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charle [14.2K]
3 years ago
10

in baseball, a pitcher can accelerate 0.15kg ball from rest to 98 mi/hr in a distance of 1.7m. (a) What is the average force exe

rted on the ball during the pitch? if the mass of the ball is increased, is the force of the required of the pitcher increased, decreased or unchanged?
Physics
2 answers:
Step2247 [10]3 years ago
3 0

Answer:

F=84.68 N

Required force increases

Explanation:

By the equation

v^{2}=u^{2} +2as\\ 98^{2}*1609.344^{2}/(60*60)^{2}=0+2*a*1.7\\a = 98^{2}*1609.344^{2}/((60*60)^{2}*2*1.7)\\a = 564.50m/s^{2} \\

By newton's second law :

F=ma\\F=0.15*564.50  \\F=84.68 N

When the mass increased the force increases as the equation.

F=ma

Serhud [2]3 years ago
3 0

a. The <u>average force exerted</u> on the ball during the pitch is equal to <u>84.68N</u>

b. Since the mass is increased, <u>the force required for the pitcher has to increase too.</u>

Why?

For the first part of the question (a), we need to convert all the units to work with the same system unit.

Converting the speed we have:

1mile=1609.34m\\1hour=3600s

98\frac{mi}{h}*\frac{1609.34}{1mi}*\frac{1h}{3600s}=43.81\frac{m}{s}

Now, calculating the force, we have:

F=mass*acceleration

To calculate the acceleration we can use the following formula:

v^{2} =v_{o}^{2} +2acceleration*distance\\\\acceleration=\frac{v^{2}-v_{o}^{2} }{2distance} \\\\acceleration=\frac{(43.81\frac{m}{s})^{2} -0}{2*1.7m}=\frac{1919.32\frac{m^{2} }{s^{2} } }{3.4m}=564.50\frac{m}{s^{2} } \\\\

Therefore, calculating the force, we have:

Force=mass*acceleration\\\\Force=0.15kg*564.50\frac{m}{s^{2}}=84.68\frac{kg.m}{s^{2}}=84.68N

Hence, we have that the force exerted on the ball during the pich is equal to 84.68 N.

For the second part (b),

We need to assume that the final speed is maintaIned, so, since the equation to calculate the exerted force is a relationship between mass and acceleration (F=ma), if the mass of the object is increased, the force required need to be increased.

Have a nice day!

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A radar station detects an airplane approaching directly from the east. At first observation, the range to the plane is 375.0 m
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Answer:

819.78 m

Explanation:

<u>Given:</u>

  • OA = range of initial position of the airplane from the point of observation = 375 m
  • OB = range of the final position of the airplane from the point of observation = 797 m
  • \theta = angle of the initial position vector from the observation point = 43^\circ
  • \alpha = angle of the final position vector from the observation point = 123^\circ
  • \vec{AB} = displacement vector from initial position to the final position

A diagram has been attached with the solution in order to clearly show the position of the plane.

\vec{OA} = OA\cos \theta \hat{i}+OA \sin \theta \hat{j}\\\Rightarrow \vec{OA} = 375\ m\cos 43^\circ \hat{i}+375\ m\sin 43^\circ \hat{j}\\\Rightarrow \vec{OA} = (274.26\ \hat{i}+255.75\ \hat{j})\ m\\\vec{OB} = OB\cos \alpha \hat{i}+OB \sin \alpha \hat{j}\\\Rightarrow \vec{OB} = 797\ m\cos 123^\circ \hat{i}+797\ m\sin 123^\circ \hat{j}\\\Rightarrow \vec{OB} = (-434.08\ \hat{i}+668.42\ \hat{j})\ m

Displacement vector of the airplane will be the shortest line joining the initial position of the airplane to the final position of the airplane which is given by:

\vec{AB}=\vec{OB}-\vec{OA}\\\Rightarrow \vec{AB} =  (-434.08\ \hat{i}+668.42\ \hat{j})\ m-(274.26\ \hat{i}+255.75\ \hat{j})\ m\\\Rightarrow \vec{AB} =  (-708.34\ \hat{i}+412.67\ \hat{j})\ m

The magnitude of the displacement vector = \sqrt{(-708.34)^2+(412.67)^2}\ m = 819.78\ m

Hence, the magnitude of the displacement of the plane is 819.67 m during the period of observation.

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