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andrey2020 [161]
2 years ago
10

A swimming pool heater has to be able to raise the temperatureof the 40,000 gallons of water in the pool by 10 degress Celsius.H

ow many kilowatt-hours of energy are required? (one gallon ofwater has a mass of approximately 3.8kg and the specific heat ofwater is 4186 J/kg C)
a.) 1960 kWh
b.) 1770 kWh
c.) 330 kWh
d.) 216 kWh
Physics
1 answer:
andrew11 [14]2 years ago
3 0

Answer:b)1770 kWh

Explanation:

Given

volume of water V=40,000 gallons

Temperature rise \Delta T=10^{\circ}C

c=4186 J/kg-^{\circ} C

also 1 kg mass is approximately is 1 gallon

therefore 40,000 gallon is equivalent to 3.8\times 40000 kg

heat Required to raise temperature is

Q=mc\Delta T

Q=3.8\times 40000\times 4186\times 10

Q=63,627.2\times 10^5 J

Q=63,672.2\times 10^2 KJ

Q=1767.42 kWh\approx 1770 kWh

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How much heat is released when 432 g of water cools down from 71'c to 18'c?
maria [59]
The heat released by the water when it cools down by a temperature difference \Delta T is
Q=mC_s \Delta T
where
m=432 g is the mass of the water
C_s = 4.18 J/g^{\circ}C is the specific heat capacity of water
\Delta T =71^{\circ}C-18^{\circ}C=53^{\circ} is the decrease of temperature of the water

Plugging the numbers into the equation, we find
Q=(432 g)(4.18 J/g^{\circ}C)(53^{\circ}C)=9.57 \cdot 10^4 J
and this is the amount of heat released by the water.
7 0
2 years ago
A photon is scattered from an initially stationary electron within a metal. How does the frequency of the photon change upon sca
sammy [17]

Answer:

The frequency of the photon decreases upon scattering

Explanation:

Here we note that when a photon is scattered by a charged particle, it is referred to as Compton scattering.

Compton scattering results in a reduction of the energy of the photon and hence an increase in the wavelength (from λ to λ') of the photon known as Compton effect.

Therefore, since the wavelength increases, we have from

λf = λ'f' = c

f = c/λ

Where:

f and f' = The frequency of the motion of the photon before and after the scattering

c = Speed of light (constant)

We have that the frequency, f, is inversely proportional to the wavelength, λ as follows;

f = c/λ

As λ = increases, and c is constant, f decreases, therefore, the frequency of the photon decreases upon scattering.

5 0
2 years ago
Water flowing through a cylindrical pipe suddenly comes to a section of pipe where the diameter decreases to 86% of its previous
Orlov [11]

Answer:

Explanation:

The speed of the water in the large section of the pipe is not stated

so i will assume 36m/s

(if its not the said speed, input the figure of your speed and you get it right)

Continuity equation is applicable for ideal, incompressible liquids

Q the flux of water that is  Av with A the cross section area and v the velocity,

so,

A_1V_1=A_2V_2

A_{1}=\frac{\pi}{4}d_{1}^{2} \\\\ A_{2}=\frac{\pi}{4}d_{2}^{2}

the diameter decreases 86% so

d_2 = 0.86d_1

v_{2}=\frac{\frac{\pi}{4}d_{1}^{2}v_{1}}{\frac{\pi}{4}d_{2}^{2}}\\\\=\frac{\cancel{\frac{\pi}{4}d_{1}^{2}}v_{1}}{\cancel{\frac{\pi}{4}}(0.86\cancel{d_{1}})^{2}}\\\\\approx1.35v_{1} \\\\v_{2}\approx(1.35)(38)\\\\\approx48.6\,\frac{m}{s}

Thus, speed in smaller section is 48.6 m/s

3 0
1 year ago
The eiffel tower has a mass of 7.3 million kilograms and a height of 324 meters. its base is square with a side length of 125 me
uranmaximum [27]

Since the tower base is square with a side length of  125 m,

Therefore,

(125\ m)^2+ (125\ m)^2=31250 m^2

Square root of 31250 = 176.776953 (Diameter) , so this is the diameter of the cylinder to enclose it, and radius, r = 88.38834765 m and height, h = 324 m.

The volume of cylinder,

=\pi r^2h=3.14(88.38834765 m)^2\times 324 m =7948168.803\ m^3

Thus, the mass of the air in the cylinder,

=1.225\ kg/m^3 \times 7948168.803\ m^3=9736506.78\ kg

Hence, the mass of the air in the cylinder is this more  than the mass of the tower.


4 0
1 year ago
Water flows without friction vertically downward through a pipe and enters a section where the cross sectional area is larger. T
djverab [1.8K]

Answer:

v_{2} will be less than v_{1} and P_{2} will be greater than P_{1}.

Explanation:

As we know from the conservation of mass, the rate at which any amount of fluid mass (m_{1}) is entering in a system is equal to the rate at which the same amount of fluid mass (m_{2}) is leaving the system.

Rate of mass flow can be written as,

m = \rho A v

where \rho is the density of the fluid, A is the area through which the fluid is flowing and v is the velocity of the fluid.

Now, according to the problem, as the density of the fluid does not change, we can write

&& m_{1} = m_{2}\\&or,& \rho A_{1} v_{1} = \rho A_{2} v_{2}\\&or,& \dfrac{v_{2}}{v_{1}} = \dfrac{A_{1}}{A_{2}}

where A_{1} and A_{2} are the cross-sectional areas through which the fluid is passing and v_{1} and v_{2} are the velocities of the fluid through the respective cross-sectional areas.

As according to the problem, A_{2} > A_{1}, so from the above formula v_{2} < v_{1}.

Also we know that fluid pressure is created by the motion of the fluid through any area. When the fluid gains speed, some of its energy is used to move faster in the fluid’s direction of motion. It causes in a lower pressure.

So, as in this case v_{2} < v_{1} the pressure in the large cross-sectional area P_{2} will be greater than the pressure  P_{1} in the small cross sectional area, i.e.,

P_{2} > P_{1}.

6 0
2 years ago
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