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viva [34]
3 years ago
15

In major league baseball, the pitcher's mound is 60 feet from the batter.If a pitcher throws a 89 mph fastball, how much time el

apses from when the ball leaves the pitcher's hand until the ball reaches the batter?
Physics
1 answer:
lianna [129]3 years ago
8 0
<span>Hitting can be broken down into three segments; SEE, REACT, SWING. You watch the ball in the pitcher's hand during the windup and you watch the ball leave the pitcher's hand when it is thrown and you watch the spin of the ball as it comes towards the plate. That is SEE. You determine what the pitch is (fastball, curveball, etc.), you determine where the ball is going to go and you determine whether it is headed towards an area that you think you can get good wood on it. That is REACT. You swing if you like the pitch or don't swing if you don't like the pitch. That is SWING. It doesn't make any difference whether you are playing with a tennis ball or a golf ball or a baseball. The theory and mechanics are the same. So, to answer your question I would say the more you practice the better you will be, regardless of the type of ball you use to practice with.</span>
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gladu [14]

Answer:

X axis

Explanation:

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3 0
3 years ago
Determine the average value of the translational kinetic energy of the molecules of an ideal gas at (a) 27.8°C and (b) 143°C. Wh
Alinara [238K]

Answer:

a) k_{avg}=6.22\times 10^{-21}

b) k_{avg}=8.61\times 10^{-21}

c)  k_{mol}=3.74\times 10^{3}J/mol

d)   k_{mol}=5.1\times 10^{3}J/mol

Explanation:

Average translation kinetic energy (k_{avg}) is given as

k_{avg}=\frac{3}{2}\times kT    ....................(1)

where,

k = Boltzmann's constant ; 1.38 × 10⁻²³ J/K

T = Temperature in kelvin

a) at T = 27.8° C

or

T = 27.8 + 273 = 300.8 K

substituting the value of temperature in the equation (1)

we have

k_{avg}=\frac{3}{2}\times 1.38\times 10^{-23}\times 300.8  

k_{avg}=6.22\times 10^{-21}J

b) at T = 143° C

or

T = 143 + 273 = 416 K

substituting the value of temperature in the equation (1)

we have

k_{avg}=\frac{3}{2}\times 1.38\times 10^{-23}\times 416  

k_{avg}=8.61\times 10^{-21}J

c ) The translational kinetic energy per mole of an ideal gas is given as:

       k_{mol}=A_{v}\times k_{avg}

here   A_{v} = Avagadro's number; ( 6.02×10²³ )

now at T = 27.8° C

        k_{mol}=6.02\times 10^{23}\times 6.22\times 10^{-21}

          k_{mol}=3.74\times 10^{3}J/mol

d) now at T = 143° C

        k_{mol}=6.02\times 10^{23}\times 8.61\times 10^{-21}

          k_{mol}=5.1\times 10^{3}J/mol

8 0
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Please help me it is for learning at home.
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Answer:

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