When you are in free fall, the force of gravity is stronger than your velocity perpendicular to where you're falling, and you move at a constant speed downwards.
Under feelings of weightlessness, you are still being pulled by gravity, but your perpendicular velocity and distance from the source can cancel each other out.
Answer:
8400m
Explanation:
The engine that falls off would have the same constant horizontal velocity as the airplane's when if falls off if we ignore air resistance. So it would have a horizontal velocity of 280m/s for 30seconds before it hits the ground.
Therefor the horizontal distance the engine travels during its fall is
280 * 30 = 8400m
Explanation:
Single slit diffraction
Diffraction is the phenomenon of spreading out of waves as they pass through an aperture or around objects. Diffraction occurs when the size of the aperture or obstacle is of the same order of magnitude as the wavelength of the incident wave. For very small aperture sizes, the vast majority of the wave is blocked. in case of large apertures the wave passes by or through the obstacle without any significant diffraction.
Answer:
F = 3.86 x 10⁻⁶ N
Explanation:
First, we will find the distance between the two particles:

where,
r = distance between the particles = ?
(x₁, y₁, z₁) = (2, 5, 1)
(x₂, y₂, z₂) = (3, 2, 3)
Therefore,

Now, we will calculate the magnitude of the force between the charges by using Coulomb's Law:

where,
F = magnitude of force = ?
k = Coulomb's Constant = 9 x 10⁹ Nm²/C²
q₁ = magnitude of first charge = 2 x 10⁻⁸ C
q₂ = magnitude of second charge = 3 x 10⁻⁷ C
r = distance between the charges = 3.741 m
Therefore,

<u>F = 3.86 x 10⁻⁶ N</u>