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dybincka [34]
3 years ago
7

. A helicopter is 8.50 m above the ground and climbing at 2.5 m/s. It drops a package from rest (relative to the helicopter). At

the moment that the package is released, what is its velocity relative to the ground?
Physics
1 answer:
Rudik [331]3 years ago
4 0

Answer:

2.5 m/s

Explanation:

The velocity of the package relative to the ground = the velocity of the package relative to the helicopter + the velocity of the helicopter relative to the ground

v = 0 m/s + 2.5 m/s

v = 2.5 m/s

At the moment it is released, the package is rising at 2.5 m/s.

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Answer:

width of slit =1.23× 10⁻⁶ m  

Explanation:

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d sin θ = n λ

λ = wavelength     θ = angle between the central maxima and 1st  minima

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for first minima  n = 1

now,

d =\dfrac{n \lambda}{sin \theta}

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d =\dfrac{1\times 460 \times 10^{-9}}{sin 22.5^0}

d = 1228 × 10⁻⁹ m  = 1.228× 10⁻⁶ m

d =  1.23× 10⁻⁶ m

width of slit =1.23× 10⁻⁶ m  

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