Answer:
The speed of the two cars after coupling is 0.46 m/s.
Explanation:
It is given that,
Mass of car 1, m₁ = 15,000 kg
Mass of car 2, m₂ = 50,000 kg
Speed of car 1, u₁ = 2 m/s
Initial speed of car 2, u₂ = 0
Let V is the speed of the two cars after coupling. It is the case of inelastic collision. Applying the conservation of momentum as :
V = 0.46 m/s
So, the speed of the two cars after coupling is 0.46 m/s. Hence, this is the required solution.
Answer:
Option 2
Explanation:
Weight of the box is being acted downwards due to gravity
However, based on Newton's third law (for every action (force) in nature there is an equal and opposite reaction), an equal force will act on the box by the table
The time for the car to drive directly south is determined as 7.15 s.
<h3>Time for the car to drive directly west</h3>
The time for the car to drive directly south is determined by applying the concept of slope.
slope = Δy/Δx
a = Δv/Δt
Δt = Δv/a
Δt = (26.8)/(3.75)
Δt = 7.15 s
Thus, the time for the car to drive directly south is determined as 7.15 s.
Learn more about acceleration here: brainly.com/question/605631
#SPJ1
Answer:
h = 618.64 m
Explanation:
First we need to calculate the height gained by rocket while the fuel is burning. We use 2nd equation of motion for that purpose:
h₁ = Vit + (1/2)at²
where,
h₁ = height gained during the burning of fuel
Vi = Initial Velocity = 0 m/s
t = time = 7 s
a = acceleration = 8 m/s²
Therefore,
h₁ = (0 m/s)(7 s) + (1/2)(8 m/s²)(7 s)²
h₁ = 196 m
Now we use 1st equation of motion to find final speed Vf:
Vf = Vi + at
Vf = 0 m/s + (8 m/s²)(7 s)
Vf = 56 m/s
Now, we calculate height covered in free fall motion. Using 3rd equation of motion:
2ah₂ = Vf² - Vi²
where,
a = - 3.71 m/s²
h₂ = height gained during free fall motion = ?
Vf = Final Velocity = 0 m/s (since, rocket will stop at highest point)
Vi = 56 m/s
Therefore,
(2)(-3.71 m/s²)h₂ = (0 m/s)² - (56 m/s)²
h₂ = 422.64 m
So the total height gained will be:
h = h₁ + h₂
h = 196 m + 422.64 m
<u>h = 618.64 m</u>
