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Svetach [21]3 years ago

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El análisis de un grafico de velocidad versus tiempo permite encontrar los resultados para las preguntas son:

a) La velocidad es v= 0,6 m/s

b) La distancia recorrida dₐ = 1,5 m

c) La distancia recorrida entre t= 5 y 10 s es $d_b$ = 3,0 m

d) La aceleración a = -0,4 m/s²

e) La distancia total recorrida es $d_{total}$ = 5,3 m

f) La velocidad media es $v_{media}$ = 0,44 m/s

La cinemática estudia el movimiento de los cuerpos, buscando relaciones entre la posición, la velocidad y la aceleración de los cuerpos.

La velocidad es definida como el cambio de la posición en función del tiempo y la aceleración es el cambio de la velocidad en el intervalo de tiempo.

. $v= \frac{\Delta x}{t} \\a = \frac{\Delta v}{t}$

Donde v y a son la velocidad y aceleración, respectivamente, t es el tiempo y Δx y Δv son la variación de la posición y la velocidad, respectivamente.

En el adjunto muestran un grafico de la velocidad en función del tiempo, donde del área bajo la curva obtenemos la distancia recorrida y la pendiente corresponde a la aceleración del cuerpo.

Respondamos las preguntas:

a) La velocidad de la bola a los 5 s.

Como es un grafico de velocidad versus tiempo, la velocidad puede ser leída directamente, en el grafico tenemos un valor de

v= 0,6 m/s

b) La distancia recorrida corresponde al área bajo la curva.

$d_a = \frac{v_f - v_o}{2} t \\d_a = \frac{0.6-0}{ 2}\ 5$

d = 1,5 m

c) La distancia en el intervalo de 5 a 10 s.

En este intervalo la velocidad es constante, el área es

$d_b = v \Delta t\\d_b = 0.6 \ 5\\$

d = 3 m

d) La aceleración en el intervalo 10 a 12 s.

La aceleracion es la pendiente del grafico

a = $\frac{\Delta v}{\Delta t}$

a = $\frac{-0.2 - 0.6}{12-10}$

a = -0,4 m/s²

El signo negativo indica que la aceleración se opone a la velocidad del cuerpo.

e) La distancia total recorrida

$d_{total} = d_a + d_b + d_c$

dₐ = 1,5 m

$d_b$ = 3,0 m

Busquemos la distancia en el ultimo intervalo

$d_c = \frac{0.6 + 0.2}{2} ( 12-10)$

$d_c$ = 0,8 m

Substituimos

$d_{total}$ = 1,5 + 3,0 + 0,8

$d_{total}$ = 5,3 m

f) La rapidez de la bola

La rapidez prmedio es la relación entre la distancia recorrida en el tiempo.

$v_{avg} = \frac{\Delta x}{\Delta t }$

$v_{avg} = \frac{5.3}{12}$ <u> </u>

$v_{avg}$ =0,44 m/s

En conclusión usando el análisis de un grafico de velocidad versus tiempo podemos encontrar los resultados para las preguntas son:

a) La velocidad es v= 0,6 m/s

b) La distancia recorrida da = 1,5 m

c) La distancia recorrida entre t= 5 y 10 s es db = 3,0 m

d) La aceleración a = -0,4 m/s²

e) La distancia total recorrida es $d_{total}$ = 5,3 m

f) La velocidad media es $v_{avg}$ = 0,44 m/s

Aprender mas aquí: brainly.com/question/22785960

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Answer:

the third stage was 480 km long

Explanation:

Stage 1:

Time = 1 hours

Speed = 80km

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Time = 2 hours

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Let the Distance at the stage 3 be x

Average speed of the train route = 100 km/h

$\frac{ \text{speed at stage 1} + \text{speed at stage 2} + \text{speed at stage 3}}{3} = 0$

$\frac{ \text{speed at stage 1} + \text{speed at stage 2} + \text{speed at stage 3}}{3} = 100$

