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Answer:
a) v = 7,207 m / s
, b) a = 42.3 m / s²
Explanation:
We will solve this exercise using the concept of mechanical energy, We will write it in two points before the car touches the springs and in point of maximum compression
Initial
Em₀ = K = ½ m v²
Final
= 2 Ke = ½ k x²
The two is placed because each barred has two springs and each does not exert the same force
Emo =
½ m v² = 2 ½ k x²
v = √(2k/m) x
v = √ (2 3,134 10⁵/4550) 0.614
v = 7,207 m / s
Let's take this speed to km / h
v = 5,096 m / s (1km / 1000m) (3600s / 1h)
v = 25.9 km / h
This speed is common in school zones
Let's use kinematics to calculate the average acceleration
vf² = v₀² - 2 a x
0 = v₀² - 2 a x
a = v₀² / 2 x
a = 7,207²/2 0.614
a = 42.3 m / s²
We buy this acceleration with the acceleration of gravity
a / g = 42.3 / 9.8
a / g = 4.3
This acceleration is well below the maximum allowed