Jack and Nas decided to have a race from their houses to the park to know who runs the fastest. Jack whose house is 15 kilometers away from the park finished the race in 2 hours while Nas finished in 1.5 hours from a distance of 10 kilometers from the park. What is Jack's speed with unit of measure also And what is Nas's speed also with the unit of measure i dont know if this is physics its just science because i cant find science

the Jack speed is greater than the Nas speed  

   v₁ = 2.08 m / s >  v2 = 1.85 m / s

Explanation:

For this exercise we can find the average speed of each of the boys, the average speed is defined as the displacement in the time interval

          v = x / t

Jack.

It tells us that it travels x = 15 km in a time of t = 2 h

let's reduce the magnitudes to the SI system

     x = 15 km (1000 m / 1 km) = 15 10³ m

     t = 2 h (3600 s / 1 h) = 7200 s

let's calculate

     v₁ = 15 10³/7200

     v₁ = 2.08 m / s

Nas travels a distance of x = 10 km in a time of t = 1.5 h

      x = 10 km = 10 10³ m

      t = 1.5 h (3600s / 1h) = 5400 s

let's calculate the speed

     v2 = 10 10³/5400

      v2 = 1.85 m / s

From these results we can see that the Jack speed is greater than the Nas speed

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.

the total distance traveled by the car is

30km + 49km = 79 km .