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3 years ago

I NEED ANSWER ASAP

Physics

2 answers:

umka21 [38]3 years ago

7 0

Answer:

97% will sink below the water

Explanation:

Waters density is 1 g/cm3

If an object's density is greater than 1g/cm3, it will sink. If it's less then it would float. 1-0.97 = 0.03. Only 3% of the ice would show while 97 will be under.

IgorLugansk [536]3 years ago

5 0

Answer:

About 97 percent of the ice will sink below water.

Explanation:

Let the volume that is submerged into the water is given as V1

so at equilibrium condition we can say that buoyancy force on it is balanced by weight of the block

so we have

$\pho_{water} V_1 g = \rho_{ice} V g$

now we have

$\frac{V_1}{V} = \frac{\rho_{ice}}{\rho_{water}}$

$\frac{V_1}{V} = \frac{0.97}{1}$

so the percentage of block that is submerged into the water is given as

volume percent submerged = 97 %

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To what volume will a sample of gas expand if it is heated from 50.0°c and 2.33 l to 500.0°c?

Lena [83]

The volume of the gas at $500^{\circ} \text{C}$ is $\fbox{\begin\ 5.576\,{\text{L}}\\\end{minispace}}$ .

Further Explanation:

Consider the pressure of the gas to be constant.

The change in the volume of the gas when the temperature of the gas is varied by keeping the pressure of the gas at a constant value is defined by the Charles' Law.

Concept:

According to the Charles law, the volume of the gas is directly proportional to the temperature of the gas at constant pressure.

The Charles' law can be stated as:

$\fbox{\begin\\V\propto T\\\end{minispace}}$

The above expression can we written as.

$\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}$

Convert the temperature of the gas into kelvin.

$T=273+T^\circ\text{C}}}$

Here, T is the temperature in kelvin and $T^\circ\text{C}}}$ is the temperature in centigrade.

The initial temperature of the gas is $50^\circ\text{C}$ . The temperature of the gas in kelvin is.

$\begin{aligned}{T_1}&=273+50\\&=323\,{\text{K}}\\\end{aligned}$

The final temperature of the gas is $500^\circ\text{C}$ . The temperature in kelvin is.

$\begin{aligned}{T_2}&=273+500\\&=773\,{\text{K}}\\\end{aligned}$

Substitute the values of temperature and volume in the expression of the Charles' Law.

$\begin{aligned}{V_2}&=\frac{{{T_2}}}{{{T_1}}}{V_1}\\&=\frac{{773\,{\text{K}}}}{{323\,{\text{K}}}}\left({2.33\,{\text{L}}}\right)\\&=5.576\,{\text{L}}\\\end{aligned}$