Fueling a Complete Game: Calculating Cereal Needs for Baseball Pitchers

  • Thread starter Thread starter LiquidIce
  • Start date Start date
  • Tags Tags
    Hard
AI Thread Summary
A baseball pitcher requires approximately 10,500 joules of energy to pitch a complete game, throwing 105 pitches at 100 joules each. Given the pitcher's body efficiency of 25%, he needs 42,000 joules of energy from food. To meet this requirement, he would need to consume about 2,625 grams, or 2.6 kilograms, of sweetened puffed corn cereal, which provides 16 kilojoules per gram. While this calculation provides a theoretical estimate, actual energy needs may vary based on individual factors. Consulting a nutritionist for tailored dietary advice is recommended.
LiquidIce
Every gram of a sweetended puffed corn cereal provides about 16kj of energy. How much cereal would a baseball pitcher need to eat to provide for the energy needs of pitching a whole complete game ? You may assume the picher uses a 100j of energy for each pitch and that he throws 105 pitches during the game. Also his body is only 25% efficient in converting food energy into useful energy.
 
Physics news on Phys.org
Do something yourself on the problem, don't just ask others to do it for you! You should be able to answer these questions:

How much energy will the pitcher exert in the entire game?

Since his body is only 25% efficient in converting food energy, how much food energy does he need to produce that much usable energy?

How much cereal does he need to eat to get that much food energy?

How dumb is he to get all of his energy from sweetened puffed corn cereal?
 


To calculate the amount of cereal a baseball pitcher would need to eat to provide enough energy for a complete game, we first need to determine the total amount of energy the pitcher will need. Based on the given information, we know that each gram of sweetened puffed corn cereal provides 16kj of energy. Additionally, we know that the pitcher will use 100j of energy for each pitch and will throw 105 pitches during the game. Therefore, the total amount of energy needed for pitching the entire game would be 100j x 105 pitches = 10,500j.

However, we also need to take into account the pitcher's body efficiency in converting food energy into useful energy. With a 25% efficiency rate, only 25% of the energy from the cereal will be converted into useful energy for pitching. This means that the pitcher will actually need 10,500j / 25% = 42,000j of energy from the cereal.

To determine the amount of cereal needed, we can divide 42,000j by the 16kj of energy provided by each gram of cereal. This gives us a total of 2,625 grams of cereal needed for the pitcher to have enough energy for a complete game. This is equivalent to approximately 2.6 kilograms or 5.7 pounds of cereal.

In conclusion, a baseball pitcher would need to eat approximately 2.6 kilograms or 5.7 pounds of sweetened puffed corn cereal to provide enough energy for a complete game. However, it is important to note that this is only an estimate and the actual amount may vary depending on individual factors such as body weight and metabolism. It is always best to consult a nutritionist or dietitian for personalized dietary recommendations.
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top