How could you have collected your data in different ways that may have more accurately identified the Seymouria?
1 answer:
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Answer:
The box stops in 0.139 seconds, after moving 7.29cm (0.0729m) backwards relative to the belt.
Explanation:
As the box is initially at rest relative to the earth, it is moving backwards with a speed of 1.05m/s relative to the belt. Then, the frictional force acts on the box to make it stop relative to the belt. So, we first have to write the equations of motion of the box in each axis:
Since the frictional force is equal to
, then we have that the acceleration is:
Now, from the definition of acceleration we get:
And, as the final velocity is zero because the box gets to a stop, we have:
(Don't worry about the negative sign. It will disappear because the initial velocity is also negative, since we take the box initially moving backwards)
Then, plugging in the given values, we calculate the time:
In words, the time the box takes to stop sliding relative to the belt is 0.139s.
The displacement of the box in this time, is given by the kinematics formula:
Finally, we calculate the displacement:
This means that the box moves 7.29cm backwards relative to the belt.
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