Lets find the speed at stage 1

Speed = $\frac{Distance }{Time}$

Speed = $\frac{80}{1}$

Speed 1= 80 km/hr

The speed at stage 2

Speed = $\frac{Distance }{Time}$

Speed = $\frac{200}{2}$

Speed 2 = 100 km/hr

The speed at stage 3

Speed = $\frac{Distance }{Time}$

Speed = $\frac{x}{4}$

Speed 3 = $\frac{x}{4}$

we kow that average is ,

$\frac{ \text{speed 1} + \text{speed 2} + \text{speed 3}}{3} = 100$

$\frac{ 80 + 100+ \frac{x}{4} }{3} = 100$

$\frac{ 180 + \frac{x}{4} }{3} = 100$

$\frac{ \frac{720 +x}{4} }{3} = 100$

$\frac{720 +x}{4} \times \frac{1}{3} = 100$

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Two Earth satellites, A and B, each of mass m, are to be launched into circular orbits about Earth's center. Satellite A is to o

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(a) 0.473

The potential energy of a satellite orbiting around Earth is given by

$U=-\frac{GMm}{R+h}$

where

G is the gravitational constant

M is the Earth's mass

m is the satellite's mass

R is the Earth's radius

h is the altitude of the satellite above the Earth's surface

So the potential energy of satellite A is

$U_A=-\frac{GMm}{R+h_A}$

while potential energy of satellite B is

$U_B=-\frac{GMm}{R+h_B}$

Therefore the ratio of the potential energy of satellite B to that of satellite A is

$\frac{U_B}{U_A}=\frac{R+h_A}{R+h_B}$

and using

hA = 5920 km

hB = 19600 km

R = 6370 km

we find

$\frac{U_B}{U_A}=\frac{6370+5920}{6370+19600}=0.473$

(b) 0.473

The kinetic energy of a satellite orbiting around Earth instead is given by

$K=\frac{GMm}{2(R+h)}$

So the kinetic energy of satellite A is

$K_A=\frac{GMm}{2(R+h_A)}$

while kinetic energy of satellite B is

$K_B=\frac{GMm}{2(R+h_B)}$

Therefore the ratio of the kinetic energy of satellite B to that of satellite A is

$\frac{K_B}{K_A}=\frac{R+h_A}{R+h_B}$

which is identical to before, so it gives

$\frac{K_B}{K_A}=\frac{6370+5920}{6370+19600}=0.473$

(c) Satellite B

The total energy of a satellite in orbit is given by

$E=U+K = -\frac{GMm}{R+h}+\frac{GMm}{2(R+h)}=-\frac{GMm}{2(R+h)}$

We see that the total energy is:

1) negative (because the satellite is on a bound orbit)

2) inversely proportional to the distance of the satellite from the Earth's center, R+h

So the magnitude of the fraction in the equation is larger for the satellite which is closer to the Earth's surface (satellite A), but since the energy is negative, this means that the total energy of this satellite is smaller than that of satellite B. So, satellite B has a greater total energy.

(d) $1.03\cdot 10^7 J$

We have to calculate the total energy of each satellite.

Given:

$G=6.67\cdot 10^{-11}$

$M=5.98\cdot 10^{24} kg$

m = 12.0 kg

$R+h_A = 6370 km+5920 km=12290 km = 12.3 \cdot 10^6 m$

$R+h_B = 6370 km+19600 km=25970 km = 26.0 \cdot 10^6 m$

We find:

$E_A = - \frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24})(12.0)}{2(12.3\cdot 10^6)}=-1.95\cdot 10^{7} J$

$E_B = - \frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24})(12.0)}{2(26.0\cdot 10^6)}=-9.2\cdot 10^{6} J$

So the difference in total energy is

$E_B-E_A = -9.2\cdot 10^6 - (-1.95\cdot 10^7) =1.03\cdot 10^7 J$

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