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Ludmilka [50]
3 years ago
8

A ball is dropped from rest from the top of a cliff that is 24 m high. From ground level, a second ball is thrown straight upwar

d at the same instant that the first ball is dropped. The initial speed of the second ball is exactly the same as that with which the first ball eventually hits the ground. In the absence of air resistance, the motions of the balls are just the reverse of each other. Determine how far below the top of the cliff the balls cross paths
Physics
1 answer:
sesenic [268]3 years ago
4 0

Answer:

6.0 m below the top of the cliff

Explanation:

We can find the velocity at which the ball dropped from the cliff reaches the ground by using the SUVAT equation

v^2-u^2 = 2gd

where

u = 0 (it starts from rest)

g = 9.8 m/s^2 (acceleration of gravity, we assume downward as positive direction)

h = 24 m is the distance covered

Solving for h,

v=\sqrt{2gh}=\sqrt{2(9.8)(24)}=21.7 m/s

So the ball thrown upward is launched with this initial velocity:

u = 21.7 m/s

From now on, we take instead upward as positive direction.

The vertical position of the ball dropped from the cliff at time t is

y_1 = h - \frac{1}{2}gt^2

While the vertical position of the ball thrown upward is

y_2 = ut - \frac{1}{2}gt^2

The two balls meet when

y_1 = y_2\\h-\frac{1}{2}gt^2 = ut - \frac{1}{2}gt^2 \\h = ut \rightarrow t = \frac{h}{u}=\frac{24}{21.7}=1.11 s

So the two balls meet after 1.11 s, when the position of the ball dropped from the cliff is

y_1 = h -\frac{1}{2}gt^2 = 24-\frac{1}{2}(9.8)(1.11)^2=18.0 m

So the distance below the top of the cliff is

d=24.0 - 18.0 = 6.0 m

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Gravitational potential energy can be calculated using the formula:

PE_{grav} =mgh

Where:
PEgrav = Gravitational potential energy
m= mass
g = acceleration due to gravity
h = height

On Earth acceleration due to gravity is a constant 9.8 but since the scenario is on Mars, the pull of gravity is different. In this case, it is 3.7, so we will use that for g.

So put in what you know and solve for what you don't know. 
m = 10kg
g = 3.7m/s^2
h = 1m

So we put that in and solve it. 
PE_{grav} =mgh
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3 years ago
Consider the train car described in the previous part. Another experiment is conducted in it: A net force of 20N is applied to a
nordsb [41]

Answer:

No

Explanation:

The supplied information about the object and train is incomplete. Acceleration is the rate at which the velocity of a body changes with time. Here the velocity and time is not given

7 0
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Un móvil se encuentra en la posición inicial x0 = 22 m, y se mueve con velocidad 5 m/s. Calcula la posición en la que se encuent
Brilliant_brown [7]

Answer:

La posición en la que se encuentra el móvil en el instante t = 30 s es 172 m.

Explanation:

El movimiento rectilíneo uniforme (MRU) es el movimiento que describe un cuerpo o partícula a través de una línea recta a velocidad constante.

La distancia recorrida,  x , por un móvil que tiene un MRU con un velocidad  v  durante el intervalo de tiempo  t  es:

x= x0 + v*t

donde x0 es la posición inicial.

En este caso:

  • x0= 22 m
  • v= 5 m/s
  • t= 30 s

Reemplazando:

x= 22 m + 5 m/s* 30 s

Resolviendo:

x= 22 m + 150 m

x= 172 m

<u><em>La posición en la que se encuentra el móvil en el instante t = 30 s es 172 m.</em></u>

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3 0
3 years ago
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A thin taut string is fixed at both ends and stretched along the horizontal x-axis with its left end at x = 0. It is vibrating i
Fofino [41]

Answer:

(a) Wavelength is 0.436 m

(b) Length is 0.872 m

(c) 11.518 m/s

Solution:

As per the question:

The eqn of the displacement is given by:

y(x, t) = (1.22 cm)sin[14.4 m^{- 1}x]cos[(166\ rad/s)t]          (1)

n = 4

Now,

We know the standard eqn is given by:

y = AsinKxcos\omega t           (2)

Now, on comparing eqn (1) and (2):

A = 1.22 cm

K = 14.4 m^{- 1}

\omega = 166\ rad/s

where

A = Amplitude

K = Propagation constant

\omega = angular velocity

Now, to calculate the string's wavelength,

(a) K = \frac{2\pi}{\lambda}

where

K = propagation vector

\lambda = \frac{2\pi}{K}

\lambda = \frac{2\pi}{14.4} = 0.436\ m

(b) The length of the string is given by:

l = \frac{n\lambda}{2}

l = \frac{4\times 0.436}{2} = 0.872\ m

(c)  Now, we first find the frequency of the wave:

\omega = 2\pi f

f = \frac{\omega}{2\pi}

f = \frac{2\pi}{166} = 26.42\ Hz

Now,

Speed of the wave is given by:

v = f\lambda

v = 26.419\times 0.436 = 11.518\ m/s

4 0
3 years ago
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