A ball is dropped from the top of a building.After 2 seconds, it’s velocity is measured to be 19.6 m/s. Calculate the accelerati
2 answers:
Explanation:
The given data is as follows.
Initial velocity; u = 0, Final velocity; v = 19.6 m/s
time; t = 2 seconds
As the relation between initial velocity, final velocity and acceleration is as follows.
v = u + at
Hence, putting the given values into the above formula as follows.
v = u + at
19.6 m/s = 0 +
a = 9.8
Thus, we can conclude that acceleration of the dropped ball is 9.8 .
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Answer:
T = 188.5 s, correct is C
Explanation:
This problem must be worked on using conservation of angular momentum. We define the system as formed by the fan and the paper, as the system is isolated, the moment is conserved
initial instant. Before the crash
L₀ = r m v₀ + I₀ w₀
the angular speed of the fan is zero w₀ = 0
final instant. After the crash
L_f = I₀ w + m r v
L₀ = L_f
m r v₀ = I₀ w + m r v
angular and linear velocity are related
v = r w
w = v / r
m r v₀ = I₀ v / r + m r v
m r v₀ = (I₀ / r + mr) v
v =
let's calculate
v =
v =
v = 0.02 m / s
To calculate the time of a complete revolution we can use the kinematics relations of uniform motion
v = x / T
T = x / v
the distance of a circle with radius r = 0.6 m
x = 2π r
we substitute
T = 2π r / v
let's calculate
T = 2π 0.6/0.02
T = 188.5 s
reduce
t = 188.5 s ( 1 min/60 s) = 3.13 min
correct is C