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Sladkaya [172]
3 years ago
10

Joe is the best baseball player in the world. He can throw a baseball faster than anyone. If the mass of the baseball was increa

sed by 50%, but he applied the same force as he did to a lighter baseball, which of these would best describe Joe's attitude? A)Joe would love to use the new ball in future baseball games.
B) Joe would be happy as the new ball would have a faster acceleration.
C) Joe would be unhappy as the new ball would have a slower acceleration.
D) Joe's attitude would not change as there would be no change in the acceleration.
Chemistry
2 answers:
GarryVolchara [31]3 years ago
8 0

We know tha

Force = Mass X acceleration

acceleration = this provides the distance to the base ball

If mass of baseball is increased keeping force constant, the acceleration will decrease, this will decrease the distance traveled by the base ball

so Joe would be unhappy as the new ball would have a slower acceleration.

storchak [24]3 years ago
4 0
Try C that is what im going with
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Answer:

Here's the conversion factor you need:

1 Kilometer   =   39370 inches

So, for your question we want to go 405,696 km....

405696 km   x   39370 inches/ 1 km  =   15972283464 inches

                       

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What pressure, in atm, would be exerted by 0.023 grams of oxygen (O2) if it occupies 31.6 mL at 91
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Answer:  A pressure of 0.681 atm would be exerted by 0.023 grams of oxygen (O_2) if it occupies 31.6 mL at 91^{o}C.

Explanation:

Given : Mass of oxygen = 0.023 g

Volume = 31.6 mL

Convert mL into L as follows.

1 mL = 0.001 L\\31.6 mL = 31.6 mL \times \frac{0.001 L}{1 mL}\\= 0.0316 L

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As molar mass of O_2 is 32 g/mol. Hence, the number of moles of O_2 are calculated as follows.

No. of moles = \frac{mass}{molar mass}\\= \frac{0.023 g}{32 g/mol}\\= 0.00072 mol

Using the ideal gas equation calculate the pressure exerted by given gas as follows.

PV = nRT

where,

P = pressure

V = volume

n = number of moles

R = gas constant = 0.0821 L atm/mol K

T = temperature

Substitute the value into above formula as follows.

PV = nRT\\P  \times 0.0316 L = 0.00072 mol \times 0.0821 L atm/mol K \times 364 K\\P = \frac{0.00072 mol \times 0.0821 L atm/mol K \times 364 K}{0.0316 L}\\= 0.681 atm

Thus, we can conclude that a pressure of 0.681 atm would be exerted by 0.023 grams of oxygen (O_2) if it occupies 31.6 mL at 91^{o}C.

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