Blood is accelerated from rest to a speed of 33.0 cm/s in a distance of 1.73 cm by the left ventricle of the heart. Determine the magnitude of the acceleration a of the blood.

Respuesta :

Answer -

[tex]314.74[/tex] [tex]\frac{m}{s^2}[/tex]

Explanation

As per the equation of Newton's third law of motion we know that -  

 [tex]V_f^2 - V_0^2 = 2aS[/tex]

where  

 [tex]V_f[/tex]

is the final speed  

and  

 [tex]V_0[/tex]

is the initial speed.  

a is the acceleration and S represents the speed.  

Given  

Final velocity  [tex]= 33[/tex]

 

centimeter per second, Initial velocity is equal to zero and the distance traveled is equal to  

 [tex]1.73[/tex]

centimeter

Now, substituting the values in above equation, we get -

[tex]33^2 - 0 = 2 * a* 1.73\\a = 314.74[/tex]

 

meter per second square

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