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mr Goodwill [35]
4 years ago
8

Se apunta un rifle horizontalmente con mira a un blanco pequeño que está a 200m en el suelo. La velocidad inicial de la bala es

de 500m/s. Calcular:
¿Qué tiempo dura la bala en el aire?
¿A qué altura del suelo estaba el rifle?
Physics
1 answer:
vfiekz [6]4 years ago
3 0

Answer:

Lo importante a tener en cuenta sobre esta pregunta es que la velocidad horizontal de la bala no hace ninguna diferencia en cuanto al tiempo que tarda en caer al suelo.

Debido a que el arma no ha aplicado ninguna fuerza vertical a la bala, la única fuerza que afecta la bala es la gravedad. Esto significa que la bala tarda tanto en caer al suelo como lo haría si se cayera, a pesar de que ahora viaja una gran distancia horizontal en la duración.

Para encontrar el tiempo de viaje antes de tocar el suelo, tenemos 3 valores:

-El desplazamiento desde el suelo que la bala debe viajar, s = 1.5m

-La aceleración que experimenta la bala. Como la gravedad está acelerando la bala hacia abajo, a = g = ~ 9.81m / s ^ 2

-La velocidad inicial de la bala verticalmente. Como la bala es estacionaria verticalmente (solo viaja horizontalmente al inicio), u = 0m

Examinamos nuestras ecuaciones de movimiento, comúnmente conocidas como ecuaciones SUVAT. Es posible que necesite aprender estos para su examen, pero algunas tablas de examen los proporcionan.

Debido a que tenemos s, u y a, y estamos buscando el tiempo t, la ecuación relevante es

s = ut + 0.5 (en ^ 2)

Completando nuestros valores tenemos:

1.5 = 0t + 0.5 (9.81 x t ^ 2)

1.5 = 4.905 x t ^ 2

Divide 1.5 entre 4.905 para encontrar t ^ 2

t ^ 2 = 0.3058 ...

Simplemente encontramos la raíz cuadrada de t ^ 2 para encontrar t, el tiempo que tarda la bala en llegar al suelo:

t = 0.553s (3 cifras significativas)

Para encontrar la distancia horizontal, d, que la bala ha viajado antes de tocar el suelo, podemos usar la ecuación que vincula el desplazamiento s con cierta velocidad v durante un tiempo t:

s = vt

La velocidad horizontal de la bala, v = 430

El tiempo antes de que la bala toque el suelo, t = 0.553

Entonces d = vt = 430 * 0.553 = 238m (3 cifras significativas)

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2 years ago
An atom of carbon has a radius of 67.0 pm and the average orbital speed of the electrons in it is about 1.3 x 10⁶ m/s.
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Answer :

The least possible uncertainty in an electron's velocity is, 4.32\times 10^{5}m/s

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According to the Heisenberg's uncertainty principle,

\Delta x\times \Delta p=\frac{h}{4\pi} ...........(1)

where,

\Delta x = uncertainty in position

\Delta p = uncertainty in momentum

h = Planck's constant

And as we know that the momentum is the product of mass and velocity of an object.

p=m\times v

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\Delta p=m\times \Delta v      .......(2)

Equating 1 and 2, we get:

\Delta x\times m\times \Delta v=\frac{h}{4\pi}

\Delta v=\frac{h}{4\pi \Delta x\times m}

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m = mass of electron = 9.11\times 10^{-31}kg

h = Planck's constant = 6.626\times 10^{-34}Js

radius of atom = 67.0pm=67.0\times 10^{-12}m     (1pm=10^{-12}m)

\Delta x = diameter of atom = 2\times 67.0\times 10^{-12}m=134.0\times 10^{-12}m

Now put all the given values in the above formula, we get:

\Delta v=\frac{6.626\times 10^{-34}Js}{4\times 3.14\times (134.0\times 10^{-12}m)\times (9.11\times 10^{-31}kg)}

\Delta v=4.32\times 10^{5}m/s

The minimum uncertainty in an electron's velocity is, 4.32\times 10^{5}m/s

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Percentage of average speed = \frac{\text{Uncertainty in speed}}{\text{Average speed}}\times 100

Uncertainty in speed = 4.32\times 10^{5}m/s

Average speed = 1.3\times 10^{6}m/s

Percentage of average speed = \frac{4.32\times 10^{5}m/s}{1.3\times 10^{6}m/s}\times 100

Percentage of average speed = 33.2 % ≈ 33 %

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3 0
3 years ago
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