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
The heat capacity is given by the expression:
When the
is measured in the calorimeter, we obtain a value, and since we know the mass of the material and we control the change in
, we can then determine the specific heat "C" by simply remplazing in the expression.
Answer:
1)
2) x = 0.28 m from cable A.
Explanation:
1 ) Let's use the first Newton to find the , because bar is in equilibrium.
In this case we just have y-direction forces.
Now, let's solve the equation for W(object).
2 ) To find the position of the heaviest weight we need to use the torque definition.
The total torque is evaluated in the axes of the object.
Let's put the heaviest weight in a x distance from the cable A. We will call this point P for instance.
First let's find the positions from each force to the P point.
; total length of the bar.
; distance between Tension A and P point.
; distance between Tension B and P point.
; distance between weight of the bar (middle of the bar) and P point.
Now, let's find the total torque in P point, assuming counterclockwise rotation as positive.
Finally we just need to solve it for x.
So the distance is x = 0.28 m from cable A.
Hope it helps!
Have a nice day! :)
Answer:
1
When second polarizer is removed the intensity after it passes through the stack is
2 When third polarizer is removed the intensity after it passes through the stack is
Explanation:
From the question we are told that
The angle of the second polarizing to the first is
The angle of the third polarizing to the first is
The unpolarized light after it pass through the polarizing stack
Let the initial intensity of the beam of light before polarization be
Generally when the unpolarized light passes through the first polarizing filter the intensity of light that emerges is mathematically evaluated as
Now according to Malus’ law the intensity of light that would emerge from the second polarizing filter is mathematically represented as
The intensity of light that will emerge from the third filter is mathematically represented as
making the subject of the formula
Note that as
is the last emerging intensity of light after it has pass through the polarizing stack
Substituting values
When the second is removed the third polarizer becomes the second and final polarizer so the intensity of light would be mathematically evaluated as
is the intensity of the light emerging from the stack
substituting values
When the third polarizer is removed the second polarizer becomes the
the final polarizer and the intensity of light emerging from the stack would be
is the intensity of the light emerging from the stack
Substituting values