Answer : whats the question
Answer:
The width of third and fifth order bright fringe is 0.00076 rad and 0.00127 rad.
Explanation:
Given that,
Distance d = 2.00 mm
Wavelength = 511 nm
Order number = 3
Order number = 5
We need to calculate the width of third-order bright fringe
Using formula of width
Put the value into the formula
We need to calculate the width of fifth-order bright fringe
Using formula of width
Put the value into the formula
Hence, The width of third and fifth order bright fringe is 0.00076 rad and 0.00127 rad.
Responder:
Explicación:
Para que podamos calcular el tiempo que le tomará al esquiador permanecer en el aire, encontraremos el tiempo de vuelo como se muestra;
T = 2Usin theta / 2
theta = 90 grados
U = 25 m / s
T = 25sin90 / 2 (9,8)
T = 25 / 19,62
T = 1,27 segundos
Por lo tanto, los cielos usarán 1.27 segundos en el aire.
La distancia horizontal es el rango;
Rango R = U√2H / g
R = 25√2 (80) /9,8
R = 25√160 / 9,8
R = 25 * √16,326
R = 25 * 4.04
R = 101,02 m
Por tanto, la distancia horizontal recorrida por el esquiador es 101,02 m
Answer:
electrostatic attraction
Explanation:
Atoms form chemical bonds with other atoms when there's an electrostatic attraction between them. This attraction results from the properties and characteristics of the atoms' outermost electrons, which are known as valence electrons.
|acceleration| = (change in speed) / (time for the change)
Change in the car's speed = (27 - 0) = 27 m/s
Time for the change = 10 sec
|acceleration| = (27 m/s) / (10 s) = 2.7 m/s² .
That's the magnitude of the car's acceleration.
We don't know anything about its direction.
