BIO A trap-jaw ant snaps its mandibles shut at very high speed, a good trait for catching small prey. But an ant can also slam its mandibles into the ground; the resulting force can launch the ant into the air for a quick escape. A 12 mg ant hits the ground with an average force of 47 mN for a time of 0.13 ms; these are all typical values. At what speed does it leave the ground

Final velocity (v) = 0.509 m/s (Approx)

Explanation:

Ant use impulse power

Given:

Mass of ant = 12 mg = 12 × 10⁻⁶ kg

Average force = 47 mN = 47 × 10⁻³ N

Initial velocity(u) = 0

Time taken = 0.13 ms = 0.13 × 10⁻³ s

Find:

Final velocity (v)

Computation:

Force × Time =  change in momentum

(47 × 10⁻³ N)(0.13 × 10⁻³ s) = mv - mu

(47 × 10⁻³ N)(0.13 × 10⁻³ s) = m(v - u)

6.11 × 10⁻⁶ = 12 × 10⁻⁶(v - 0)

6.11 = 12 v

Final velocity (v) = 0.509 m/s (Approx)

answer: the exterior temperature of commercial jet planes as they leave the stratosphere and return to the ground increases steadily.

explanation:

the lower most layer of atmosphere is troposphere which extends from 0 to 12 km above the surface of the earth. the temperature initially decreases as distance is increased from the earth's surface. commercial jets fly in lower part of the stratosphere. the stratosphere extends upto 50 km. the temperature in stratosphere starts rising again due to presence of ozone which traps the uv rays.

thus, when the planes leave the stratosphere and return to the ground, the exterior temperature increases steadily.