To solve this problem it is necessary to apply the concepts related to the concept of overlap and constructive interference.
For this purpose we have that the constructive interference in waves can be expressed under the function

Where
a = Width of the slit
d = Distance of slit to screen
m = Number of order which represent the number of repetition of the spectrum
Angle between incident rays and scatter planes
At the same time the distance on the screen from the central point, would be

Where y = Represents the distance on the screen from the central point
PART A ) From the previous equation if we arrange to find the angle we have that



PART B) Equation both equations we have


Re-arrange to find a,


Wow ! This is not simple. At first, it looks like there's not enough information, because we don't know the mass of the cars. But I"m pretty sure it turns out that we don't need to know it.
At the top of the first hill, the car's potential energy is
PE = (mass) x (gravity) x (height) .
At the bottom, the car's kinetic energy is
KE = (1/2) (mass) (speed²) .
You said that the car's speed is 70 m/s at the bottom of the hill,
and you also said that 10% of the energy will be lost on the way
down. So now, here comes the big jump. Put a comment under
my answer if you don't see where I got this equation:
KE = 0.9 PE
(1/2) (mass) (70 m/s)² = (0.9) (mass) (gravity) (height)
Divide each side by (mass):
(0.5) (4900 m²/s²) = (0.9) (9.8 m/s²) (height)
(There goes the mass. As long as the whole thing is 90% efficient,
the solution will be the same for any number of cars, loaded with
any number of passengers.)
Divide each side by (0.9):
(0.5/0.9) (4900 m²/s²) = (9.8 m/s²) (height)
Divide each side by (9.8 m/s²):
Height = (5/9)(4900 m²/s²) / (9.8 m/s²)
= (5 x 4900 m²/s²) / (9 x 9.8 m/s²)
= (24,500 / 88.2) (m²/s²) / (m/s²)
= 277-7/9 meters
(about 911 feet)
I'm not sure but I had this question on a benchmark I think its the density of the wire you need to find the density or the mass I'm not sure but i do remember this question
Observer A is moving inside the train
so here observer A will not be able to see the change in position of train as he is standing in the same reference frame
So here as per observer A the train will remain at rest and its not moving at all
Observer B is standing on the platform so here it is a stationary reference frame which is outside the moving body
So here observer B will see the actual motion of train which is moving in forward direction away from the platform
Observer C is inside other train which is moving in opposite direction on parallel track. So as per observer C the train is coming nearer to him at faster speed then the actual speed because they are moving in opposite direction
So the distance between them will decrease at faster rate
Now as per Newton's II law
F = ma
Now if train apply the brakes the net force on it will be opposite to its motion
So we can say
- F = ma

so here acceleration negative will show that train will get slower and its distance with respect to us is now increasing with less rate
It is not affected by the gravity because the gravity will cause the weight of train and this weight is always counterbalanced by normal force on the train
So there is no effect on train motion
Answer:
The potential difference across the plates is 226 V.
Explanation:
Given;
area of the capacitor plate, A = 0.2 m²
separation, d = 0.1 mm = 0.1 x 10⁻³ m
charge on each plate, Q = 4 x 10⁻⁶ C
Charge on the capacitor is given by;
Q = CV
Where;
C is the capacitance of the capacitor, given as;
C = ε₀A / d
Then, the potential difference across the plates is given by;

Therefore, the potential difference across the plates is 226 V.