Ask question

Ask question

All categories

3 years ago

12

The volume flow rate of the water supplied by a well is 2.0×10−4m3/s.The well is 40.0 m deep. (a) What is the power output of th

e pump-in other words, at what rate does the well do work on the water? (b) Find the pressure difference the pump must maintain. (c) Can the pump be at the top of the well or must it be at the bottom? Explain.

1 answer:

MaRussiya [10]3 years ago

4 0

Answer:

a). $P=78.4W$

b). $P=392kPa$

c.) It must be at the bottom

Explanation:

Given:

Volume flow $V_f=2.0x10^{-4}m^3/s$

Well depp $h=40.m$

a.

The power output of the pum

$W=F*d$

$F=m*g$

$m=p*V=1000kg/m^3*2.0x10^{-4}m^3}=0.2Kg$

$W=m*g*d=0.2kg*9.8m/s^2*40.0m=78.4kg*m^2/s^2$

$W=78.4J$

$P=\frac{W}{t}=\frac{78.4J}{1s}=78.4W$

b.

The pressure of difference the pum

Δ $P=p*g*y'$

Δ $P=1000kg/m^3*9.8m/s^2*40.0m=392x10^3Pa$

$P=392kPa$

c.

It must be at the bottom since the pressure difference is greater than atmospheric pressure, so it wouldn't be able to lift the water all the way

You might be interested in

A tennis ball is tossed upward with a speed of 3.0\,\dfrac{\text m}{\text s}3.0 s m 3, point, 0, start fraction, start text, m

MrRa [10]

Answer:

The velocity of the ball is 0.92 m/s in the downward direction (-0.92 m/s).

Explanation:

Hi there!

The equation for the velocity of an object thrown upward is the following:

v = v0 + g · t

Where:

v = velocity of the ball.

v0 = initial velocity.

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

t = time.

To find the velocity of the ball at t = 0.40 s, we have to replace "t" by 0.40 s in the equation:

v = v0 + g · t

v = 3.0 m/s - 9.8 m/s² · 0.40 s

v = -0.92 m/s

The velocity of the ball is 0.92 m/s in the downward direction (-0.92 m/s).

8 0

2 years ago

Read 2 more answers

Bryce, a mouse lover, keeps his four pet mice in a roomy cage, where they spend much of their spare time (when they are not slee

user100 [1]

Answer:

I₁ = (7.78 i ^ - 6.71 j ^) 10⁻³ J s , I₂ = (-12.5 i ^ -14.6 j ^) 10⁻³ J s , I₃ = (19.1i ^ + 18.6 j ^) 10⁻³ J s and I₄ = (-9.14i ^ + 7.24 j ^) 10⁻³ J s

Explanation:

The impulse is equal to the variation of the moment, to apply this relationship to our case, we will assume that initially the mouse was at rest

I = Δp = m $v_{f}$ -m v₀

I = m ( $v_{f}$ -v₀)

Bold indicates vector quantities, let's calculate the momentum of each mouse in for the x and y axes

We recommend bringing all units to the SI system

Mouse 1.

It has a mass of 22.3 g = 22.3 10⁻³ kg, a final velocity of (v = 0.349 i ^ - 0.301 j ^) m / s with an initial velocity of zero

Iₓ = m ( $v_{fx}$ - v₀ₓ)

Iₓ = 22.3 10⁻³ (0.349 -0)

Iₓ = 7.78 10⁻³ J s

$I_{y}$ = m ( $v_{fy}$ - $v_{oy}$ )

$I_{y}$ = 22.3 10⁻³ (-0.301)

$I_{y}$ = -6.71 10⁻³ J s

I₁ = (7.78 i ^ - 6.71 j ^) 10⁻³ J s

Mouse 2

Mass 17.9 g = 17.9 10⁻³ kg

Speed (-0.699 i ^ - 0.815 j ^) m / s

Iₓ = m ( $v_{fx}$ - v₀ₓ)

Iₓ = 17.9 10⁻³ (-0.699 -0)

Iₓ = -12.5 10⁻³ J s

$I_{y}$ = 17.9 10⁻³ (-0.815 - 0)

