Answer
given,
v = (6 t - 3 t²) m/s
we know,
position of the particle
integrating both side
x = 3 t² - t³
Position of the particle at t= 3 s
x = 3 x 3² - 3³
x = 0 m
now, particle’s deceleration
a = 6 - 6 t
at t= 3 s
a = 6 - 6 x 3
a = -12 m/s²
distance traveled by the particle
x = 3 t² - t³
at t = 0 x = 0
t = 1 s , x = 3 (1)² - 1³ = 2 m
t = 2 s , x = 3(2)² - 2³ = 4 m
t = 3 s , x = 0 m
total distance traveled by the particle
D = distance in 0-1 s + distance in 1 -2 s + distance in 2 -3 s
D = 2 + 4 + 2 = 8 m
average speed of the particle
Answer: 1477.78 N
Explanation:
Let's assume that the cross sectional area of the smaller piston be A1
let's also assume the cross sectional area of the larger piston be A2
We assume the force applied to the smaller piston be F1
We also assume the force applied to the larger piston be F2
we then use the formula
F1/A1 = F2/A2
From our question,
The radius of the smaller piston is 5 cm = 0.05 m
The radius of the larger piston is 15 cm = 0.15 m
The force of the larger piston is 13300 N
The force of the smaller piston is unknown = F
A1 = πr² = 3.142 * 0.05² = 0.007855 m²
A2 = πr² = 3.142 * 0.15² = 0.070695 m²
F1/0.007855 = 13300/0.070695
F1 = (13300 * 0.007855) / 0.070695
F1 = 104.4715 / 0.070695
F1 = 1477.78 N
Thus, the force the compressed air must exert is 1477.78 N
Answer:
225 meters
Explanation:
45*5 or 225 meters.
Hope this helps plz hit the crown :D
