Answer:
U = 30 m/s
a = 3 m/s²
Explanation:
"A car accelerating uniformly from rest reaches a maximum speed of U in 10 s. It then moves with that speed for an additional 20 s. The distance covered by the car in the 30 s interval is 750 m. Find U and the acceleration of the car in the first 10 s."
During the first 10 s:
v₀ = 0 m/s
v = U m/s
t = 10 s
The distance covered in this time is the average velocity times time:
Δx = ½ (v + v₀) t
Δx = ½ (U + 0) (10)
Δx = 5U
The distance covered in the next 20 seconds is speed times time:
Δx = 20U
The total distance is 750 m:
5U + 20U = 750
25U = 750
U = 30 m/s
The acceleration during the first 10 seconds is the change in speed over change in time.
a = Δv / Δt
a = (30 m/s − 0 m/s) / 10 s
a = 3 m/s²
Answer:
the drag coefficient = 0.2703126
Explanation:
see the attached file
Answer: h = 20.92 m
Explanation: By using the law of conservation of energy, the kinetic energy of the ball equals it potential energy.
Kinetic energy =mv^2/2
Potential energy = mgh
Where m = mass of the object, v = velocity of object = 23.5 m/s
g = acceleration due gravity = 9.8 m/s^2
mv^2/2 = mgh
m cancels out each other on both sides , hence we have that
v^2 = 2gh.
We want the ball to move towards the wall (horizontal motion), hence we need the horizontal component of the velocity since the velocity is inclined at an angle of 30.5 to the ground (horizontal).
Hence v = 23.5 × cos 30.5, v = 20.248 m/s
Recall that v^2 = 2gh
(20.248)^2 = 2×9.8×h
409.98 = 19.6 h
h = 409.98/ 19.6
h = 20.92 m
He identified triads of elements that had similar properties
