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Klio2033 [76]
3 years ago
13

A child slides down a hill on a toboggan with an acceleration of 1.8 m/s^2. If she starts at rest, how far has she traveled in :

(a) 1.0 s, (b) 2.0 s, and (c) 3.0 s?
Physics
1 answer:
DENIUS [597]3 years ago
4 0

Explanation:

It is given that,

The acceleration of the toboggan, a=1.8\ m/s^2

Initial speed of the toboggan, u = 0

We need to find the distance covered by the toboggan. Using the second equation of motion as :

s=ut+\dfrac{1}{2}at^2

At t = 1 s

s=\dfrac{1}{2}\times 1.8\times 1^2

s_1=0.9\ m

At t = 2 s

s=\dfrac{1}{2}\times 1.8\times 2^2

s_2=3.6\ m

At t = 3 s

s=\dfrac{1}{2}\times 1.8\times 3^2

s_3=8.1\ m

Hence, this is the required solution.

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What fraction of an iceberg is submerged? (ρice = 917 kg/m3, ,ρsea = 1030 kg/m3.)
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Choice d. Approximately 89\% of the volume of this iceberg would be submerged.

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Let V_\text{ice} denote the total volume of this iceberg. Let V_\text{submerged} denote the volume of the portion that is under the liquid.

The mass of that iceberg would be \rho_\text{ice} \cdot V_\text{ice}. Let g denote the gravitational field strength (g \approx 9.81\; \rm N \cdot kg^{-1} near the surface of the earth.) The weight of that iceberg would be: \rho_\text{ice} \cdot V_\text{ice} \cdot g.

If the iceberg is going to be lifted out of the sea, it would take water with volume V_\text{submerged} to fill the space that the iceberg has previously taken. The mass of that much sea water would be \rho_\text{sea} \cdot V_\text{submerged}.

Archimedes' Principle suggests that the weight of that much water will be exactly equal to the buoyancy on the iceberg. By Archimedes' Principle:

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The buoyancy on the iceberg should balance the weight of this iceberg. In other words:

\underbrace{\rho_\text{ice} \cdot V_\text{ice} \cdot g}_\text{weight of iceberg} =  \underbrace{\rho_\text{sea} \cdot V_\text{submerged} \cdot g}_\text{buoyancy on iceberg}.

Rearrange this equation to find the ratio between V_\text{submerged} and V_\text{ice}:

\begin{aligned} &\frac{V_\text{submerged}}{V_\text{ice}} \\&= \frac{\rho_\text{ice} \cdot g}{\rho_\text{sea} \cdot g}\\ &= \frac{\rho_\text{ice}}{\rho_\text{sea}}\ = \frac{917\; \rm kg \cdot m^{-3}}{1030\; \rm kg \cdot m^{-3}} \approx 0.89 \end{aligned}.

In other words, 89\% of the volume of this iceberg would have been submerged for buoyancy to balance the weight of this iceberg.

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