17.food that is both nutrient- and calorie-dense can be a good choice in what circumstance?
1 answer:
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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
Explanation:
Given that,
Mass = 0.254 kg
Spring constant [tex[\omega_{0}= 10.0\ N/m[/tex]
Force = 0.5 N
y = 0.628
We need to calculate the A and d
Using formula of A and d
.....(I)
....(II)
Put the value of in equation (I) and (II)
From equation (II)
Put the value of in equation (I) and (II)
From equation (II)
Put the value of in equation (I) and (II)
From equation (II)
Put the value of in equation (I) and (II)
From equation (II)
Hence, This is the required solution.