A navigational beacon in deep space broadcasts at a radio frequency of 50 MHz. A spaceship approaches the beacon with a relative
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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.
We have that the spring constant is mathematically given as
Generally, the equation for angular velocity is mathematically given by
Where
k=spring constant
And
Therefore
Hence giving spring constant k
Generally
Mass of earth
Period for on complete resolution of Earth around the Sun
Therefore
In conclusion
The effective spring constant of this simple harmonic motion is
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