Compare the inertia of a car to the inertia of a bicycle
2 answers:
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the solution for the oscillatory movement allows to find the result for the amplitude of the initial displacement is:
- The range of motion is: A = 4.44 cm
The oscillatory periodic motion of a system occurs when there is a recovered force, in the special case that this force is proportional to the displacement is called simple harmonic motion, which is described by the expression.
x = A cos (wt + Ф)
w² = k/m
where x is the displacement, A the amplitude, w the angular velocity, t the time, k the spring constant, m the mass, and Ф a phase constant determined by the initial conditions.
Let's find the angular velocity/
w=
w = 25.8 rad/s
Let's look for the constant Ф, as the system is released from rest its initial velocity is zero, for zero time. The definition of speed is:
v=
v= - A w sin (wt +Ф)
0 = -A w sin Ф
Ф= 0
They indicate that at a given instant of the time the velocity is v= 50.0 cm/s and it is in a position x= 4.00 cm, let us write the equations for this time
Position.
4.00 = A cos 25.8t
Speed.
50.0 = - At 25.8 sin 25.8t
To solve the system, ;et's square and add.
Cos² 25.8t =
sin² 25.8t =
1 =
A =
A= 4.44 cm
In conclusion using the solution for the oscillatory movement we can find the result for the amplitude of the initial displacement is:
The range of motion is: A = 444 cm
Learn more about oscillatory motion here: brainly.com/question/14311816
Answer:
a) -2.516 × 10⁻⁴ V
b) -1.33 × 10⁻³ V
Explanation:
The electric field inside the sphere can be expressed as:
The potential at a distance can be represented as:
V(r) - V(0) =
V(r) - V(0) = ₀
V(r) = ₀
Given that:
q = +3.83 fc = 3.83 × 10⁻¹⁵ C
r = 0.56 cm
= 0.56 × 10⁻² m
R = 1.29 cm
= 1.29 × 10⁻² m
E₀ = 8.85 × 10⁻¹² F/m
Substituting our values; we have:
= -2.15 × 10⁻⁴ V
The difference between the radial distance and center can be expressed as:
V(r) - V(0) =
V(r) - V(0) =
V(r) =
V(r) =
V(r)
V(r) = -0.00133
V(r) = - 1.33 × 10⁻³ V