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Alexxx [7]
3 years ago
6

A girl stands on the edge of a merry-go-round of radius 1.71 m. If the merry go round uniformly accerlerates from rest to 20 rpm

in 6.73s. what is the magnitiude of the girls average acceleration.
Physics
1 answer:
Mashutka [201]3 years ago
8 0

Answer:

a = 0.53 m/s^2

Explanation:

initially the merry go round is at rest

after 6.73 s the merry go round will accelerates to 20 rpm

so final angular speed is given as

\omega = 2\pi f

\omega = 2\pi ( \frac{20}{60})

\omega = 2.10 rad/s

so final tangential speed is given as

v = r\omega

v = 1.71 (2.10) = 3.58 m/s

now average acceleration of the girl is given as

a = \frac{v_f - v_i}{\Delta t}

a = \frac{3.58 - 0}{6.73}

a = 0.53 m/s^2

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3 0
2 years ago
Si tenemos tres cubos del mismo tamaño (hierro, madera e icopor). ¿Qué diferencias puede encontrar entre ellos?
cricket20 [7]

Answer: La diferencia es el peso (o la masa), siendo que el cubo de hierro es el mas pesado, después viene el de madera y después el de icopor.

Explanation:

Ok, los 3 cubos tienen el mismo tamaño, lo que implica que tienen el mismo volumen.

Ahora es útil recordar la relación:

Densidad = masa/volumen.

Masa = densidad*volumen.

Nosotros sabemos que la densidad del hierro es mas grande que la de la madera, y la densidad de la madera es mas grande que la del icopor.

Entonces, por la relación anterior, y sabiendo que todos los cubos tienen el mismo volumen, podemos reconocer que el cubo de hierro tiene mayor masa, después viene el de madera y después viene el de icopor.

Y sabiendo que:

masa*gravedad = peso

podemos saber que el cubo mas pesado es el de hierro, después el de madera y después el de icopor.

Además de esta diferencia, también hay otras que no dependen tanto del tamaño del objeto, como pueden ser las capacidades caloríficas, el como reaccionan a campos eléctricos y cosas así que son triviales, pues son diferentes para casi todos los materiales.

5 0
3 years ago
A rocket moves upward, starting from rest with an acceleration of 25.4 m/s^2 for 3.39 s. It runs out of fuel at the end of the 3
kolbaska11 [484]

Explanation:

Initial speed of the rocket, u = 0

Acceleration of the rocket, a=25.4\ m/s^2

Time taken, t = 3.39 s

Let v is the final velocity of the rocket when it runs out of fuels. Using the equation of kinematics as :

v=u+at

v=25.4\times 3.39=86.10\ m/s    

Let x is the initial position of the rocket. Using third equation of kinematics as :

v^2=u^2+2ax_o

x_o=\dfrac{v^2}{2a}

x_o=\dfrac{86.10^2}{2\times 25.4}=145.92\ m  

Let x_o is the position at the maximum height. Again using equation of motion as :

v^2-u^2=2a(x-x_o)

Now a=-g and v and u will interchange

u^2=2g(x-x_o)

x=x_o+\dfrac{u^2}{2g}

x=145.92+\dfrac{(86.10)^2}{2\times 9.8}

x = 524.14 meters

Hence, this is the required solution.

5 0
3 years ago
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