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A 2.5 g marshmallow is placed in one end of a 40 cm pipe, as shown in the figure above. A person blows into the left end of the pipe to eject the marshmallow from the right end. The average net force exerted on the marshmallow while it is in the pipe is 0.7 N. The speed of the marshmallow as it leaves the pipe is most nearly: Ans: 15m/s.

User Wahwahwah
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Answer:

Step-by-step explanation:

To determine the speed of the marshmallow as it leaves the pipe, we can apply the principle of conservation of energy.

The average net force exerted on the marshmallow can be related to the work done on it. The work done on an object is equal to the change in its kinetic energy. In this case, the work done on the marshmallow is equal to the product of the net force and the distance over which the force is applied:

Work = Force × Distance

Given that the average net force exerted on the marshmallow is 0.7 N and the distance over which the force is applied is 40 cm (0.4 m), we can calculate the work done on the marshmallow:

Work = 0.7 N × 0.4 m

= 0.28 J

The work done on the marshmallow is equal to its change in kinetic energy. Assuming the marshmallow starts from rest, the initial kinetic energy is zero. Therefore, the work done on the marshmallow is equal to its final kinetic energy:

0.28 J = (1/2) × mass × velocity^2

We are given the mass of the marshmallow as 2.5 g (0.0025 kg), so we can rearrange the equation to solve for velocity:

velocity^2 = (2 × 0.28 J) / 0.0025 kg

velocity^2 = 224 m^2/s^2

Taking the square root of both sides gives us the velocity of the marshmallow as it leaves the pipe:

velocity = √(224 m^2/s^2)

velocity ≈ 14.97 m/s

Rounding to the nearest meter per second, the speed of the marshmallow as it leaves the pipe is approximately 15 m/s.

User Vidstige
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