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umka21 [38]
2 years ago
9

A uniform soda can of mass 0.140 kg is 12.0 cm tall and filled with 0.354 kg of soda (Fig. 9-41). Then small holes are drilled i

n the top and bottom (with negligi- ble loss of metal) to drain the soda. What is the height h of the com of the can and contents
a. initially
b. after the can loses all the soda?
c. What happens to h as the soda drains out?
d. If x is the height of the remaining soda at any given instant, find x when the com reaches its lowest point .
Physics
1 answer:
Ugo [173]2 years ago
3 0

Answer:

Explanation:

a ) The centre of mass of a can filled completely with any liquid of any mass lies at the center of the can or at height h /2 where h is height of the can . So centre of mass of can of mass .140 kg of height 12 cm will lie at height 6 cm . Similarly center of mass of its content will also lie at height 6 cm that is center  point of the content .

Hence overall center of mass of can along with its content will lie at height 6 cm initially  .

b )

After the can loses all the soda , can becomes empty . Hence its center of mass will lie at its center again  .

c ) Its center of mass remains unchanged as the soda drains out completely .

d ) When soda starts draining out of the can , its lower part becomes heavier than the upper part . Hence its center of mass moves downwards . Its height is lowered . As height of soda is lowered in the can , center of mass also is lowered . It reaches its lowest point when soda in the can reaches half the height or when the can is half filled .

After that center of mass again starts rising upwards until the can is completely empty .

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Where the bold letters indicate vectors, F is the force, m the mass and the acceleration of the body.

They indicate that the drag force on the filters is

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Where b is a positive constant that depends on the shape and area of ​​the filter and v is the speed of the filter.

Let's write Newton's second law.

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In the experiment the students indicate that they can measure the position, velocity and acceleration of the body.

Based on the above, if we place several filter weights and measure their speed and acceleration in each case, we can make a graph of velocity versus acceleration, we can take the value of the constant b from the slope.

Let's analyze the different answers:

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B) False. The constant depends on the area, so it must be kept constant.

C) May be. In this case, since we have the terminal velocity, the acceleration is zero, Newton's second law remains.

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The problem with the method is the difficulty of measuring the compression terminal velocity.

D) True. According to the discussion of the velocity versus acceleration. graph, the constant b is equal to the slope of the graph.

E) May be. The problem with this method is finding a reliable value for the terminal velocity.

In conclusion, using Newton's second law and graphical analysis we can find the correct answer for the measure of the constant b is:

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Question 26 suppose that a constant force is applied to an object. newton's second law of motion states that the acceleration of
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<span>The answer is 6 kg the mass of the second object. By using Inversely proportional formula it means that (14 kg) (3 m/s</span>²<span>) = M (7 m/s</span>²<span>). Where M is the mass of the second object. For the Newton’s second law of motion formula which is: Force = mass x acceleration, we have:</span>

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Answer:

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