Is impossible say how many stars are in the sky haha
The ball rolled for 13.2 s
<h3>Further explanation</h3>
Speed is scalar and no direction

A bowling ball rolls 33 m, with average speed = 2.5 m/s
So elapsed time :

No, you can't aim where you see the fish, because refraction will change the position of the fish and make it appear in a different position from where it actually is.
Hope this helps!
When the body is at rest, its speed is zero, and the graph lies on the x-axis.
When the body is in uniform motion, the speed is constant, and the graph is a horizontal line, parallel to the x-axis and some distance above it.
It's impossible to tell, based on the given information, how these two parts of the
graph are connected. There must be some sloping (accelerated) portion of the graph
that joins the two sections, but it cannot be accounted for in either the statement
that the body is at rest or that it is in uniform motion, since acceleration ... that is,
any change of speed or direction ... is not 'uniform' motion'.
Answer:
a) K = 0.63 J, b) h = 0.153 m
Explanation:
a) In this exercise we have a physical pendulum since the rod is a material object, the angular velocity is
w² =
where d is the distance from the pivot point to the center of mass and I is the moment of inertia.
The rod is a homogeneous body so its center of mass is at the geometric center of the rod.
d = L / 2
the moment of inertia of the rod is the moment of a rod supported at one end
I = ⅓ m L²
we substitute
w =
w =
w =
w = 4.427 rad / s
an oscillatory system is described by the expression
θ = θ₀ cos (wt + Φ)
the angular velocity is
w = dθ /dt
w = - θ₀ w sin (wt + Ф)
In this exercise, the kinetic energy is requested in the lowest position, in this position the energy is maximum. For this expression to be maximum, the sine function must be equal to ±1
In the exercise it is indicated that at the lowest point the angular velocity is
w = 4.0 rad / s
the kinetic energy is
K = ½ I w²
K = ½ (⅓ m L²) w²
K = 1/6 m L² w²
K = 1/6 0.42 0.75² 4.0²
K = 0.63 J
b) for this part let's use conservation of energy
starting point. Lowest point
Em₀ = K = ½ I w²
final point. Highest point
Em_f = U = m g h
energy is conserved
Em₀ = Em_f
½ I w² = m g h
½ (⅓ m L²) w² = m g h
h = 1/6 L² w² / g
h = 1/6 0.75² 4.0² / 9.8
h = 0.153 m