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Paul [167]
3 years ago
8

Mary and her younger brother Alex decide to ride the 17-foot-diameter carousel at the State Fair. Mary sits on one of the horses

in the outer section at a distance of 8 feet from the center. Alex decides to play it safe and chooses to sit in the inner section at a distance of 5 feet from the center.(a) What is Mary's angular speed ?M compared to that of Alex's angular speed ?A? Give your answer as a multiple of ?A.(b) What is Mary's tangential speed vM compared to that of Alex's tangential speed vA? Give your answer as a multiple of vA.
Physics
1 answer:
lorasvet [3.4K]3 years ago
5 0

Answer:

\omege_A=\omega_M

v_M=1.6v_A

Explanation:

A denotes Alex

M denotes Mary

r = Distance from center

Mary and Alex will have the equal displacements in equal interval of time as they are in uniform circular motion. So,

\omega_A=\omega_M

Tangential speed speed is given by

\dfrac{v_M}{v_A}=\dfrac{r_M\omega_M}{r_A\omega_A}\\\Rightarrow \dfrac{v_M}{v_A}=\dfrac{r_M}{r_A}\\\Rightarrow \dfrac{v_M}{v_A}=\dfrac{8}{5}\\\Rightarrow \dfrac{v_M}{v_A}=1.6\\\Rightarrow v_M=1.6v_A

The tangential speed of Mary is v_M=1.6v_A

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A spring with a mass of 5 Kg has natural length 0.5m. A force of 35.6 N is required to maintain it stretched to a length of 0.5m
aleksandrvk [35]

Answer:

Explanation:

force constant of spring  k = force / extension

= 35.6 / 0.5

k = 71.2 N / m

angular frequency ω of oscillation by spring mass system

\omega = \sqrt{\frac{k}{m} }

where m is mass of the body attached with spring

Putting the values

\omega = \sqrt{\frac{71.2}{5} }

ω = 3.77 radian / s

The oscillation of the mass will be like SHM having amplitude of 0.5 m and angular frequency of 3.77 radian /s . Initial phase will be π / 2

so the equation for displacement from equilibrium position that is middle point can be given as follows

x = .5 sin ( ω t + π / 2 )

= 0.5 cos ω t

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x = 0.5 cos 3.77 t .

8 0
3 years ago
A projectile is fired at an angle of 53° to the horizontal with a speed of 80. meters per second. What is the vertical component
ElenaW [278]

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Explanation:❤️

3 0
3 years ago
Un objeto de 200 gramos está amarrado del extremo de una cuerda y gira describiendo un círculo horizontal de 1.20 m de radio a r
vesna_86 [32]

Answer:

La tensión es 85.3 N.

Explanation:

Cuando el objeto gira en dirección horizontal, la sumatoria de fuerzas se puede calcular usando la segunda ley de Newton:  

\Sigma F_{x} =ma_{c}

T = ma_{c}  

Dado que el movimiento es horizontal, el peso (que está en el eje y) no contribuye en la sumatoria de fuerzas en el eje x. Por lo que la única fuerza actuando sobre el objeto en la dirección del movimiento es la tensión.  

En donde:                                          

m: es la masa del objeto = 200 g = 0.200 kg

a_{c}: es la aceleración centrípeta

La aceleración centrípeta viene dada por:  

a_{c} = \omega^{2} r

En donde:    

ω: es la velocidad angular del objeto = 3 rev/s

r: es el radio = 1.20 m

Entonces, la tensión es:

T = m\omega^{2} r = 0.200 kg(3\frac{rev}{s}*\frac{2\pi rad}{1 rev})^{2}*1.20 m = 85.3 N

   

Por lo tanto, la tensión es 85.3 N.  

Espero que te sea de utilidad!                                                                          

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5 0
3 years ago
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A pH scale goes from 0 - 14.
7 is in the middle so it is neutral
7 0
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