<span>We can assume that the horizontal surface has no friction and the pulley is massless. We can use Newton's second law to set up an equation.
F = Ma
F is the net force
M is the total mass of the system
a is the acceleration
a = F / M
a = (mb)(g) / (ma + mb)
a = (6.0 kg)(9.80 m/s^2) / (6.0 kg + 14.0 kg)
a = 58.8 N / 20 kg
a = 2.94 m/s^2
The magnitude of the acceleration of the system is 2.94 m/s^2</span>
Hi there!

Use the equation:

Where m2 and v2 deal with the larger object, and m1 and v1 with the smaller object. Plug in the given values:
v2 = ?
m1 = 0.048 kg (converted)
m2 = 2.95
v1 = 391


Answer:
sin 2θ = 1 θ=45
Explanation:
They ask us to prove that the optimal launch angle is 45º, for this by reviewing the parabolic launch equations we have the scope equation
R = Vo² sin 2θ / g
Where R is the horizontal range, Vo is the initial velocity, g the acceleration of gravity and θ the launch angle. From this equation we see that the sine function is maximum 2θ = 90 since sin 90 = 1 which implies that θ = 45º; This proves that this is the optimum angle to have the maximum range.
We calculate the distance traveled for different angle
R = vo² Sin (2 15) /9.8
R = Vo² 0.051 m
In the table are all values in two ways
Angle (θ) distance R (x)
0 0 0
15 0.051 Vo² 0.5 Vo²/g
30 0.088 vo² 0.866 Vo²/g
45 0.102 Vo² 1 Vo²/g
60 0.088 Vo² 0.866 Vo²/g
75 0.051 vo² 0.5 Vo²/g
90 0 0
See graphic ( R Vs θ) in the attached ¡, it can be done with any program, for example EXCEL