Answer:
θ = ω0 t + 1/2 ω t^2 = 1/2 α t^2 since it starts from rest
α = (15000 / 60) / 240 = 250 / 240 = 1.04 rev/sec^2
Note that α is in rev/sec^2
T(otal) = 1/2 * 1.04 * 240^2 = 30,000 revolutions
Check:
Average RPM = (15000/2) = 7500 rev/min = 125 / sec
125 / sec * 240 sec = 30,000 rev
Complete question is: While running, a person dissipates about 0.60 J of mechanical energy per step per kilogram of body mass. If a 60-kg person develops a power of 70 W during a race, how fast is the person running? (Assume a running step is 1.5 m long).
Answer: The person running at a speed of 2.91 m/s.
Explanation:
Given: Mass of runner = 60 kg
Runner dissipates = 0.6 J/kg per step
Average power = 70 W
1 step = 1.50 m
Energy dissipated by the runner is as follows.
Formula used to calculate the value of one step 'S' is as follows.
It is known that average velocity is equal to the total distance divided by time interval.
So, total distance for the given situation is as follows.
Hence, speed of the person is calculated as follows.
Thus, we can conclude that the person running at a speed of 2.91 m/s.
Answer:
This is Newton’s third law (forces due to friction). There would be 15N being pushed back at Jane.
Explanation:
Newton’s third law states that for every action there is an equal and opposite reaction. So if Jane pushed with a force of 15N, 15N would be pushed back at her. Hope this helps! :)
Fromula for calculating mass when density and volume are given is as m=ρV mass of blood = 1.05 × 2 mass of bag of blood = 2.15 kg
Answer:
O I'm a goofy goober yeah you're a goofy goober yeah we're all goofy goobers yeah goofy goofy goobers yeah
Explanation: