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The froghopper (Philaenus spumarius), the champion leaper of the insect world, has a mass of 12.3 mg. It can leave the ground with a speed as high as 4.0 m/s in the vertical direction. The jump itself lasts a mere 1.0 ms before the insect is clear of the ground. Assuming constant acceleration, find the force that the ground exerts on the froghopper during its jump.

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Answer:

0.0492 N force

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

Equation of motion

[tex]v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow a=\frac{4-0}{0.001}\\\Rightarrow a=4000\ m/s^2[/tex]

F=ma

[tex]F=12.3\times 10^{-6}\times 4000\\\Rightarrow F=0.0492\ N[/tex]

The ground exerts 0.0492 N force on the froghopper during its jump.

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