [the negative sign shows that the truck decelerates]

Substitute these values into equation (i) as follows;

0² = 18² + 2 (-3.66)s

0 = 324 - 7.32s

7.32s = 324

s = (324)/(7.32)

s = 44.26m

The distance from where the truck starts decelerating to where it eventually stops is 44.26m which is past the stop sign (which is at 80ft = 24.38m).  This means that the truck stops, 44.26m - 24.38m = 19.88m, after the stop sign. Therefore, the truck will not be able to stop in time.

In same Fig. A-3, a force F, having a slope of 2 vertical 3 horizontal, produces a clockwise moment of 330 ft-lb about A and a counterclockwise moment of 420 ft-lb about B. Compute the moment of F about C.​


The moment of force at the given slope about point C is 210 ft-lb.

The given parameters:

  • Vertical slope, = 2
  • Horizontal slope = 3
  • Clockwise moment = 330 ft-lb
  • Counterclockwise moment = 420 ft-lb

The magnitude of the two moments are in the following simple ratio;

330:420 = 11:14 (divide through by 30)

  • the A coordinate = (0, 5)
  • the B-coordinate = (5,0)

The line of action of the force passes line AB at the final following coordinates;

total ratio of 11:14 = 11 + 14 = 25

= (11 )/(25) * 5, \ \ (14)/(25) * 5\n\n= (2.2, \ 2.8)

The position of C = (3, 1)

The resultant position of point C = (3 - 2.2,  2.8-1) = (0.8, 1.8)

The moment of force at the given slope about point C is calculated as;

3(1.8) + 2(0.8) = 7

Recall that this is the simplest form of the moment produced by the force.

Moment about C = 7 x 30 = 210 ft-lb

Thus, the moment of force at the given slope about point C is 210 ft-lb.

Learn more about moment of force here: brainly.com/question/6278006

5. Switch a in the circuit has been open for a long time and switch b has been closed for a long time. Switch a is closed at t = 0. After remaining closed for 1s, switches a and b are opened simultaneously and remain open indefinitely. Determine the expression for the inductor current i that is valid when (a) 0 ≤ t ≤ 1s and (b) t ≥ 1s



(a) 1/L∫Vdt; integral t [0,1]

(b) 1/L∫Vdt; integral t [ 1, infinity]


An Inductor current I, flowing through an inductor depends on the voltage, V, across the inductor and the inductance, L, of the inductor. The switch 1, 2 timing varies the voltage V with time t

The expression for inductor current is given as:

I= 1/L∫Vdt,

where I is equal to the current flowing through the inductor, L is equal to the inductance of the inductor, and V is equal to the voltage across the inductor.

The formula can also be written as:

I= I0 + 1/L∫Vdt, where I is inductor current at time t, and io is inductor current at t = 0. Time can be varied by controlling the switch

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