Answer:
The slowdown took 2.05 seconds and the roller coaster made a distance 20.5 during that time.
Explanation:
Start with the kinematic equation for velocity. Let v1 and v2 be the initial and the new velocity, respectively. Let s denote the distance made duringthe slowdown. Let m denote the mass of the roller coaster and "a" its acceleration (negative = deceleration). the equation is as follows:
Let us first determine the distance made:
The roller coaster made 20.5 meters during slowing down to 2 m/s. Now calculate the time this took.
It took 2.05 seconds to achieve the slowdown from 18 to 2 m/s
Answer:
It is also called law of Malus,the law states that the intensity of a beam of plane-polarized light after passing through a rotatable polarizer varies as square of the cosine of the angle through which the polarizer is rotated from the position that gives maximum intensity expand.
Before u can know it u have to understand it, so go through it over and over again and u will get it.
Answer:
Explanation:
The capacitor of a parallel-plate capacitor is given by:
where
A is the area of each plate
d is the separation between the plates
is the vacuum permittivity
The energy stored in a capacitor instead is given by
where
Q is the charge stored in each plate
Substituting the expression we found for C inside the last formula,
And re-arranging it
Now if we substitute
We find the charge stored on the capacitor:
Answer:
the minimum wall thickness that will enhance the reflection of light is 146.9 nm
Explanation:
Given the data in the question;
At the first interface, a phase shift occurs as the incident light is in air that has less refractive index compare to the thin film of soap bubble.
At the second interface, no shift occurs,
condition for constructive interference;
t = ( m + 1/2) × λ/2n
where m = 0, 1, 2, 3 . . . . . .
now, the condition for the constructive interference;
t = mλ/2n
where t is the thickness of the soap bubble, λ is the wavelength of light and n is the refractive index of soap bubble.
so the minimum thickness of the film which will enhance reflection of light will be;
t = ( m + 1/2) × λ/2n
we substitute
t = ( 0 + 1/2) × 711 /2(1.21)
t = 0.5 × 711/2.42
t = 0.5 × 293.80165
t = 146.9 nm
Therefore, the minimum wall thickness that will enhance the reflection of light is 146.9 nm