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3 years ago

An eletric pump in the ground floor

Physics

1 answer:

ELEN [110]3 years ago

6 0

Answer:

The input power is $44.4\times10^{3}\ kW$

Explanation:

Given that,

Time = 15 min

Volume of water = 30 m³

Height = 40 m

Efficiency = 30%

Density of water = 1000 kg/m³

Suppose, acceleration due to gravity = 10 m/s²

We need to calculate the mass of water pumped

Using formula of mass

$Mass = Volume\times density$

Put the value into the formula

$Mass=30\times1000$

$Mass=3\times10^{4}\ kg$

We need to calculate the output power

Using formula of power

$P_{out}=\dfrac{W}{t}$

$P_{out}=\dfrac{mgh}{t}$

Put the value into the formula

$P_{out}=\dfrac{3\times10^{4}\times10\times40}{15\times60}$

$P_{out}=\dfrac{4}{3}\times10^{4}\ Watt$

We need to calculate the input power

Using formula of efficiency

$\eta=\dfrac{P_{out}}{P_{in}}$

$P_{in}=\dfrac{P_{out}}{\eta}$

Put the value into the formula

$P_{in}=\dfrac{4\times100\times10^{4}}{3\times30}$

$P_{in}=\dfrac{4\times10^{5}}{9}\ Watt$

$P_{in}=44.4\times10^{3}\ kW$

Hence, The input power is $44.4\times10^{3}\ kW$

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When you jump off the earth, your momentum changes, but the Earth does not move. 1)If momentum is always conserved, why do we no

Mazyrski [523]

This question is off-base and misleading from the beginning.

When you jump off the Earth, your momentum changes, and the Earth moves away from you with an equal change of momentum in the opposite direction.

1). Momentum is conserved when you jump. But we don't feel the Earth moving. Since the Earth's mass is a bazillion times greater than YOUR mass, the speed with which the Earth moves away from you is only one bazillionth of your speed. That way, the product of (mass) x (speed) is the SAME for you and for the Earth, and momentum is conserved.

2). Of course ! If everyone jumped at the same time, the Earth's momentum would change. In answer-(1), I explained that the Earth's momentum changes whenever ONE PERSON jumps. So 7 billion people all jumping at the same time would certainly make it change.

8 0

3 years ago

A hula hoop is rolling along the ground with a translational speed of 26 ft/s. It rolls up a hill that is 16 ft high. Determine

nikklg [1K]

Answer:12.8 ft/s

Explanation:

Given

Speed of hoop $v=26\ ft/s$

height of top $h=16\ ft$

Initial energy at bottom is

$E_b=\frac{1}{2}mv^2+\frac{1}{2}I\omega ^2$

Where m=mass of hoop

I=moment of inertia of hoop

$\omega$ =angular velocity

for pure rolling $v=\omega R$

$I=mR^2$

$E_b=\frac{1}{2}mv^2+\frac{1}{2}mR^2\times (\frac{v}{R})^2$

$E_b=mv^2=m(26)^2=676m$

Energy required to reach at top

$E_T=mgh=m\times 32.2\times 16$

$E_T=512.2m$

Thus 512.2 m is converted energy is spent to raise the potential energy of hoop and remaining is in the form of kinetic and rotational energy

$\Delta E=676m-512.2m=163.8m$

Therefore

$163.8 m=mv^2$

$v=\sqrt{163.8}$

$v=12.798\approx 12.8\ ft/s$

7 0

3 years ago

A boy reaches out of a window and tosses a ball straight up with a speed of 10 m/s. The ball is 20 m above the ground as he rele

kirill115 [55]

Explanation:

It is given that,

Speed of the ball, v = 10 m/s

Initial position of ball above ground, h = 20 m

(a) Let H is the maximum height reached by the ball. It can be calculated using the conservation of energy as :

$\dfrac{1}{2}mv^2=mgh'$

$h'=\dfrac{v^2}{2g}$

$h'=\dfrac{(10)^2}{2\times 9.8}$

h' = 5.1 m

The maximum height above ground,

H = 5.1 + 20

H = 25.1 meters

So, the maximum height reached by the ball is 25.1 meters.

(b) The ball's speed as it passes the window on its way down is same as the initial speed i.e. 10 m/s.

Hence, this is the required solution.

6 0

3 years ago

Two positive charges of 20 micro coulomb and 100 micro coulomb and the distance between them is 150cm.What will be the electrica

Roman55 [17]

Answer:

0.8 N

Explanation:

From coulomb's law,

Formula:

F = kqq'/r²........................ Equation 1

Where F = Force of repulsion, k = coulomb's constant, q = first positive charge, q' = second positive charge, r = distance between the charge.

