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Eva8 [605]
3 years ago
8

Jack and George meet in the middle of a lake while paddling in their canoes. They come to a complete stop and talk for a while.

When they are ready to leave, Jack pushes George's canoe with a force F to separate the two canoes. What is correct to say about the final momentum and kinetic energy of the system if we can neglect any resistance due to the water?
Physics
1 answer:
spayn [35]3 years ago
4 0

Answer:

momentum =0

kinetic energy >0, positive

Explanation:

The final momentum is zero but the final kinetic energy is positive. The momentum is zero because momentum is the sum of the mass×velocity for each component. momentum is conserved

The KE is positive because KE is the sum of the 1/2×mass×velocity^2.

Whereas velocity can be positive and negative since it is (it's what makes the momentum zero because on is positive and one is negative), the velocity squared will always be positive.

Adding together two positives will always be a positive number.

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3 years ago
An object weighs 63.8 N in air. When it is suspended from a force scale and completely immersed in water the scale reads 16.8 N.
I am Lyosha [343]

Answer:

The density of this object is approximately 1.36\; {\rm kg \cdot L^{-1}}.

The density of the oil in this question is approximately 0.600\; {\rm kg \cdot L^{-1}}.

(Assumption: the gravitational field strength is g =9.806\; {\rm N \cdot kg^{-1}})

Explanation:

When the gravitational field strength is g, the weight (\text{weight}) of an object of mass m would be m\, g.

Conversely, if the weight of an object is (\text{weight}) in a gravitational field of strength g, the mass m of that object would be m = (\text{weight}) / g.

Assuming that g =9.806\; {\rm N \cdot kg^{-1}}. The mass of this 63.8\; {\rm N}-object would be:

\begin{aligned} \text{mass} &= \frac{\text{weight}}{g} \\ &= \frac{63.8\; {\rm N}}{9.806\; {\rm N \cdot kg^{-1}}} \\ &\approx 6.506\; {\rm kg}\end{aligned}.

When an object is immersed in a liquid, the buoyancy force on that object would be equal to the weight of the liquid that was displaced. For instance, since the object in this question was fully immersed in water, the volume of water displaced would be equal to the volume of this object.

When this object was suspended in water, the buoyancy force on this object was (63.8\; {\rm N} - 16.8\; {\rm N}) = 47.0\; {\rm N}. Hence, the weight of water that this object displaced would be 47.0 \; {\rm N}.

The mass of water displaced would be:

\begin{aligned}\text{mass} &= \frac{\text{weight}}{g} \\ &= \frac{47.0\: {\rm N}}{9.806\; {\rm N \cdot kg^{-1}}} \\ &\approx 4.793\; {\rm kg}\end{aligned}.

The volume of that much water (which this object had displaced) would be:

\begin{aligned}\text{volume} &= \frac{\text{mass}}{\text{density}} \\ &\approx \frac{4.793\; {\rm kg}}{1.00\; {\rm kg \cdot L^{-1}}} \\ &\approx 4.793\; {\rm L}\end{aligned}.

Since this object was fully immersed in water, the volume of this object would be equal to the volume of water displaced. Hence, the volume of this object is approximately 4.793\; {\rm L}.

The mass of this object is 6.50\; {\rm kg}. Hence, the density of this object would be:

\begin{aligned} \text{density} &= \frac{\text{mass}}{\text{volume}} \\ &\approx \frac{6.506\; {\rm kg}}{4.793\; {\rm L}} \\ &\approx 1.36\; {\rm kg \cdot L^{-1}} \end{aligned}.

(Rounded to \text{$3$ sig. fig.})

Similarly, since this object was fully immersed in oil, the volume of oil displaced would be equal to the volume of this object: approximately 4.793\; {\rm L}.

The weight of oil displaced would be equal to the magnitude of the buoyancy force: 63.8\; {\rm N} - 35.6\; {\rm N} = 28.2\; {\rm N}.

The mass of that much oil would be:

\begin{aligned}\text{mass} &= \frac{\text{weight}}{g} \\ &= \frac{28.2\: {\rm N}}{9.806\; {\rm N \cdot kg^{-1}}} \\ &\approx 2.876\; {\rm kg}\end{aligned}.

Hence, the density of the oil in this question would be:

\begin{aligned} \text{density} &= \frac{\text{mass}}{\text{volume}} \\ &\approx \frac{2.876\; {\rm kg}}{4.793\; {\rm L}} \\ &\approx 0.600\; {\rm kg \cdot L^{-1}} \end{aligned}.

(Rounded to \text{$3$ sig. fig.})

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Answer:

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