Answer: The Moon appears to change shape because we see different amounts of the lit part as the Moon orbits Earth. ... As it moves around Earth, more and more of the lit side comes into view. Then it begins to disappear again. It takes 29.5 days for the Moon to orbit Earth.
Explanation:look it up lol
The position of the particle is given by:
x(t) = t³ - 12t² + 21t - 9
Differentiate x(t) with respect to t to find the velocity x'(t):
x'(t) = 3t² - 24t + 21
Differentiate x'(t) with respect to t to find the acceleration x''(t):
x''(t) = 6t - 24
Answer : Magnitude
Explanation :
In a value, the magnitude is represented by its units. It can be adopted by convention or by law.
Magnitude of any unit is used to measure the same kind of quantity.
For example: The unit of length which is a physical quantity is meter (m).
So, magnitude is correct answer.
Answer:
Explanation:
Given
acceleration is given by

where 

Also acceleration is given by








at 





when air drag is neglected maximum height reached is


Thick lens will have shorter and consequently thin lens will have greater focal length. Because, For a thick lens, the optical path length of the light is more, than for a thin lens, thus, the bending of light will be more in case of a thicker lens. Consequently, it has a shorter focal length.