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Kay [80]
3 years ago
9

What is the wavelength associated with an electron with a velocity of 4.8X10s m/s? (Mass of the electron is 9.1X10-31kg)?

Physics
1 answer:
amid [387]3 years ago
4 0

Answer:

1.52 nm

Explanation:

Using the De Broglie wavelength equation,

λ = h/p where λ = wavelength associated with electron, h = Planck's constant = 6.63 × 10⁻³⁴ Js and p = momentum of electron = mv where m = mass of electron = 9.1 × 10⁻³¹ kg and v = velocity of electron = 4.8 × 10⁵ m/s

So, λ = h/p

λ = h/mv

substituting the values of the variables into the equation, we have

λ = h/mv

λ = 6.63 × 10⁻³⁴ Js/(9.1 × 10⁻³¹ kg × 4.8 × 10⁵ m/s)

λ = 6.63 × 10⁻³⁴ Js/(43.68 × 10⁻²⁶ kgm/s)

λ = 0.1518 × 10⁻⁸ m

λ = 1.518 × 10⁻⁹ m

λ = 1.518 nm

λ ≅ 1.52 nm

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Calculate the approximate force on a square meter (1.00 m3) of sail, given the horizontal velocity of the wind is 6.00 m/s paral
KiRa [710]

Answer:

The force exerted on square meter of cubic sail is F = 15.3 N  

Explanation:

Given:-

- The face area of cubic sail, A = 1 m^2

- The velocity at frontal face, v1 = 6.0 m/s

- The velocity at back face, v2 = 3.5 m/s

- The density of air. ρ = 1.29 kg/m^3

Find:-

Calculate the approximate force on a square meter (1.00 m3) of sail

Solution:-

- We will apply the Bernoulli's equation to the flow of air around the cubic sail. Assuming that elevation changes are negligible. The constant elevation Bernoulli's equation is:

                           P1 + ρ*v1^2 / 2 = P2 + ρ*v2^2 / 2  

- The force (F) applied by any fluid is given by:

                           F = ( P2 - P1 )*A

- Re-arranging bernoulli's expression:

                          P2 - P1 = ρ/ 2 [ v2^2 - v1^2 ]

- Multiple the equation by area A:

                          A*[P2 - P1] = A*ρ/ 2 [ v1^2 - v2^2 ]

                          F = A*ρ/ 2 [ v1^2 - v2^2 ]

- Plug in the values:

                         F = (1)*(1.29/2)*[ 6^2 - 3.5^2 ]

                         F = 15.3 N  

7 0
4 years ago
If the reading of a linear scale is 4 mm and no of division of the circular scale is 50, then what will be the diameter of the w
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Answer:

Diameter of Wire = 4.5 mm

Explanation:

First, we need to find the fractional part of the reading. The fractional part o the reading can be given by the following formula:

Fractional Part = Circular Scale Reading x Least Count

where,

Circular Scale Reading = 50

Least Count = 0.01 mm

Therefore,

Fractional Part = (50)(0.01 mm)

Fractional Part = 0.5 mm

Now, the diameter of the wire can be given by using the following formula:

Diameter of Wire = Linear Scale Reading + Fractional Part

Diameter of Wire = 4 mm + 0.5 mm

<u>Diameter of Wire = 4.5 mm</u>

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