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
The object will move 5 meters per second (5m/s)
Answer:

Explanation:
We have the following data:
- distance covered by the child: d = 2 m (length of the slide)
- time taken to cover this distance: t = 3 s
- initial velocity of the child: 0 m/s (he starts from rest)
So we can find the acceleration by using the equation:

Where a is the acceleration.
Substituting the values and solving for a,

The answer is the FIRST OPTION
Work occurs when a force is applied to an object and the object moves in the direction of the force applied <span />
Answer:
The pressure exerted by the brick on the table is 18,933.3 N/m².
Explanation:
Given;
height of the brick, h = 0.1 m
density of the brick, ρ = 19,300 kg/m³
acceleration due to gravity, g = 9.81 m/s²
The pressure exerted by the brick on the table is calculated as;
P = ρgh
P = (19,300)(9.81)(0.1)
P = 18,933.3 N/m²
Therefore, the pressure exerted by the brick on the table is 18,933.3 N/m².