Answer:
Part A) the angular acceleration is α= 44.347 rad/s²
Part B) the angular velocity is 195.13 rad/s
Part C) the angular velocity is 345.913 rad/s
Part D ) the time is t= 7.652 s
Explanation:
Part A) since angular acceleration is related with angular acceleration through:
α = a/R = 10.2 m/s² / 0.23 m = 44.347 rad/s²
Part B) since angular acceleration is related
since
v = v0 + a*(t-t0) = 51.0 m/s + (-10.2 m/s²)*(3.4 s - 2.8 s) = 44.88 m/s
since
ω = v/R = 44.88 m/s/ 0.230 m = 195.13 rad/s
Part C) at t=0
v = v0 + a*(t-t0) = 51.0 m/s + (-10.2 m/s²)*(0 s - 2.8 s) = 79.56 m/s
ω = v/R = 79.56 m/s/ 0.230 m = 345.913 rad/s
Part D ) since the radial acceleration is related with the velocity through
ar = v² / R → v= √(R * ar) = √(0.23 m * 9.81 m/s²)= 1.5 m/s
therefore
v = v0 + a*(t-t0) → t =(v - v0) /a + t0 = ( 1.5 m/s - 51.0 m/s) / (-10.2 m/s²) + 2.8 s = 7.652 s
t= 7.652 s
Answer: Speeding Up
Explanation:
Force is proportional to the acceleration of an object. The greater the force, the more acceleration will be produced.
When the force on the tire is equal to the weight of the car, the car is reaching a stability as a result of increase in motion.
But when the force of the load on the tire is quite large(most likely several times the weight of the car) and is directed forward, then, the car is at high speed.
Explanation:
The runner was 8.6km away from the finish line when the bird starts flying.
Therefore it takes the bird 8.6/14.4 = 0.60 hours for the bird to fly to the finish line.
In that 0.60 hours, the runner would have ran an extra 3.6km/h * 0.6h = 2.16km.
Now, the runner and the bird are flying towards each other. The distance between them is 8.6 - 2.16 = 6.44km and their combined speed is 18.0km.
Hence, they will meet in 6.44/18.0 = 0.36 hours.
Overall, the bird flew for 0.60 + 0.36 = 0.96 hours, and flew 14.4km/h * 0.96h = 13.8km.
The answer is diffraction or interference
