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LekaFEV [45]
3 years ago
8

Si se arroja una ròca en sentido horizontal desde un barranco de 100 m de altura. Choca

Physics
1 answer:
klasskru [66]3 years ago
7 0

x = 20 m

Calcular las componentes rectangulares de la velocidad inicial

En el lanzamiento horizontal la velocidad inicial vertical (Voy) es igual a cero, por lo que:

Vx = 20 m/s

Voy = 0

Paso No. 2: Anotar los datos para X y para Y. Recuerde que las velocidades y los desplazamientos

Para “X”Para “Y”Vx = 20 m/st =X =Voy = 0g= -9.81 m/s2Y = -5 m

Paso No. 3: Selección de las ecuaciones a utilizar

Recuerde que “X” que es la distancia horizontal que recorre un proyectil y para calcularla es necesario saber el valor de t (tiempo). Observe que en “Y ” tiene datos suficientes para calcular “t”. Paso 4: Resolver la ecuación considerando que Voy = 0, por lo que el primer término se anula.

Y= gt^2 / 2

Resolviendo para “ t “ :

t = 1.009637 s

Calculo de “ t “

 

Paso5: Calcular “ X “ utilizando la ecuación:  

Recuerde que “X” que es la distancia horizontal que recorre un proyectil y para calcularla es necesario saber el valor de t (tiempo). Observe que en “Y ” tiene datos suficientes para calcular “t”.

Resolviendo para “ X “ : X=Vx (t)

X = (20 m/s)(1.09637s)

X = 20 m

¡Por favor, marca Brainliest!

Gracias :)

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A cylindrical capacitor has an inner conductor of radius 2.7 mmmm and an outer conductor of radius 3.1 mmmm. The two conductors
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Answer:

(A) Capacitance per unit length = 4.02 \times 10^{-10}

(B) The magnitude of charge on both conductor is Q = 4.22 \times 10^{-19} C and the sign of charge on inner conductor is +Q and the sign on outer conductor is -Q

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Radius of inner part of conductor  (R_{1}) = 2.7 \times 10^{-3} m

Radius of outer part of conductor  (R_{2}) = 3.1 \times 10^{-3} m

The length of the capacitor (l) = 3 \times 10^{-3} m

(A)

Capacitance is purely geometrical property. It depends only on length, radius of conductor.

From the formula of cylindrical capacitor,      

     C = \frac{2\pi\epsilon_{o} l }{ln\frac{R_{2} }{R_{1} } }

Where, \epsilon_{o} = 8.85 \times 10^{-12}

But we need capacitance per unit length so,

     \frac{C}{l}  = \frac{2\pi\epsilon_{o}  }{ln\frac{R_{2} }{R_{1} } }

capacitance per unit length = \frac{6.28 \times 8.85 \times 10^{-12} }{ln(1.148)} = 4.02 \times 10^{-10}

(B)

The charge on both conductors is given by,

     Q = C \Delta V

Where, C = capacitance of cylindrical capacitor and value of C = 12.06 \times 10^{-13} F, \Delta V = 350 \times 10^{-3} V

∴ Q = 4.22 \times 10^{-19} C

The magnitude of charge on both conductor is same as above but the sign of charge is different.

Charge on inner conductor is +Q and Charge on outer conductor is -Q.

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3 years ago
The Millersburg Ferry (m = 13000.0 kg loaded) is travelling at 11 m/s when the engines are put in reverse. The engineproduces a
Pie

Explanation:

It is given that,

Mass of Millersburg Ferry, m = 13000 kg

Velocity, v = 11 m/s

Applied force, F = 10⁶ N

Time period, t = 20 seconds

(a) Impulse is given by the product of force and time taken i.e.

J=F.\Delta t

J=10^6\ N\times 20\ s

J=2\times 10^7\ N-s

(b) Impulse is also given by the change in momentum i.e.

J=\Delta p=p_f-p_i

J=p_f-p_i

p_f=J+p_i

p_f=2\times 10^7\ N-s+13000\ kg\times 11\ m/s

p_f=20143000\ kg-m/s

(c) For new velocity,

v_f=\dfrac{p_f}{m}

v_f=\dfrac{20143000\ kg-m/s}{13000\ kg}

v_f=1549.46\ m/s

Hence, this is the required solution.

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