Your odometer of your car says 54 123 miles when you start your trip, your watch says 9 45 AM When you get to your
1 answer:
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Answer:
V₁ = √ (gy / 3)
Explanation:
For this exercise we will use the concepts of mechanical energy, for which we define energy n the initial point and the point of average height and / 2
Starting point
Em₀ = U₁ + U₂
Em₀ = m₁ g y₁ + m₂ g y₂
Let's place the reference system at the point where the mass m1 is
y₁ = 0
y₂ = y
Em₀ = m₂ g y = 2 m₁ g y
End point, at height yf = y / 2
= K₁ + U₁ + K₂ + U₂
= ½ m₁ v₁² + ½ m₂ v₂² + m₁ g
+ m₂ g
Since the masses are joined by a rope, they must have the same speed
= ½ (m₁ + m₂) v₁² + (m₁ + m₂) g
= ½ (m₁ + 2m₁) v₁² + (m₁ + 2m₁) g
How energy is conserved
Em₀ =
2 m₁ g y = ½ (m₁ + 2m₁) v₁² + (m₁ + 2m₁) g
2 m₁ g y = ½ (3m₁) v₁² + (3m₁) g y / 2
3/2 v₁² = 2 g y -3/2 g y
3/2 v₁² = ½ g y
V₁ = √ (gy / 3)
Answer:
The deceleration is
Explanation:
From the question we are told that
The distance of the car from the crossing is
The speed is
The reaction time of the engineer is
Generally the distance covered during the reaction time is
=>
=>
Generally distance of the car from the crossing after the engineer reacts is
=>
=>
Generally from kinematic equation
Here v is the final velocity of the car which is 0 m/s
So
=>