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3 years ago

Describe in short the working system of a water pump

1 answer:

5 0

The pump is powered by an electric motor that drives an impeller, or centrifugal pump. The impeller moves water, called drive water, from the well through a narrow orifice, or jet, mounted in the housing in front of the impeller.

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Convert the volume 8.06 in.3 to m3, recalling that1in. =2.54cmand100cm=1m. Answer in units of m3.

galina1969 [7]

1 in=2.54 cm=(2.54 cm)(1 m/100 cm)=0.0254 m
Therefore:
1 in=0.0254 m
1 in³=(0.0254 m)³=1.6387064 x 10⁻⁵ m³

Therefore:

8.06 in³=(8.06 in³)(1.6387064 x 10⁻⁵ m³ / 1 in³)≈1.321 x 10⁻⁴ m³.

Answer: 8.06 in³=1.321 x 10⁻⁴ m³

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3 years ago

Solve 3.4 = 5.1(3.7x + 4.7) for x.

solniwko [45]

Answer:

Explanation:

3.4 = 18.87x + 23.97

collecting like terms

3.4 - 23.97 = 18.87x

-20.57 = 18.87x

dividing both sides by 18.87

x = -20.57/18.87

x= -1.09

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3 years ago

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3 years ago

A century ago, swords were made from various metals, especially steel. The property that makes steel a good choice for sword mak

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An object in space has altitude of 2210 km, velocity of 7000 m/s and flight path angle of 20 degrees. Find the eccentricity, sem

Dmitry [639]

Answer:

Eccentricity = 0.0557

Semi-major axis = 9,095 km

Angular momentum/mass = 60,116 $\frac{km^{2} }{Sec}$

Kinetic energies/mass = 24,500 KJ

Potential energies/mass = 21,673 KJ

Explanation:

Eccentricity

To find the eccentricity use the following formula

Eccentricity = [ Altitude from the earth x ( $Velocity^{2}$ / Gravitational parameter for the Earth ) ] - 1

Where

Altitude from the earth radius = Radius of the earth + Altitude of the earth from radius = 6,378 km + 2,210 km = 8,588 km

Velocity = 7,000 m/s

Gravitational parameter for the Earth = 3.986004418 × $10^{14}$

Eccentricity = ?

Placing values in the formula

Eccentricity = [ ( 8,588 km x 1000 ) x ( $7000^{2}$ / 3.986004418 × $10^{14}$ ) ] - 1

Eccentricity = 0.0557

Semi-major axis

Total Distance = Semi-major axis x ( 1 - Eccentricity )

Where

Total Distance = 8,588 km

Eccentricity = 0.0557

8,588 x 1,000 m = Semi-major axis x ( 1 - 0.0557 )

8,588,000 m = Semi-major axis x 0.9443

Semi-major axis = 8,588,000 m / 0.9443

Semi-major axis = 9,094,567.40 m

Semi-major axis = 9,094.56740 km

Semi-major axis = 9,095 km

Angular momentum/mass

L = MVR

L/M = VR

Where

V = Velocity = $\frac{7,000 m/s}{1000}$ = 7 km/s

R = Total Radius = Radius of Earth + Altitude = 6,378 km + 2,210 km = 8,588 km

Placing values in the formula

L/M = 7 km/s x 8,588 km = 60,116 $\frac{km^{2} }{Sec}$

Kinetic energies/mass

Ke = I $W^{2}$

Ke = $\frac{1}{2} mr^{2}$ $W^{2}$ ( Where I = $mr^{2}$ )

$\frac{ke}{m}$ = $\frac{r^2}{2}$ $W^{2}$

$\frac{ke}{m}$ = $\frac{r^2}{2}$ $\frac{V^2}{r^2}$ ( $W^{2}$ = $\frac{V^2}{r^2}$ )

$\frac{ke}{m}$ = $\frac{V^2}{2}$

$\frac{ke}{m}$ = $\frac{7000^2}{2}$

$\frac{ke}{m}$ = 24,500,000 J

$\frac{ke}{m}$ = 24,500 KJ

Potential energies/mass

PE = mgh

$\frac{PE}{m}$ = gh

Where

g = 9.807 $\frac{m}{s^2}$

h = 2,210 km x 1,000 = 2,210,000 m

Placing values in the formula

$\frac{PE}{m}$ = 9.807 $\frac{m}{s^2}$ x 2,210,000 m

$\frac{PE}{m}$ = 21,673,470 J

$\frac{PE}{m}$ = 21,673.470 KJ

$\frac{PE}{m}$ = 21,673 KJ

8 0

3 years ago

Describe in short the working system of a water pump ​

Describe in short the working system of a water pump