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3 years ago

What is the marble's range if it is fired horizontally from 1.6 m above the ground?

Physics

1 answer:

cupoosta [38]3 years ago

5 0

Answer:

The answer is 6.40 meters.

Explanation:

The speed v = √(2gh)

v = √( 2×9.8×6.4) = 11.2 m/s

After, finding the time it takes to hit the ground from a height of 1.6 meters.

time = √(2H÷g)

time = √(2×1.6÷9.8)

time = 0.5714 seconds.

Horizontal distance is speed × time = 11.2 × 0.5714 = 6.40 meters.

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4 years ago

The inner and outer surfaces of a cell membrane carry a negative and positive charge, respectively. Because of these charges, a

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Electric field, E = $6.36\times 10^{6}\ V/m$

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The potential difference across the membrane, $\Delta V=0.063\ V$

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We need to find the magnitude of the electric field in the membrane. We know that the relation between electric field and electric potential is given by :

$E=\dfrac{-\Delta V}{\Delta x}\\\\E=\dfrac{0.063\ V}{9.9\times 10^{-9}\ m}\\\\E=6.36\times 10^{6}\ V/m$

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4 years ago

A model rocket is launched straight upward with an initial speed of 52.0 m/s. It accelerates with a constant upward acceleration

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Explanation:

(a) After the engines stop, the rocket reaches a maximum height at which it will stop and begin to descend in free fall due to gravity.

(b) We must separate the motion into two parts, when the rocket's engines is on and when the rocket's engines is off.

First we must find the rocket speed when the engines stop:

$v_f^2=v_0^2+2ay_1\\v_f^2=(52\frac{m}{s})^2+2(1\frac{m}{s^2})(160m)\\v_f^2=3024\frac{m^2}{s^2}\\v_f=\sqrt{3024\frac{m^2}{s^2}}=54.99\frac{m}{s}$

This final speed is the initial speed in the second part of the motion, when engines stop until reach its maximun height. Therefore, in this part the final speed its zero and the value of g its negative, since decelerates the rocket:

$v_f^2=v_0^2+2gy_{2}\\y_{2}=\frac{v_f^2-v_0^2}{2g}\\y_{2}=\frac{0^2-(54.99\frac{m}{s})^2}{2(-9.8\frac{m}{s^2})}=154.28m$

So, the maximum height reached by the rocket is:

$h=y_1+y_2\\h=160m+154.28m=314.28m$

$v_f=v_0+at_1\\t_1=\frac{v_f-v_0}{a}\\t_1=\frac{54.99\frac{m}{s}-52\frac{m}{s}}{1\frac{m}{s^2}}\\t_1=2.99s$

And in the second part:

$t_2=\frac{v_f-v_0}{g}\\t_2=\frac{0-54.99\frac{m}{s}}{-9.8\frac{m}{s^2}}\\t_2=5.61s$

So, the time it takes to reach the maximum height is:

$t_3=t_1+t_2\\t_3=2.99s+5.61s=8.60s$

(d) We already know the time between the liftoff and the maximum height, we must find the rocket's time between the maximum height and the ground, therefore, is a free fall motion:

$v_f^2=v_0^2+2ay\\v_f^2=0^2+2(9.8\frac{m}{s^2})(314.28m)\\v_f=\sqrt{6159.888\frac{m^2}{s^2}}=78.48\frac{m}{s}$

$t_4=\frac{v_f-v_0}{g}\\t_4=\frac{78.48\frac{m}{s}-0}{9.8\frac{m}{s^2}}\\t_4=8.01s$

So, the total time is:

$t=t_3+t_4\\t=8.60s+8.01s\\t=16.61s$

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4 years ago