Given: q = 20 μC = 20×10⁻⁶ C, q' = 100 μC = 100×10⁻⁶ C, r = 150 cm = 1.5 m.

Constant: k = 9×10⁹ Nm²/C²

Substitute these values into equation 1

F = (20×10⁻⁶ )( 100×10⁻⁶)(9×10⁹)/1.5²

F = 1800×10⁻³/2.25

F = 1.8/2.25

F = 0.8 N

3 0

3 years ago

4.77 Augment the rectifier circuit of Problem 4.70 with a capacitor chosen to provide a peak-to-peak ripple voltage of (i) 10% o

goblinko [34]

The question incomplete! The complete question along with answer and explanation is provided below.

Question:

Augment the rectifier circuit of Problem 4.68 with a capacitor chosen to provide a peak-to-peak ripple voltage of (i) 10% of the peak output and (ii) 1% of the peak output. In each case:

(a) What average output voltage results?

(b) What fraction of the cycle does the diode conduct?

(d) What is the peak diode current?

Problem 4.68:

A half-wave rectifier circuit with a 1-kΩ load operates from a 120-V (rms) 60-Hz household supply through a 10-to-1 step-down transformer. It uses a silicon diode that can be modeled to have a 0.7-V drop for any current.

Given Information:

Input voltage = 120 Vrms

10 to 1 step-down transformer

Voltage drop at diode = 0.7 V

Load resistance = R = 1 kΩ

Required Information:

(i) 10% of the peak output and (ii) 1% of the peak output. In each case:

(a) What average output voltage results?

(b) What fraction of the cycle does the diode conduct?

(d) What is the peak diode current?

Answer:

Case (i)

Vavg = 15.45 V

Conduction of diode = 7.11 %

Iavg = 0.232 A

Ip = 0.449 A

Case (ii)

Vavg = 16.18 V

Conduction of diode = 2.25 %

Iavg = 0.735 A

Ip = 1.453 A

Explanation:

Voltage at the secondary side of the transformer is

Vrms = Vpri/turn ratio

Vrms = 120/10 = 12 V

The relation between rms voltage and peak voltage is

Vp = Vrms/√2

Vp = 12√2 = 16.97 V

Vd = 0.7 V

First we will calculate all the required parameters for the 10% ripple voltage and then for 1% ripple voltage.

case (i) 10% of the peak output:

(a) What average output voltage results?

Average output voltage = Vavg = Vp - Vd - 0.5Vr

Where Vp is the peak output voltage Vd is the voltage drop of diode and Vr is the ripple voltage which is given as a percentage of Vp

Vavg = Vp - Vd - 0.5Vr

Vavg = 16.97 - 0.7 - 0.5[0.1(16.97 - 0.7)]

Vavg = 15.45 V

(b) What fraction of the cycle does the diode conduct?

ω = √2Vr/Vp - Vd

ω = √2*0.1(Vp-Vd)/Vp - Vd

ω = √2*0.1(16.97-0.7)/16.97 - 0.7

ω = 0.447 rad

Conduction of diode = (ω/2π)*100

Conduction of diode = (0.447/2π)*100

Conduction of diode = 7.11 %

(c) What is the average diode current?

Average current = Iavg = Vavg/R[ 1 + π( √2(Vp - Vd)/0.1(Vp-Vd))]

Average current = Iavg = 15.45/1000[ 1 + π( √2(16.97 - 0.7)/0.1(16.97-0.7))]

Average current = Iavg = 0.232 A

(d) What is the peak diode current?

Peak current = Ip = Vavg/R[ 1 + 2π( √2(Vp - Vd)/0.1(Vp-Vd))]

Peak current = Ip = 15.45/1000[ 1 + 2π( √2(16.97 - 0.7)/0.1(16.97-0.7))]

Peak current = Ip = 0.449 A

case (ii) 1% of the peak output:

(a) What average output voltage results?

Vavg = 16.97 - 0.7 - 0.5[0.01(16.97 - 0.7)]

Vavg = 16.18 V

(b) What fraction of the cycle does the diode conduct?

ω = √2*0.01(Vp-Vd)/Vp - Vd

ω = √2*0.01(16.97-0.7)/16.97 - 0.7

ω = 0.1417 rad

Conduction of diode = (0.1417/2π)*100

Conduction of diode = 2.25 %

(c) What is the average diode current?

Average current = Iavg = 16.18/1000[ 1 + π( √2(16.97 - 0.7)/0.01(16.97-0.7))]

Average current = Iavg = 0.735 A

(d) What is the peak diode current?

Peak current = Ip = 16.18/1000[ 1 + 2π( √2(16.97 - 0.7)/0.01(16.97-0.7))]

Peak current = Ip = 1.453 A

3 0

3 years ago