The de Broglie wavelength of a 0.56 kg ball moving with a constant velocity of 26 m/s is 4.55×10⁻³⁵ m.
<h3>De Broglie wavelength:</h3>
The wavelength that is incorporated with the moving object and it has the relation with the momentum of that object and mass of that object. It is inversely proportional to the momentum of that moving object.
λ=h/p
Where, λ is the de Broglie wavelength, h is the Plank constant, p is the momentum of the moving object.
Whereas, p=mv, m is the mass of the object and v is the velocity of the moving object.
Therefore, λ=h/(mv)
λ=(6.63×10⁻³⁴)/(0.56×26)
λ=4.55×10⁻³⁵ m.
The de Broglie wavelength associated with the object weight 0.56 kg moving with the velocity of 26 m/s is λ=4.55×10⁻³⁵ m.
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A trace gas is a gas which makes up less than 1% by volume of the Earth's atmosphere, and it includes all gases except nitrogen (78.1%) and oxygen (20.9%). The most abundant trace gas at 0.934% is argon.
Answer:

Explanation:
Given,
Red light wavelength = 633 nm
width of slit = 0.320 mm
distance,d = 2.60 m
Condition of first maximum


m = 1


Width of the first minima



Now, width of the central region



Answer:
The diameter of the camera aperture must be greater than or equal to 1.49m
Explanation:
Let the distance separating two objects, x = 6.0 cm = 0.06 m
The distance between the observer and the two objects, d = 160 km = 160000 m
Let ∅ = minimum angular separation between the two objects that the satellite can resolve
tan( ∅) = x/d
Since there is minimum angular separation, tan( ∅) ≈∅
∅ = x/d
∅ = 0.06/160000
∅ = 3.75 * 10⁻⁷rad
For the satellite to be able to resolve the objects,
D ≥ 1.22λ/∅
λ = 560 nm = 560 * 10⁻⁹
D ≥ 1.22 * (560 * 10⁻⁹)/(3.75 * 10⁻⁷)
D ≥ 149.33 * 10⁻² m
D ≥ 1.49 m
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Explanation: