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.
Answer:
a) 141.6m
b) 8.4m/s
Explanation:
a) to find the total displacement you use the following formula for each trajectory. Next you sum the results:

hence, the total distance is 141.6m
b) the mean velocity of the total trajectory is given by:

hence, the mean velocity is 8.4 m/s
I know that protons and neutrons are located at the center of an atom, so the correct answer is D
Answer:
75k
Explanation:
You can see solution in the picture
5.08
hope you got it right !