Answer:
Explanation:
We know that the volume V for a sphere of radius r is

If we got an uncertainty
the formula for the uncertainty of V is:

We can calculate this uncertainty, first we obtain the derivative:


And using it in the formula:



The relative uncertainty is:



Using the values for the problem:

This is, a percent uncertainty of 4.77 %
Answer:
Gravitational force. Magnetic force. Electrostatics. Nuclear force.
Explanation:
Apple falling from a tree
raindrops falling from the sky
The properties of the wave don't determine its speed. The properties of the medium do. You can FIND the speed by measuring the wave's frequency and wavelength.
Answer:
C. My soup will cool down because of radiation.
Explanation:
Since, there is vacuum between hot soup and its outside wall, then heat can not flow through conduction and convection.
The heat then only flows through radiation.
Therefore, the soup will not cool down because of convection or conduction by because of radiation.
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