Heating water so that it boils is an example of _____.
2 answers:
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Answer:
41°
Explanation:
Kinetic energy at bottom = potential energy at top
½ mv² = mgh
½ v² = gh
h = v²/(2g)
h = (2.4 m/s)² / (2 × 9.8 m/s²)
h = 0.294 m
The pendulum rises to a height of above the bottom. To determine the angle, we need to use trigonometry (see attached diagram).
L − h = L cos θ
cos θ = (L − h) / L
cos θ = (1.2 − 0.294) / 1.2
θ = 41.0°
Rounded to two significant figures, the pendulum makes a maximum angle of 41° with the vertical.
Answer:
No
Explanation:
The force of tension exerted by the string on the rock acts as centripetal force, so its direction is always towards the centre of the circle.
However, the direction of motion of the rock is always tangential to the circle: this means that the force is always perpendicular to the direction of motion of the rock.
As we know, the work done by a force on an object is
where
F is the magnitude of the force
d is the displacement of the object
is the angle between the force and the displacement
In this situation, F and d are perpendicular, so , therefore
and the work done is zero:
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