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jeka94
3 years ago
5

A swimming pool has the shape of a box with a base that measures 30 m by 10 m and a uniform depth of 2.5 m. How much work is req

uired to pump the water out of the pool when it is​ full? Use 1000 kg divided by m cubed for the density of water and 9.8 m divided by s squared for the acceleration due to gravity.

Physics
2 answers:
alisha [4.7K]3 years ago
8 0

Explanation:

Below is an attachment containing the solution.

pav-90 [236]3 years ago
4 0

Answer:

18,375 kJ

Explanation:

Parameters given:

Density of water: 1000 kg/m³

Acceleration due to gravity = 9.8 m/s²

Dimensions of pool = 30m × 10m × 2.5m

Depth of pool = 2.5 m

First we need to find the mass of the water in the pool:

Density = mass/volume

Mass = density * volume

Since the pool is full of water, the volume of the water is equal to the volume of the pool:

Volume = 30 * 10 * 2.5 = 750 m³

Mass = 1000 * 750

Mass = 750,000 kg

We can now find the force required to pump the water out of the pool:

F = m * g

Where m = mass of water

g = acceleration due to gravity

F = 750,000 * 9.8 = 7,350,000 N

The work needed to be done is the product of the force required by the distance the water would need to move to be pumped out (depth of pool):

Work = F * d

Work = 7,350,000 * 2.5

Work = 18,375,000 J = 18,375 kJ

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Citrus2011 [14]

A. Lithium

The equation for the photoelectric effect is:

E=\phi + K

where

E=\frac{hc}{\lambda} is the energy of the incident light, with h being the Planck constant, c being the speed of light, and \lambda being the wavelength

\phi is the work function of the metal (the minimum energy needed to extract one photoelectron from the surface of the metal)

K is the maximum kinetic energy of the photoelectron

In this problem, we have

\lambda=190 nm=1.9\cdot 10^{-7}m, so the energy of the incident light is

E=\frac{hc}{\lambda}=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{1.9\cdot 10^{-7} m}=1.05\cdot 10^{-18}J

Converting in electronvolts,

E=\frac{1.05\cdot 10^{-18}J}{1.6\cdot 10^{-19} J/eV}=6.5 eV

Since the electrons are emitted from the surface with a maximum kinetic energy of

K = 4.0 eV

The work function of this metal is

\phi = E-K=6.5 eV-4.0 eV=2.5 eV

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B. cesium, potassium, sodium

The wavelength of green light is

\lambda=510 nm=5.1\cdot 10^{-7} m

So its energy is

E=\frac{hc}{\lambda}=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{5.1\cdot 10^{-7} m}=3.9\cdot 10^{-19}J

Converting in electronvolts,

E=\frac{3.9\cdot 10^{-19}J}{1.6\cdot 10^{-19} J/eV}=2.4 eV

So, all the metals that have work function smaller than this value will be able to emit photoelectrons, so:

Cesium

Potassium

Sodium

C. 4.9 eV

In this case, we have

- Copper work function: \phi = 4.5 eV

- Maximum kinetic energy of the emitted electrons: K = 2.7 eV

So, the energy of the incident light is

E=\phi+K=4.5 eV+2.7 eV=7.2 eV

Then the copper is replaced with sodium, which has work function of

\phi = 2.3 eV

So, if the same light shine on sodium, then the maximum kinetic energy of the emitted electrons will be

K=E-\phi = 7.2 eV-2.3 eV=4.9 eV

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Answer:

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b)153.06kg

Explanation:

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g(earth) =6g(moon)

F(gearth) = mg(earth)

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