A 3.0 kg pendulum swings from point A of height ya = 0.04 m to point B of height yb = 0.12 m, as seen in the diagram below.
2 answers:
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The correct answer is:
<span>Point charges must be in a vacuum.
In fact, the usual form for of the Coulomb's law is:
</span>
<span>where
</span>
is the permittivity of free space
<span>q1 and q2 are the two charges
q is the separation between the two charges
However, this formula is valid only if the charges are in vacuum. If they are in a material medium, the law is modified as follows:
</span>
where
is the relative permittivity, which takes into account the dielectric effects of the material.
<span>Point charges must be in a vacuum.
In fact, the usual form for of the Coulomb's law is:
</span>
<span>where
</span>
<span>q1 and q2 are the two charges
q is the separation between the two charges
However, this formula is valid only if the charges are in vacuum. If they are in a material medium, the law is modified as follows:
</span>
where
Answer:
block K = 29.39 J and spring #1 Ke = 360 J
Explanation:
In this problem we have that the elastic energy of the spring becomes part kinetic energy and the part in work against the force of friction, so, to use the law of conservation of energy, the decrease in energy is the rubbing force work
= Ef - E₀
Let's look for the energies
Initial
E₀ = Ke = ½ k₁ x₁²
Final, this is just before starting to compress the spring
Ef = Ke = ½ m v²
The work of the rubbing force is
= -fr x
Let's write Newton's second law the y axis
N-W = 0
N = W
fr = μ N
fr = μ mg
Let's replace
-μ mg x = ½ m v² - ½ k₁ x₁²
v² = 2/m (½ k₁ x1₁² -μ mg x)
v² = 2/6 (½ 2000 0.6²2 - 0.5 6 9.8 1) = 1/3 (360 - 29.4)
v = 3.13 m / s
With this value we calculate the energy of the block
K = ½ m v²
K = ½ 6 3.13²
K = 29.39 J
Calculate eenrgy of the spring ke 1
Ke = ½ k₁ x₁²
Ke = ½ 2000 0.60²
Ke = 360 J