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gavmur [86]
3 years ago
8

Water is pumped steadily out of a flooded basement at a speed of 5.4 m/s through a uniform hose of radius 0.83 cm. The hose pass

es out through a window to a street ditch 3.5 m above the waterline. What is the power of the pump?
Physics
1 answer:
Gala2k [10]3 years ago
6 0

To solve this problem it is necessary to apply the concepts related to the flow as a function of the volume in a certain time, as well as the potential and kinetic energy that act on the pump and the fluid.

The work done would be defined as

\Delta W = \Delta PE + \Delta KE

Where,

PE = Potential Energy

KE = Kinetic Energy

\Delta W = (\Delta m)gh+\frac{1}{2}(\Delta m)v^2

Where,

m = Mass

g = Gravitational energy

h = Height

v = Velocity

Considering power as the change of energy as a function of time we will then have to

P = \frac{\Delta W}{\Delta t}

P = \frac{\Delta m}{\Delta t}(gh+\frac{1}{2}v^2)

The rate of mass flow is,

\frac{\Delta m}{\Delta t} = \rho_w Av

Where,

\rho_w = Density of water

A = Area of the hose \rightarrow A=\pi r^2

The given radius is 0.83cm or 0.83 * 10^{-2}m, so the Area would be

A = \pi (0.83*10^{-2})^2

A = 0.0002164m^2

We have then that,

\frac{\Delta m}{\Delta t} = \rho_w Av

\frac{\Delta m}{\Delta t} = (1000)(0.0002164)(5.4)

\frac{\Delta m}{\Delta t} = 1.16856kg/s

Final the power of the pump would be,

P = \frac{\Delta m}{\Delta t}(gh+\frac{1}{2}v^2)

P = (1.16856)((9.8)(3.5)+\frac{1}{2}5.4^2)

P = 57.1192W

Therefore the power of the pump is 57.11W

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Suppose the electrons and protons in 1g of hydrogen could be separated and placed on the earth and the moon, respectively. Compa
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Answer:

The gravitational force is 3.509*10^17 times larger than the electrostatic force.

Explanation:

The Newton's law of universal gravitation and Coulombs law are:

F_{N}=G m_{1}m_{2}/r^{2}\\F_{C}=k q_{1}q_{2}/r^{2}

Where:

G= 6.674×10^−11 N · (m/kg)2

k =  8.987×10^9 N·m2/C2

We can obtain the ratio of these forces dividing them:

\frac{F_{N}}{F_{C}}=\frac{Gm_{1}m_{2}}{kq_{1}q_{2}}=0.742\times10^{-20}\frac{C^{2}}{kg^{2}}\frac{m_{1}m_{2}}{q_{1}q_{2}}   --- (1)

The mass of the moon is 7.347 × 10^22 kilograms

The mass of the earth is  5.972 × 10^24 kg

And q1=q2=Na*e=(6.022*10^23)*(1.6*10^-19)C=9.635*10^4 C

Replacing these values in eq1:

\frac{F_{N}}{F_{C}}}}=0.742\times10^{-20}\frac{C^{2}}{kg^{2}}\frac{7.347\times5.972\times10^{46}kg^{2}}{(9.635\times10^{4})^{2}}

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7 0
3 years ago
A physics teacher performing an outdoor demonstration suddenly falls from rest off a high cliff and simultaneously shouts help.
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Answer:

H = 532 m

Explanation:

When teacher falls into the cliff then she shout for Help at the same time

so here we know that sound will go down and reflect back up

so here in 3 s distance traveled by the sound

d = vt

d = 340 \times 3

d = 1020 m

now in the same time the distance that teacher will fall down is given as

d_1 = \frac{1}{2}gt^2

d_1 = \frac{1}{2}(9.81)(3^2)

d_1 = 44.1 m

now total distance traveled by teacher and sound in 3 s

d_{total} = d + d_1

d_{total} = 1020 + 44.1

this total distance must be equal to twice the height of the cliff

2H = 1064.1 m

H = 532 m

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