$I_{y}$ = -14.6 10⁻³ J s

I₂ = (-12.5 i ^ -14.6 j ^) 10⁻³ J s

Mouse 3

Mass 19.1 g = 19.1 10⁻³ kg

Speed (0.745i ^ + 0.975 j ^) m / s

Iₓ = 19.1 10⁻³ (0.745 -0)

Iₓ = 14.2 10⁻³ J s

$I_{y}$ = 19.1 10⁻³(0.975 -0)

$I_{y}$ = 18.6 10⁻³ J s

I₃ = (19.1i ^ + 18.6 j ^) 10⁻³ J s

Mouse 4

Mass 10.1 g = 10.1 10⁻³ kg

Speed (-0.905i ^ + 0.717j ^) m / s

Iₓ = 10.1 10⁻³ (-0.905 -0)

Iₓ = -9.14 10⁻³ J s

$I_{y}$ = 10.1 10⁻³ (0.717 -0)

$I_{y}$ = 7.24 10⁻³ J s

I₄ = (-9.14i ^ + 7.24 j ^) 10⁻³ J s

8 0

3 years ago

QUICK ITS DUE IN 6 MINUTES. create a scenario that applies the 3 laws of motion. Explain in complete sentences how each law if d

yKpoI14uk [10]

Answer:

In the first law, an object will not change its motion unless a force acts on it. In the second law, the force on an object is equal to its mass times its acceleration. In the third law, when two objects interact, they apply forces to each other of equal magnitude and opposite direction.

5 0

2 years ago

A 1.0 m string with a 5 g stopper on the end is whirled in a vertical circle. The speed of the stopper is 8 m/s at the top of th

Andrew [12]

Answer:

Explanation:

A )

At the bottom of the circle , the potential energy of the stopper is converted into kinetic energy

1/2 m V² = mg x 2r + 1/2 mv²

m is mass of stopper , V is velocity at the bottom , r is radius of the circular path which is length of the string , v is velocity at the top

1/2 V² = g x 2r + 1/2 v²

V² = g x 4r + v²

V² = 9.8 x 4 + 8²

V² = 103.2

V = 10.16 m/s

B )

If T be the tension at the top

Net downward force

= mg + T . This force provides centripetal force for the circular motion

mg +T = mv² / r

T = mv²/r -mg

= m ( v²/r - g )

= .005 ( 8²/1 -g )

= .005 x 54.2

= .27 N .

C ) At the bottom

Net force = T - mg , T is tension at the bottom , V is velocity at bottom

T-mg = mV²/r

T = m ( V²/r +g )

= .005 ( 10.16²/1 +9.8)

= .005 x 113

= .56 N .

3 0

3 years ago

A capacitor consists of two parallel plates, each with an area of 17.0 cm2 , separated by a distance of 0.150 cm . The material

Llana [10]

Answer:

a) $C = 40.138\,pF$ , b) $q = 16.056\,nC$ , c) $U = 3.212\,\mu J$

Explanation:

a) The capacitance of two parallel plates capacitor with dielectric is given by the following expression:

$C = K\cdot \epsilon_{o}\cdot \frac{A}{d}$

Where:

$K$ - Dielectric constant.

$\epsilon_{o}$ - Vaccum permitivity.

$A$ - Plate area.

$d$ - Distance between plates.

Hence, the capacitance of the system is:

$C = (4.00)\cdot (8.854\times 10^{-12}\,\frac{F}{m} )\cdot \left(\frac{17\times 10^{-4}\,m^{2}}{0.150\times 10^{-2}\,m}\right)$

$C = 4.014\cdot 10^{-11}\,F$

$C = 40.138\,pF$

b) The charge can be found by using the definition of capacitance:

$q = C\cdot V_{batt}$

$q = (4.014\times 10^{-11}\,F)\cdot (400\,V)$

$q = 1.606\times 10^{-8}\,C$

$q = 16.056\,nC$

c) The energy stored in the charged capacitor is:

$U=\frac{1}{2}\cdot Q\cdot V_{batt}$

$U=\frac{1}{2}\cdot (1.606\times 10^{-8}\,C)\cdot (400\,V)$

$U = 3.212\times 10^{-6}\,J$

$U = 3.212\,\mu J$

3 0

2 years ago

Read 2 more answers