An unusual lightning strike has a vertical portion with a current of −400 A downwards. The Earth's magnetic field at that locati
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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.
The speed of the mass v = 0.884 m/s.
<u>Explanation</u>:
Let
K1 represents the kinetic energy of the mass when it is released,
U1 represents the potential energy of the spring when the mass is released,
K2 represents the kinetic energy of the mass when the spring returns to relaxed length,
U2 represents the potential energy of the spring when the spring returns to relaxed length
The spring is stretched by 0.27 - 0.12 = 0.15 m
K1 = 0
U1 = (1/2) 0.8
(0.15)^2
= 0.009 J
U2 = 0
By conservation of energy,
K2 + U2 = K1 + U1
K2 + 0 = 0 + 0.009 J
K2 = 0.009 J
Let v = speed of the mass
K2 = 1/2 m
v^2
m = 23 g = 0.023 kg
0.009 = 1/2 0.023
v^2
0.009 = 0.0115 v^2
v = √(0.009 / 0.0115)
v = 0.884 m/s.