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snow_tiger [21]
3 years ago
12

Two men are standing on a frictionless ice surface holding opposite ends of a rope. One man (mass = 80 kg) pulls on the rope wit

h a force of 250 N. The other has a mass of 60 kg. What is the acceleration of each man?
Physics
1 answer:
ankoles [38]3 years ago
7 0

Answer:

The acceleration of man 1 and 2 is 3.125\ m/s^2 and 4.167\ m/s^2.

Explanation:

Mass of man 1, m₁ = 80 kg

Mass of man 2, m₂ = 60 kg

One man pulls on the rope with a force of 250 N.

Let a₁ is acceleration of man 1,

F = m₁a₁

a_1=\dfrac{F}{m_1}\\\\a_1=\dfrac{250}{80}\\\\a_1=3.125\ m/s^2

Let a₂ is acceleration of man 1,

F = m₂a₂

a_2=\dfrac{F}{m_2}\\\\a_2=\dfrac{250}{60}\\\\a_2=4.167\ m/s^2

So, the acceleration of man 1 and 2 is 3.125\ m/s^2 and 4.167\ m/s^2.

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Consider an electron in an infinite well of width 0.7 nm . What is the wavelength of a photon emitted when the electron in the i
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Explanation:

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They ask for the transition from the first excited state n = 2 to the base state n = 1

       E₂ - E₁ = = (h² / 8mL²) (n₂² - n₁²)

Let's calculate

       E₂-E₁ = (6.63 10⁻³⁴)² / (8 9.1 10⁻³¹ (0.7 10⁻⁹)²) (2² -1²)

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Let's use the Planck equation

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Let's reduce

      λ  = 5.380 10⁻⁷ m (10 9 nm / 1 m)

      λ  = 538.0 nm

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4 years ago
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