The amount of work that has been done on the projectile by air friction is 7,569.2 J.
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Work done by air friction on the projectile</h3>
The work done by air friction on the projectile is calculated as follows;
W = ΔK.E
W = ¹/₂mvf² - ¹/₂mv₀²
where;
- vf is the final velocity = 85 m/s
- v₀ is the initial vertical velocity = 100 x sin(20) = 34.2 m/s
W = ¹/₂m(vf² - v₀²)
W = ¹/₂ x 2.5 (85² - 34.2²)
W = 7,569.2 J
Thus, the amount of work that has been done on the projectile by air friction is 7,569.2 J.
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Complete Question
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Answer:
The pressure difference of the first bubble is
The pressure difference of the second bubble is
The pressure difference on the second bubble is higher than that of the first bubble so when the valve is opened pressure from second bubble will cause air to flow toward the first bubble making is bigger
Explanation:
From the question we are told that
The radius of the first bubble is
The radius of the second bubble is
The surface tension of the soap solution is
Generally according to the Laplace's Law for a spherical membrane the pressure difference is mathematically represented as
Now the pressure difference for the first bubble is mathematically evaluated as
substituting values
Now the pressure difference for the second bubble is mathematically evaluated as
One way is to apply a surface the produces more friction (like sandpaper)