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
Turn lights off, unplug electronics, and use solar energy
Acceleration = (0.2 x g) = 1.96m/sec^2.
<span>Accelerating force on 1kg. = (ma) = 1.96N. </span>
<span>1kg. has a weight (normal force) of 9.8N. </span>
<span>Coefficient µ = 1.96/9.8 = 0.2 minimum. </span>
<span>Coefficient is a ratio, so holds true for any value of mass to find accelerating force acting. </span>
<span>e.g. 75kg = (75 x g) = 735N. </span>
<span>Accelerating force = (735 x 0.2) = 147N</span>
Answer:
Capacitance, C = 26.1 picofarad
Explanation:
It is given that,
Side of square, x = 4.3546 cm = 0.043546 m
Distance between electrodes, d = 0.6408 mm = 0.0006408 m
Voltage, V = 73.68 V
Capacitance of parallel plates is given by :
or
C = 26.1 picofarad
So, the capacitance of the capacitor is 26.1 picofarad. Hence, this is the required solution.