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Nadusha1986 [10]
3 years ago
6

A water pump is a positive displacement-type pump true or false

Physics
1 answer:
Kay [80]3 years ago
6 0

Answer: True

A water pump belong to a positive displacement pump that provides constant flow of water at fixed speed, regardless of changes in the counter pressure. The two main types of positive displacement pump are rotary pumps and reciprocating pumps.

Moreover, water pump is a reciprocating positive displacement pump that have an expanding cavity on the suction side and a decreasing cavity on the discharge side. In water pumps, the liquid flows into the pumps as the cavity on the suction side expands and then the liquid flows out of the discharge as the cavity collapses providing water in a pail.

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Ok i apologise for the messy working but I'll try and explain my attempt at logic

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I also noted the "initial" and "final" height of the swing with hi and hf respectively.

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So then the equation reads something like Ep = (rearranged Ek equation for m) × g (which is -9.81) × change in height (hf - hi).

Then i simplify the equation a little. When i multiply both sides by v^2 i can clearly see that there is one E on each side (at that stage i don't need to clarify which type of energy it is because Ek = Ep so they're just the same anyway). So i just canceled them out and square rooted both sides.

The answer i got was that the max velocity would be 4.85m/s 3sf, assuming no losses (eg energy lost to friction).

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I sort of understand but what does it mean by.... Another?
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<em>WAS</em><em> </em><em>THIS</em><em> </em><em>ANSWER</em><em> </em><em>HELPFUL</em><em> </em><em>?</em>

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4 0
2 years ago
The tub of a washer goes into its spin-dry cycle, starting from rest and reaching an angular speed of 6.0 rev/s in 11.0 s. At th
sleet_krkn [62]

Answer

given,

ω₁ = 0 rev/s

ω₂ = 6 rev/s

t = 11 s

Using equation of rotational motion

The angular acceleration is

  ωf - ωi = α t

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The angular displacement

  θ₁= ωi t + (1/2) α t²

  θ₁= 0 + (1/2) (0.545)(11)^2

  θ₁= 33 rev

case 2

ω₁ = 6 rev/s

ω₂ = 0 rev/s

t = 14 s

Using equation of rotational motion

The angular acceleration is

  ωf - ωi = α t

  14 α = 0 - 6

        = - 0.428 rev/s²

The angular displacement

  θ₂= ωi t + (1/2) α t²

  θ₂= 6 x 14 + (1/2) (-0.428)(14)^2

  θ₂= 42 rev

total revolution in 25 s is equal to

θ =  θ₁ +  θ₂

θ =  33 + 42

θ = 75 rev

3 0
3 years ago
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