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liberstina [14]
3 years ago
5

An astronaut on Earth notes that in her soft drink an ice cube floats with 9/10 of its volume submerged. If she were instead in

a lunar module parked on the Moon where the gravitation force is 1/6 that of Earth, the ice in the same soft drink would float:A) with more than 9/10 submerged. B) with 9/10 submerged.C) with 6/10 submerged.D) totally submerged.
Physics
2 answers:
riadik2000 [5.3K]3 years ago
8 0

Answer:

B) with 9/10 submerged

Explanation:

m = mass of ice cube

\rho = density of soft drink

V = Volume of soft drink displaced

ice cube floats in the soft drink when the force of buoyancy on it balances its weight. Force of buoyancy acting on the cube in upward direction is same as the weight of the soft drink displaced. hence we can write

weight of ice cube = weight of soft drink displaced

mg = \rho V g

m = \rho V

we see that the acceleration due to gravity cancel out both side and hence it does affect as astronaut is on earth on in a lunar module.

Stells [14]3 years ago
3 0

Answer:

Explanation:

Volume of ice block submerged om earth = 9 / 10

According to the floating principle

The weight of the body is balanced by the buoyant force acting on the block.

Let V be the total volume of the block and d be the density of ice and ρ be the density of soft drink.

So, V x d x g = 9 / 10 V x ρ x g

So g cancels out from both the sides.

If we place the ice cube in soft drink on the surface of moon, then the volume submerged remains same.

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Explanation:

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          \frac{1}{f} = (n - 1) [\frac{1}{R_{1}} - \frac{1}{R_{2}}]

Since, the lens is immersed in liquid with index of refraction n_{1}. Therefore, focal length obeys the following.  

            \frac{1}{f_{1}} = \frac{n - n_{1}}{n_{1}} [\frac{1}{R_{1}} - \frac{1}{R_{2}}]  

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and,       \frac{n_{1}}{f(n - n_{1})} = \frac{1}{R_{1}} - \frac{1}{R_{2}}

or,          f_{1} = \frac{fn_{1}(n - 1)}{(n - n_{1})}

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Using thin lens equation, we will find the focal length as follows.

             \frac{1}{f} = \frac{1}{s_{o}} + \frac{1}{s_{i}}

Hence, image distance can be calculated as follows.

       \frac{1}{s_{i}} = \frac{1}{f} - \frac{1}{s_{o}} = \frac{s_{o} - f}{fs_{o}}

              s_{i} = \frac{fs_{o}}{s_{o} - f}

             s_{i} = \frac{32.4 \times 100}{100 - 32.4}

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An astronaut finds herself in a predicament in which she has become untethered from her shuttle. She figures that she could get
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Solving to find the final speed, after throwing the object we have

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