A bowling ball is pushed with a force of 22.0 N and accelerates at 5.5m/s. What is the mass of the bowling ball

Answers

Answer 1

Answer: Hello!

Data:

F (force) = 22.0 N
m (mass) = ? (in Kg)
a (acceleration) = 5.5 m/s²

We apply the data to the Resultant Force formula, we have:

$F =m*a$

$22 = m*5.5$

$5.5 m = 22$

$m = (22)/(5.5)$

$\boxed{\boxed{m = 4\:Kg}}\end{array}}\qquad\checkmark$

I hope this helps. =)

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A tank has 30 gallons of gasoline how many liters of gasoline are there in the tank

Answers

Answer:

$113.55 l$

Explanation:

1 gallon equals 3,785 liters.

How many liters are in 30 gallons?

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Answer: I think it is: 113.5624 Liters of gas

Explanation:

Explain what happens to car tires and the air inside in each of these scenarios.a. Hot summer day:

b. Cold winter day:

Answers

Answer:

They get flat and start to deflate

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2C4H10 + 13O2 -----> 8CO2 + 10H2OIf 3000 grams of C4H10 react according to this equation, and the actual yield for water is a mass of 1300 grams, what is the percent yield?

Answers

Answer:

27·92 %

Explanation:

Given chemical equation is

2C4H10 + 13O2 -----> 8CO2 + 10H2O

So according to the above balanced chemical equation for 2 moles of C4H10, 10 moles of water is produced

Molecular weight of C4H10 is 58 g

Molecular weight of water is 18 g

So for 116 g of C4H10, 180 g of water is produced and therefore for 1 g of C4H10 (180 ÷ 116) g of water is produced

∴ For 1 g of C4H10 1·552 g of water is produced

For 3000 g of C4H10, (3000 × 1·552) g of water is produced

∴ Number of grams of water produced for 3000 g of C4H10 is 4656 g

Percent yield =((experimental amount) ÷ (theoretical amount)) × 100

Here experimental amount is 4656 g

theoretical amount is 1300 g

∴ Percent yield = (1300 ÷ 4656) × 100 = 27·92 %

∴ Percent yield = 27·92 %

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Answers

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