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iren2701 [21]
2 years ago
12

Please help.

Chemistry
1 answer:
denpristay [2]2 years ago
5 0

Answer:

The bowling ball has more kinetic energy than the tennis ball

Explanation:

Using the formula 1/2 mass × acceleration we found that the tennis ball had a kinetic energy of 0.75 while the bowling ball had a kinetic energy of 10.5 hence the bowling ball has the ability to do more work

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The half life of oxygen is 2 minutes. What fraction of a sample of 0.15 will remain after 5 half lives?​
Natasha_Volkova [10]

Answer:

3.13%.

Explanation:

The following data were obtained from the question:

Original amount (N₀) = 0.15

Half life (t½) = 2 mins

Number of half-life (n) = 5

Fraction of sample remaining =.?

Next, we shall determine the amount remaining (N) after 5 half-life. This can be obtained as follow:

Amount remaining (N) = 1/2ⁿ × original amount (N₀)

NOTE: n is the number of half-life.

N = 1/2ⁿ × N₀

N = 1/2⁵ × 0.15

N = 1/32 × 0.15

N = 0.15/32

N = 4.69×10¯³

Therefore, 4.69×10¯³ is remaining after 5 half-life.

Finally, we shall the fraction of the sample remaining after 5 half-life as follow:

Original amount (N₀) = 0.15

Amount remaining (N) = 4.69×10¯³

Fraction remaining = N/N₀ × 100

Fraction remaining = 4.69×10¯³/0.15 × 100

Fraction remaining = 3.13%

3 0
3 years ago
In the EXPLORE section of your lesson 4.08 on Potential energy there were several animations to watch that provided a graphic il
Lunna [17]

Answer:

This is because no energy is being created or destroyed in this system

Explanation:

I think this is correct? I hope it helps.        

7 0
3 years ago
Read 2 more answers
What is the PTP for 11/4?<br> serving size<br> portion size<br> big mac size
vagabundo [1.1K]
The answer is Big Mac size
4 0
3 years ago
KClO3 ---&gt; KCl + O2 Balance the decomposition chemical reaction. A) KClO3 ---&gt; KCl + O2 B) 2KClO3 ---&gt; KCl + 3O2 C) 2KC
Damm [24]
And the answer is C.
5 0
3 years ago
How many joules are needed to change the temperate of 22g of water from 18°C to 33°C?
Ira Lisetskai [31]

Answer:

Q =  1379.4 J

Explanation:

Given data:

Mass of water = 22  g

Initial temperature = 18°C

Final temperature = 33°C

Heat absorbed = ?

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Specific heat capacity of water is 4.18 J/g. °C

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 33°C - 18 °C

ΔT =  15°C

Q = 522 g ×4.18 J/g.°C× 15°C

Q =  1379.4 J

5 0
3 years ago
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