This is a Fraunhofer single slit experiment, where the light passing through the slit produces an interference pattern on the screen, and where the dark bands (minima of diffraction) are located at a distance of
from the center of the pattern. In the formula, m is the order of the minimum,
the wavelenght,
the distance of the screen from the slit and
the width of the slit.
In our problem, the distance of the first-order band (m=1) is
. The distance of the screen is D=86 cm while the wavelength is
. Using these data and re-arranging the formula, we can find a, the width of the slit:
The answer is B frequency. When frequency increases more wave crests pass a fixed point each second. That means the wavelength shortens. So, as frequency increases, wavelength decreases
( (77/4) + 76/2 )/2 = 28.625 km/h is what i got
Answer:
0.187 m
Explanation:
We'll begin by calculating the acceleration of the ball. This can be obtained as follow:
Mass (m) = 0.450 Kg
Force (F) = 38 N
Acceleration (a) =?
F = m × a
38 = 0.450 × a
Divide both side by 0.450
a = 38 / 0.450
a = 84.44 m/s²
Finally, we shall determine the distance. This can be obtained as follow:
Initial velocity (u) = 2.20 m/s.
Final velocity (v) = 6 m/s
Acceleration (a) = 84.44 m/s²
Distance (s) =?
v² = u² + 2as
6² = 2.2² + (2 × 84.44 × s)
36 = 4.4 + 168.88s
Collect like terms
36 – 4.84 = 168.88s
31.52 = 168.88s
Divide both side by 168.88
s = 31.52 / 168.88
s = 0.187 m
Thus, the distance is 0.187 m
Depending on the height of the building they can break due to impact on the floor.
