Answer:
his acceleration rate is -0.00186 m/s²
Explanation:
Given;
initial position of the car, x₀ = 100 miles = 160, 900 m ( 1 mile = 1609 m)
time of motion, t₀ = 60 minutes = 60 mins x 60 s = 3,600 s
final position of the car, x₁ = 150 miles = 241,350 m
time of motion, t₁ = 100 minutes = 100 mins x 60 s = 6,000 s
The initial velocity is calculated as;
u = 160, 900 m / 3,600 s
u = 44.694 m/s
The final velocity is calculated as;
v = 241,350 m / 6,000 s
v = 40.225 m/s
The acceleration is calculated as;
Therefore, his acceleration rate is -0.00186 m/s²
The tension in the first and second rope are; 147 Newton and 98 Newton respectively.
Given the data in the question
- Mass of first block;
- Mass of second block,
- Tension on first rope;
- Tension on second rope;
To find the Tension in each of the ropes, we make use of the equation from Newton's Second Laws of Motion:
Where F is the force, m is the mass of the object and a is the acceleration ( In this case the block is under gravity. Hence ''a" becomes acceleration due to gravity 
)
For the First Rope
Total mass hanging on it; 
So Tension of the rope;
Therefore, the tension in the first rope is 147 Newton
For the Second Rope
Since only the block of mass 10kg is hang from the second, the tension in the second rope will be;
Therefore, the tension in the second rope is 98 Newton
Learn More, brainly.com/question/18288215
Answer:
h2 = 0.092m
Explanation:
From a balance of energy from point A to point B, we get speed before the collision:
Solving for Vb:
Since the collision is elastic, we now that velocity of bead 1 after the collision is given by:
Now, by doing another balance of energy from the instant after the collision, to the point where bead 1 stops, we get the distance it rises:
Solving for h2:
h2 = 0.092m
