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Novay_Z [31]
3 years ago
15

Water drips from the nozzle of a shower onto the floor 193 cm below. The drops fall at regular (equal) intervals of time, the fi

rst drop striking the floor at the instant the fourth drop begins to fall. When the first drop strikes the floor, how far below the nozzle are the (a) second and (b) third drops?
Physics
1 answer:
Law Incorporation [45]3 years ago
8 0

Answer: 108.81 cm and 48.66 cm

Explanation:

In this, we have to make sure to keep in mind the Gravity effects on the drops. The drops will accelerate when they fall making them travel faster. This means, the velocity is not constant.

What is know:

Height (h) = 193 cm

Gravity (g) =  981 cm/s^{2}

Initial Velocity = 0

First, we can know how long it take to the drop to travel to the floor. It can be done with the following equation:

x = V_{0} t + \frac{1}{2} at^{2}    (1)

Where:

x is the distance which is 193cm

Vo is the Initial Velocity  which is zero

t is the time the time it takes the drop to travel from the shower to the floor

a is the aceleration, which in this case is the gravity.

With the Initial Velocity equals zero the equations simply:

193 cm = \frac{1}{2}gt^{2}

To search for the time:

t =\sqrt{\frac{2*193cm}{981cm/s^{2} } }

t = 0.627 s

This is the time it takes a drop to fall to the floor, with this time and knowing other 3 drops have driped from the shower by this time. We can calculate how much time it takes the shower to drip each drop.

Time for Drip = t/4

Time for Drip = 0.156

This time is the difference between each drop, using the same equation we can calculate where was each drop, because now it is know how much time had each drop after being drip from the shower.

Our first is already on the floor (193 cm) with 0.627 s, The second drops have been falling for (0.627s - 0.156) 0.471 s and our third drop for (0.627s - 0.156 - 0.156) 0.315 s

We can use (1) to know how far have each drop traveled on these times. We know the Initials Velocity are 0, know we need ot know the distances.

For the second drop:

x = \frac{1}{2} (981cm/s^{2})(0.471s)^{2}

x =108.81 cm

For the third drop:

x = \frac{1}{2} (981cm/s^{2})(0.315s)^{2}

x = 48.66 cm

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PLEASE HELP
Sergeu [11.5K]

The vertical component of the initial velocity is v_0_y = \frac{y}{t} + \frac{1}{2} gt

The horizontal component of the initial velocity is v_0_x = \frac{x}{t}

The horizontal displacement when the object reaches maximum height is X = \frac{xy}{gt^2} + \frac{x}{2}

The given parameters;

the horizontal displacement of the object, = x

the vertical displacement of the object, = y

acceleration due to gravity, = g

time of motion, = t

The vertical component of the initial velocity is given as;

y = v_0_yt - \frac{1}{2} gt^2\\\\v_0_yt = y + \frac{1}{2} gt^2\\\\v_0_y = \frac{y}{t} + \frac{1}{2} gt

The horizontal component of the initial velocity is calculated as;

x = v_0_xt\\\\v_0_x = \frac{x}{t}

The time to reach to the maximum height is calculated as;

T = \frac{v_f_y -v_0_y}{-g} \\\\T = \frac{-v_0_y}{-g} \\\\T = \frac{v_0_y}{g} \\\\T =  \frac{1}{g}  (v_0_y)\\\\T = \frac{1}{g} (\frac{y}{t} + \frac{1}{2} gt)\\\\T = \frac{y}{gt} + \frac{1}{2} t

The horizontal displacement when the object reaches maximum height is calculated as;

X= v_0_x \times T\\\\X= \frac{x}{t} \times (\frac{y}{gt} + \frac{1}{2} t)\\\\X = \frac{xy}{gt^2} + \frac{x}{2}

Learn more here: brainly.com/question/20689870

8 0
3 years ago
Ejercicio 1: Un cuerpo gira en un círculo de 8cm de diámetro con una rapidez constante
Lera25 [3.4K]

Answer:

Exercise 1;

The centripetal acceleration is approximately 94.52 m/s²

Explanation:

1) The given parameters are;

The diameter of the circle = 8 cm = 0.08 m

The radius of the circle = Diameter/2 = 0.08/2 = 0.04 m

The speed of motion = 7 km/h = 1.944444 m/s

The centripetal acceleration = v²/r = 1.944444²/0.04 ≈ 94.52 m/s²

The centripetal acceleration ≈ 94.52 m/s²

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