Suppose the coefficient of static friction between the road and the tires on a car is 0.638 and the car has no negative lift. Wh
1 answer:
Answer:
12.6332454263 m/s
Explanation:
m = Mass of car
v = Velocity of the car
= Coefficient of static friction = 0.638
g = Acceleration due to gravity = 9.81 m/s²
r = Radius of turn = 25.5 m
When the car is on the verge of sliding we have the force equation
The speed of the car that will put it on the verge of sliding is 12.6332454263 m/s
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Answer:
= +3,394 103 m / s
Explanation:
We will solve this problem with the concept of the moment. Let's start by defining the system that is formed by the complete rocket before and after the explosions, bone with the two stages, for this system the moment is conserved.
The data they give is the mass of the first stage m1 = 2100 kg, the mass of the second stage m2 = 1160 kg and its final velocity v2f = +5940 m / s and the speed of the rocket before the explosion vo = +4300 m / s
The moment before the explosion
p₀ = (m₁ + m₂) v₀
After the explosion
pf = m₁ + m₂
p₀ = [texpv_{f}[/tex]
(m₁ + m₂) v₀ = m₁ + m₂
Let's calculate the final speed (v1f) of the first stage
= ((m₁ + m₂) v₀ - m₂
) / m₁
= ((2100 +1160) 4300 - 1160 5940) / 2100
= (14,018 10 6 - 6,890 106) / 2100
= 7,128 106/2100
= +3,394 103 m / s
come the same direction of the final stage, but more slowly