How much heat is required to convert 0.3 kilogram of ice at 0°C to water at the same temperature? A. 334,584 J B. 167,292 J C. 100,375 J D. 450,759 J


Answer 1


Option C is the correct answer.


Heat required to melt solid in to liquid is calculated using the formula

            H = mL, where m is the mass and L is the latent heat of fusion.

Latent heat of fusion for water = 333.55 J/g

Mass of ice = 0.3 kg = 300 g

Heat required to convert 0.3 kilogram of ice at 0°C to water at the same temperature

          H = mL = 300 x 333.55 = 100,375 J

Option C is the correct answer.

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" A sound wave moving through air consists of alternate regions of high pressure and low pressure. If the frequency of the sound is increased, what happens, if anything, to the distance between successive high-pressure regions, and why


Answer: wavelength will reduce

Explanation: The region of low pressure is know as the rarefraction region while the region of high pressure is the compression region.

The distance between 2 successive rarefraction or compression region is known as the wavelength.

Now the question is concerned about what an increase in frequency will cause to wavelength.

The speed of sound in air is a constant and it is approximately 343 m/s.

But recall that v = fλ

By assuming a fixed value for speed (v), we have that

k = fλ

Hence, f = k/ λ

This implies that at a fixed wave speed, the wavelength and frequency have an inverse relationship.

An increase in frequency will bring about a reduction in wavelength.

We showed that the length of the pendulum of period 2.000 seconds on the Earth’s surface was 0.99396 meters. What period would this same pendulum have on the surface of Mars? What length would the pendulum be in order to have a period of 2.000 seconds?


To solve this problem it is necessary to apply the concepts related to the Period based on gravity and length.

Mathematically this concept can be expressed as

T= 2\pi \sqrt{(l)/(g)}


l = Length

g = Gravitational acceleration

First we will find the period that with the characteristics presented can be given on Mars and then we can find the length of the pendulum at the desired time.

The period on Mars with the given length of 0.99396m and the gravity of the moon (approximately 1.62m / s ^ 2) will be

T= 2\pi \sqrt{(l)/(g)}

T= 2\pi \sqrt{(0.99396)/(1.62)}

T = 4.921seg

For the second question posed, it would be to find the length so that the period is 2 seconds, that is:

T= 2\pi \sqrt{(l)/(g)}

2= 2\pi \sqrt{(l)/(1.62)}

l = 0.16414m

Therefore, we can observe also that the shorter distance would be the period compared to the first result given.

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Describe the objects that make up Saturn's rings. Your answer should include the range of sizes of objects in the rings, and the composition of the at least the outer layers of the objects.


Saturn's rings are made of billions of pieces of ice, dust and rocks. Some of these particles are as small as a grain of salt, while others are as big as houses.

An electron has a velocity of 3.2 x 10^6 m/s. What is its’ momentum? (b) What is its’ wavelength? (c) What other objects/materials have this space/size? (d) Assuming that we can measure the velocity to an accuracy of 10%. Use the Heisenberg uncertainty principle to calculate the uncertainty in the position.



P = 2.91*10^{-24} kg m/s

\lambda = 2.73 *10^(-10) m

size of atom hat lie in range of 1 to 5 Angstrom

\Delta x = 0.2272 Angstrom



p = mv

where m is mass of electron

so momentum p can be calculated as

p = 9.11*10^{-31} *3.2*10^{6}

P = 2.91*10^{-24} kg m/s

b) wavelength

\lambda = (h)/(mv)

where h is plank constant

so\lambda = (6.626*10^(-34))/(2.91*10^(-24))

\lambda = 2.73 *10^(-10) m

c) size of atom hat lie in range of 1 to 5 Angstrom

d) from the information given in the question we have

(\Delta v)/(v) = 0.1

\Delta v = 0.1 v

we know that

\Delta p *\Delta x = (h)/(4\pi)

m \Delta v \Delta x =(h)/(4\pi)

\Delta x = (h)/(m \Delta v)

\Delta x  = (2.272)/(0.1)                      [\Delta v = 0.1 v]

\Delta x = 0.2272 Angstrom

Overnight a thin layer of ice forms on the surface of a 40-ft-wide river that is essentially of rectangular cross-sectional shape. Under these conditions, the flow depth is 3 ft. During the following day the sun melts the ice cover. Determine the new depth if the flowrate remains the same and the surface roughness of the ice is essentially the same as that for the bottom and sides of the river.



the new depth is 2.3 ft


the solution is in the attached Word file

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