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Contact [7]
3 years ago
10

I REALLY NEED HELP PLEASE!!! CAN SOMEONE ANSWER THIS? At 0 ºC, some amount of energy is required to change 1 kg of water from a

solid into a liquid. If you had a 2 kg piece of ice, what effect would this have on the amount of thermal energy required to change the water from a solid to a liquid? A)It would require energy to be removed from the 2 kg piece of ice. The larger piece of ice already has more total energy than the smaller piece of ice, so energy must be removed for the ice to become liquid. B)It would require less energy to change solid water into liquid water because the energy would spread through the ice more quickly and the ice already has a larger total amount of energy because it is larger than a 1 kg piece of ice. C)It would require more energy to change solid water into liquid water because there are more molecules in this larger piece of ice. D)It would still require the same amount of energy to change solid water into liquid water because the entire piece of ice would still gain the same amount of energy in each case.
Physics
2 answers:
Nezavi [6.7K]3 years ago
8 0

Answer:

C)It would require more energy to change solid water into liquid water because there are more molecules in this larger piece of ice.

Charra [1.4K]3 years ago
4 0

Explanation:

c. it would require more energy to change ........

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A) x=\pm \frac{A}{2\sqrt{2}}

The total energy of the system is equal to the maximum elastic potential energy, that is achieved when the displacement is equal to the amplitude (x=A):

E=\frac{1}{2}kA^2 (1)

where k is the spring constant.

The total energy, which is conserved, at any other point of the motion is the sum of elastic potential energy and kinetic energy:

E=U+K=\frac{1}{2}kx^2+\frac{1}{2}mv^2 (2)

where x is the displacement, m the mass, and v the speed.

We want to know the displacement x at which the elastic potential energy is 1/3 of the kinetic energy:

U=\frac{1}{3}K

Using (2) we can rewrite this as

U=\frac{1}{3}(E-U)=\frac{1}{3}E-\frac{1}{3}U\\U=\frac{E}{4}

And using (1), we find

U=\frac{E}{4}=\frac{\frac{1}{2}kA^2}{4}=\frac{1}{8}kA^2

Substituting U=\frac{1}{2}kx^2 into the last equation, we find the value of x:

\frac{1}{2}kx^2=\frac{1}{8}kA^2\\x=\pm \frac{A}{2\sqrt{2}}

B) x=\pm \frac{3}{\sqrt{10}}A

In this case, the kinetic energy is 1/10 of the total energy:

K=\frac{1}{10}E

Since we have

K=E-U

we can write

E-U=\frac{1}{10}E\\U=\frac{9}{10}E

And so we find:

\frac{1}{2}kx^2 = \frac{9}{10}(\frac{1}{2}kA^2)=\frac{9}{20}kA^2\\x^2 = \frac{9}{10}A^2\\x=\pm \frac{3}{\sqrt{10}}A

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Hello there.

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sp2606 [1]

Answer:

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Density varies with temperature

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Density may have units of grams per milliliter (g/mL)

Explanation:

Density D is a characteristic property of a substance or material and is defined as the relationship between the mass m of a body or substance and the volume V it occupies:

D=\frac{m}{V}

This means the density is inversely proportional to the volume.

On the other hand, density is a scalar quantity and according to the International System of Units its unit is D=\frac{kg}{m^{3}} , although it can be also expressed in \frac{g}{ml}.

It should be noted that the density of a body is related to its buoyancy, a substance or body will float on another fluid if its density is lower. In addition, if the pressure of the substance remains constant, as the temperature increases, the density decreases; this means density varies with the temperature as well.

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Assuming the accleration applied was constant, we have

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F=ma=(0.45\,\mathrm{kg})\left(200\,\dfrac{\mathrm m}{\mathrm s^2}\right)

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