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
Answer:
f = 19,877 cm and P = 5D
Explanation:
This is a lens focal length exercise, which must be solved with the optical constructor equation
1 / f = 1 / p + 1 / q
where f is the focal length, p is the distance to the object and q is the distance to the image.
In this case the object is placed p = 25 cm from the eye, to be able to see it clearly the image must be at q = 97 cm from the eye
let's calculate
1 / f = 1/97 + 1/25
1 / f = 0.05
f = 19,877 cm
the power of a lens is defined by the inverse of the focal length in meters
P = 1 / f
P = 1 / 19,877 10-2
P = 5D
Answer:
<h3>
The charge transferred from the cloud to earth is 1 Coulomb.</h3>
Explanation:
Given :
Current 
A
Time 
sec
We know that the current is the rate of flow of charge.
From the formula of current,
<h3>
</h3>
Where 
charge transfer between cloud and earth.
C
Hence, the charge transferred from the cloud to earth is 1 Coulomb.
This would most likely be considered speed.
