Se dispara un proyectil desde el nivel del suelo con una velocidad inicial de 50 m/s a 38° porencima de la horizontal. Sin considerar la resistencia del aire. Determineel tiempo que tarda el proyectil en alcanzar su punto más alto,b)¿Cuál es su altura máxima?, c) Cual es el alcance horizontal del proyectil.
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Ответ:
Explanation:
Wavelength of light λ = 610 nm
Screen distance D = 3.00 m
Slit separation be d
position of first bright fringe = λ D / d
Putting the values given
4.84 x 10⁻³ =![\frac{610\times10^{-9}\times3}{d}](/tpl/images/0646/6412/43093.png)
d =![\frac{610\times10^{-9}\times3}{4.84\times10^{-3}}](/tpl/images/0646/6412/e813c.png)
= 378.1 x 10⁻⁶ m
Let the required wave length be λ₁
Equation for position of dark fringe is given by following relation
x =![\frac{(2n+1)\lambda\times D}{2d}](/tpl/images/0646/6412/9315b.png)
Putting the value in the equation
4.84 x 10⁻³ =
, for first dark fringe n = 0
λ₁ = 1220 x 10⁻⁹
= 1220 nm .