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Lesechka [4]
3 years ago
12

Crickets Chirpy and Milada jump from the top of a vertical cliff. Chirpy drops downward and reaches the ground in 2.70 s, while

Milada jumps horizontally with an initial speed of 95.0 cm/s. How far from the base of the cliff will Milada hit the ground? Ignore air resistance.
Physics
1 answer:
Vinvika [58]3 years ago
7 0

Answer:

Explanation:

Given

Time taken to reach ground is t=2.7\ s

Malda initial velocity u=95\ cm/s

Let h be the height of Cliff

using h=ut+\frac{1}{2}at^2

where, u=initial velocity

t=time

In first case chirpy drop downward thus u=0

h=0+\frac{1}{2}(9.8)(2.7)^2

h=35.72\ m

For Milada there is horizontal velocity u=95 cm/s=0.95 m/s[/tex]

time taken to reach the ground will be same so distance traveled in this time with 0.95 m/s horizontal velocity is given by

R=u\times t

R=0.95\times 2.7=2.43\ m    

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A volcano erupts and launches a chunk of hot magma horizontally with a speed of 252 m/s. The magma travels a horizontal distance
ArbitrLikvidat [17]

Answer:

The value is v_y  =  -48.61 \ m/s

Explanation:

From the question we are told that

   The horizontal speed is  u_x  = 252 \  m/s

    The horizontal distance is  d = 1250 \ m

Generally the time taken by the hot magma in air before landing is mathematically represented as

       t = \frac{d}{u_x}

=>    t = \frac{ 1250 }{252}

=>    t = 4.96 \  s

Generally the initial vertical velocity of the magma when it was lunched is  

    u_y = 0 \ m/ s

Then the final velocity of the magma when it hits the ground is mathematically represented s

       - v_y  =  u_y + gt

Here the negative sign mean that the direction of the velocity is towards the negative y -axis

So  

        - v_y  =  48.61 \ m/s

=>     v_y  =  -48.61 \ m/s

7 0
3 years ago
List any similarities between discovered exoplanets and planets in our solar system.
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3 0
3 years ago
Se lanza verticalmente una esfera con una rapidez de 30m/se. Determinar la rapidez de la esfera a una altura de 40m (g=10m/s2)
sergeinik [125]

v^2-{v_0}^2=2a(x-x_0)

dónde v es la velocidad de la esfera, v_0 es suya velocidad inicial, a=-g la aceleración debida a la gravedad, x la posición, y x_0 la posición inicial. Tomamos x_0=0\,\mathrm m a referirse a la posición de la esfera en el momento que la esfera fue lanzada.

Entonces

v^2-\left(30\,\dfrac{\mathrm m}{\mathrm s}\right)^2=2\left(-10\,\dfrac{\mathrm m}{\mathrm s^2}\right)(40\,\mathrm m)

\implies v^2=100\,\dfrac{\mathrm m^2}{\mathrm s^2}\implies v=\pm10\,\dfrac{\mathrm m}{\mathrm s}

Esto nos dice que la esfera alcanza una altura de 40 m en dos momentos - una vez hacia arriba y una vez hacia abajo. Sin embargo, independientemente del signo de la velocidad, sabemos que suya magnitud es 10 m/s, y así tenemos una rapidez de 10 m/s también en ambos momentos.

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