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ddd [48]
3 years ago
10

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.

Physics
2 answers:
lakkis [162]3 years ago
5 0

i dont know soory pls give brainlyist

juin [17]3 years ago
4 0

Answer:

3.0−0.12=2.88 or 2.88÷0.04=72

0.04×3.0=0.12 and 0.04+3.0=3.04

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Because the disturbances are in opposite directions for this superposition, the resulting amplitude is zero for pure destructive interference

Explanation:

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What is Planck’s law?
Lyrx [107]

Answer:

The wavelength of the emitted radiation is inversely proportional to its frequency, or λ = c/ν. The value of Planck's constant is defined as 6.62607015 × 10−34 joule∙second.

Explanation:

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Hope this helps!

3 0
3 years ago
Which condition is required for Coulomb's law to hold true?
AleksAgata [21]
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>F= \frac{1}{4 \pi \epsilon_0}  \frac{q_1 q_2}{r^2}
<span>where
</span>\epsilon_0 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>F= \frac{1}{4 \pi \epsilon_0 \epsilon_r} \frac{q_1 q_2}{r^2}
where \epsilon_r is the relative permittivity, which takes into account the dielectric effects of the material.
7 0
3 years ago
If you double the voltage in a circuit and reduce the resistance by a factor of four what will happen to the current? A) it will
KatRina [158]
By Ohm's Law, we can relate current, voltage and resistance. It is expressed as V=IR. That is, there is a direct relationship between voltage and resistance and voltage and current.

V = IR
V1/2V1 = I1R1 / I2R1/4
1/2 = 4I1/I2
I2 = 8I1

Therefore, <span> it will increase by a factor of 8. Hope this answers the question.</span>
5 0
3 years ago
A block (6 kg) initially compresses spring #1 (k = 2000 N/m) by 60 cm from its equilibrium point. When the block is released, it
mart [117]

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

        W_{fr}= 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

       W_{fr}= -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

4 0
3 years ago
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