A 5.00 kg crate is suspended from the end of a short vertical rope of negligible mass. An upward force F(t) is applied to the en
1 answer:
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Answer:
t = 96.1 nm
Explanation:
For strong reflection through liquid layer we know that the path difference between two reflected light rays must be integral multiple of wavelength
now we know that the path difference of two reflected light from thin liquid layer is given as
here we know that
t = thickness of layer
N = 0 (for minimum thickness of layer)
now we have
Answer:
Theta1 = 12° and theta2 = 168°
The solution procedure can be found in the attachment below.
Explanation:
The Range is the horizontal distance traveled by a projectile. This diatance is given mathematically by Vo cos(theta) t. Where t is the total time of flight of the projectile in air. It is the time taken for the projectile to go from starting point to finish point. This solution assumes the projectile finishes uts motion on the same horizontal level as the starting point and as a result the vertical displacement is zero (no change in height).
In the solution as can be found below, the expression to calculate the range for any launch angle theta was first derived and then the required angles calculated from the equation by substituting the values of the the given quantities.
