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:
Answer:
The 2 light bulbs can be connected in parallel to each other to avoid disconnection when one bulb burns out.
Explanation:
The parallel connection is required not series. A parallel connection is the connection of electronic components (e.g bulbs, LED, resistors, capacitors etc) in such a way that the same voltage is supplied across the ends of the components. While in a series connection, the components are connected to each other end-to-end.
As regard the question, parallel connection ensures that the brightness any of the bulbs is not affected with respect to the other bulbs. And other bulbs continue to function when any burns out. The 2 light bulbs should be connected in parallel to the baterry to avoid disconnection of all the bulbs.
It would be not be able to move yet it would be in the air
Answer:
a= 3.49 m/s^2
Explanation:
magnitude of total acceleration = sqrt{radial acceleration^2+tangential acceleration^2}.
we know that tangential acceleration a_t= change in velocity /time taken
now 90 km/h = 25 m/s
a_t = 25/17 = 1.47 m/s^2.
radial acceleration a_r = v^2/r
v= a_t×t = 1.47×13 = 19.11 m/s
a_r = 19.11^2/115= 3.175
now,
a= 3.49 m/s^2
