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lord [1]
3 years ago
12

A baseball pitcher throws a baseball horizontally at a linear speed of 42.5 m/s (about 95 mi/h). Before being caught, the baseba

ll travels a horizontal distance of 17.4 m and rotates through an angle of 44.1 rad. The baseball has a radius of 3.67 cm and is rotating about an axis as it travels, much like the earth does. What is the tangential speed of a point on the "equator" of the baseball
Physics
1 answer:
Xelga [282]3 years ago
4 0

Answer:

v_t=46.4532\ m.s^{-1}

Explanation:

Given:

  • Linear horizontal speed of the ball, v_x=42.5\ m.s^{-1}
  • distance travelled by the ball, x=17.4\ m
  • angle rotated by the ball, \theta=44.1\ rad
  • radius of the ball, r=3.67\ cm=0.0367\ m

<u>the time for which the ball travels:</u>

t=\frac{x}{v_x}

t=\frac{17.4}{42.5}

t=0.4094\ s

<u>Angular speed of the ball:</u>

\omega=\frac{\theta}{t}

\omega=\frac{44.1}{0.4094}

\omega=107.7155\ rad.s^{-1}

<u>Now the tangential speed at the equator of the ball:</u>

v_t=r.\omega+v_x (the tangential speed due to rotation and the linear speed are having the same direction)

v_t=0.0367\times 107.7155+42.5

v_t=46.4532\ m.s^{-1}

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x_{9.20\,\mathrm s}=(19,500\,\mathrm m)\cos32.0^\circ

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So we have enough information to solve for the components of the acceleration vector, a_x and a_y:

x_{9.20\,\mathrm s}=\left(1810\,\dfrac{\mathrm m}{\mathrm s}\right)\cos20.0^\circ(9.20\,\mathrm s)+\dfrac12a_x(9.20\,\mathrm s)^2\implies a_x=21.0\,\dfrac{\mathrm m}{\mathrm s^2}

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Answer:

The temperature of the metal is  T_m  =  376.8 ^o C

Explanation:

From the question we are told that

     The mass of the metal is  M =  60 \ kg

     The specific heat of the metal is  c_p  =  0.1027 kcal/(kg \cdot ^oC)

       The mass of the oil is M_o  =  810 \ kg

       The temperature of the oil is  T_o  =  35^oC

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       The equilibrium temperature is T_e  =  39 ^oC

According to the law of energy conservation

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So  

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=>            Q =  -Mc_p (T_m  -  T_c)

Where T_ m  the temperature of metal before immersion

The negative sign show heat lost

The quantity  of gained t by the metal is mathematically represented as      

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substituting values

          - 60 * 0.1027 (T_m  - 39)   =   810 * 0.7167 *  (39 - 35)

=>       T_m  =  376.8 ^o C

         

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