Answer
Given,
the angle formed between two polarizing filters is 45°
What is the reduction in intensity due to the filters =?
a) I = I₀ Cos² θ
where θ = 45°
The intensity of the light is diminished by 0.500
The correct answer among the given choices is D
b) the new polarization direction of the transmitted light is,
θ = 45°
To tackle this question we will apply the kinematic equations that describe the motion of a projectile, where both maximum height and distance traveled are defined. This scenario demonstrates a lion achieving a height (H) of 3m and covering a horizontal distance (R) of 10m. The equations governing this kind of motion are expressed as follows:
By dividing the two equations, we determine:
Plugging in the values for H and R yields:
After substituting \theta into the relevant equation, we find:
In conclusion, the mountain lion's launch speed upon takeoff is approximately 9.98m/s at an angle of 50.2°.
Answer:
C = 4,174 10³ V / m^{3/4}, E = 7.19 10² / ∛x, E = 1.5 10³ N/C
Explanation:
In this problem, we are tasked with determining the constant value and the generated electric field.
We will begin with computing the constant C:
V = C
C = V / x^{4/3}
C = 220 / (11 10⁻²)^{4/3}
C = 4,174 10³ V / m^{3/4}
Next, we will find the electric field by utilizing the formula:
V = E dx
E = dx / V
E = ∫ dx / C x^{4/3}
E = 1 / C x^{-1/3} / (- 1/3)
E = 1 / C (-3 / x^{1/3})
We consider the evaluation from the lower limit x = 0 where E = E₀ = 0 to the upper limit x = x, resulting in E = E:
E = 3 / C (0- (-1 / x^{1/3}))
E = 3 / 4,174 10³ (1 / x^{1/3})
E = 7.19 10² / ∛x
Substituting x = 0.110 cm:
E = 7.19 10² /∛0.11
E = 1.5 10³ N/C