Answer:
t = 16.94 s
Explanation:
t is the time passes before police catch the speeder
speed of speeder Vo = V = 23.3 m/s
T = t
Police Info
Vo = 0 m/s
a = 2.75 m/s^2
t = t
Now,
displacement of the police car = displacement of the speeder.
x_{police} = Vo *t + 1/2 at^2
since Vo = 0
x police = 1/2 at^2
x police = 1/2 (2.75)(t)^2
Now the displacement of speeder is
x_{speeder} = Vt
x_{speeder} = 23.3 t
x_{speeder} = x_{police}
23.3 t = 1/2 * 2.75 t^2
23.3 t = 1.375 t^2
t = 23.3\1.375
t = 16.94
t = 16.94 s
Answer: 3.4s
Explanation:
There are three stages in the motion of the ball, so you have to calculate the times for every stage.
1) Ball dropping from 9.5m: free fall
d = Vo + gt² / 2
Vo = 0 ⇒ d = gt² / 2 ⇒ t² = 2d / g = 2 × 9.5 m / 9.81 m/s² = 1.94 s²
⇒ t = √ (1.94 s²) = 1.39s
2) Ball rising 5.7m (vertical rise)
i) Determine the initial speed:
Vf² = Vo² - 2gd
Vf² = 0 ⇒ Vo² = 2gd = 2 × 9.81 m/s² × 5.7m = 111.8 m²/s²
⇒ Vo = 10.6 m/s
ii) time rising
Vf = Vo - gt
Vf = 0 ⇒ Vo = gt ⇒
t = Vo / g = 10.6 m/s / 9.81 m/s² = 1.08 s
3) Ball dropping from 5.7 m to 1.20m above the pavement (free fall)
i) d = 5.7m - 1.20m = 4.5m
ii) d = gt² / 2 ⇒ t² = 2d / g = 2 × 4.5 m / 9.81 m/s² = 0.92 s²
⇒ t = √ (0.92 s²) = 0.96s
4) Total time
t = 1.39s + 1.08s + 0.96s = 3.43s ≈ 3.4s
Well i had the same question on my test, and when the graph in 2.5 seconds goes up from the equilibrium it reaches a positive maximum, so that would be your answer.
b. positive maximum
Infrared radiation <span>is sensed by human skin as warmth.</span>
