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boyakko [2]
3 years ago
10

Harry Potter is chasing his nemesis Draco Malfoy during a quidditch match. Initially, Harry is 35m behind Draco, who has just sp

otted the Golden Snitch hovering 75m away at the edge of the playing field. Draco Flies at a constant speed of 40 m/s straight towards the Golden Snitch. If Harry is originally flying at 50 m/s, what acceleration does he need to reach the Snitch 175 m ahead of Draco? How fast will Harry be moving when he passes Draco?
Physics
1 answer:
Sophie [7]3 years ago
3 0

Answer:

the acceleration of harry is equal to 66.126 m/s²

Explanation:

given,

harry is 35 m behind Draco

speed of Draco = 40 m/s

original speed of harry = 50 m/s

acceleration = ?

time taken by the Draco

    t =\dfrac{r}{u} =\dfrac{75}{40}

     t = 1.875 s

distance covered by Harry

  d = 35 + 175 = 210 m

to calculate the acceleration of harry

s = u t+ \dfrac{1}{2}at^2

210 = 50\times 1.875+ \dfrac{1}{2}\times a\times 1.875^2

a × 3.516 × 0.5 = 116.25

a = 66.126 m/s²

hence, the acceleration of harry is equal to 66.126 m/s²

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An electron (restricted to one dimension) is trapped between two rigid walls 1.40 nm apart. The electron's energy is approximate
Bumek [7]

Answer:

a)    n = 9.9       b)      E₁₀ = 19.25 eV

Explanation:

Solving the Scrodinger equation for the electronegative box we get

         Eₙ = (h² / 8m L²2) n²

where l is the distance L = 1.40 nm = 1.40 10⁻⁹ m and n the quantum number

 In this case En = 19 eV let us reduce to the SI system

          En = 19 eV (1.6 10⁻¹⁹ J / 1 eV) = 30.4 10⁻¹⁹ J

          n = √ (In 8 m L² / h²)

let's calculate

          n = √ (8  9.1 10⁻³¹ (1.4 10⁻⁹)²  30.4 10⁻¹⁹ / (6.63 10⁻³⁴)²

          n = √ (98)            n = 9.9

since n must be an integer, we approximate them to 10

b) We substitute for the calculation of energy

        In = (h² / 8mL2² n²

        In = (6.63 10⁻³⁴) 2 / (8 9.1 10⁻³¹  (1.4 10⁻⁹)² 10²

        E₁₀ = 3.08 10⁻¹⁸ J

we reduce eV

      E₁₀ = 3.08 10⁻¹⁸ j (1ev / 1.6 10⁻¹⁹J)

      E₁₀ = 1.925 101 eV

      E₁₀ = 19.25 eV

the result with significant figures is

        E₁₀ = 19.25 eV

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