An airplane requires a much less force in order to get off the ground than a rocket does, since a rocket needs to exit Earth’s atmosphere.
Planes typically travel at much slower speeds that traditional rockets, as they are faced with varying restrictions that affect how quickly they fly.
Both forms of transportation, although very different in certain aspects, are similar in the fact that they both need to adhere to similar constraints. Although most airplanes aren’t faced with the challenge of exiting the atmosphere, they do need to focus on their fuel, their safety, etc. just as rockets do as well.
The vertical acceleration of the yo yo cent of mass is 0.66 m/s².
We know that, T = M*a
And T is given as 2M/3
2M/3 = M*a
So, a = 0.66 m/s²
What is Acceleration ?
Acceleration is the general term for any process where the velocity changes. There are only two ways to accelerate either by increasing the speed or decreasing direction, or both. The reason for this is that velocity includes both a speed and a direction.
You cannot possibly be accelerating if you don't also change you direction and speed, regardless of how swiftly you are travelling. Due to this, a jet experiences no acceleration even when it is moving at a high speed. In this case, 800 miles per hour, because its velocity is constant.
When it lands, the jet will accelerate as it slows down and quickly come to a stop.
Learn more about Acceleration from given link
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Answer:
Therefore the answer is the precision in the speed DECREASES
Explanation:
In quantum mechanics, we have the uncertainty principle that establishes that when the accuracy of the position increases the accuracy the speed decreases, being related by the expression
Δx Δv ≥ h'/ 2
h' = h/2π
Therefore the answer is the precision in the speed DECREASES

Wavelength, Wavespeed and frequency depends on each other in the following way :



And we know the Reciprocal relationship between frequency and period ~
So, let's find it's period •


Answer:
1.06 m
Explanation:
Since the charge is at the centre of two concentric spheres, we use the formula for electric potential due to a point charge. V = kq/r. Let r₁ be the radius of the sphere with potential, V₁ = 200 V and r₂ be the radius of the sphere with potential, V₂ = 82.0 V. From V = kq/r, r = kq/V. So that r₁ = kq/V₁ and r₂ = kq/V₂. The magnitude of the difference r₁ - r₂ is the distance between the two surfaces. q the charge equals 1.63 × 10⁻⁸ C
r₂ - r₁ = kq/V₂ - kq/V₁ = kq(1/V₂ - 1/V₁) = 1.63 × 10⁻⁸ × 9 × 10⁹ (1/82 -1/200) m = 1.63 × 10⁻⁸ × 9 × 10⁹ (0.0122 - 0.005) = 1.63 × 10⁻⁸ × 9 × 10⁹(0.0072) m = 1.06 m
The distance between them is 1.06 m