Question

the air level in a moped's tire decreases by 8.56×10−9 psi per second due to an air leak. About how many psi does the air in the tire decrease in 20 hours or 7.2×104 seconds? Drag and drop the values into the boxes to express the answer in scientific notation.

Answer 1

• 6.2 × 10^-4 psi

Given :

• Rate of change in the air level per second : 8.56×10^−9 psi

To find :

• Decrease in the air level after 20 hours or 7.2×10^4 seconds

Solution :

To find the value which would be the estimated decrease in the air level after 20 hours , we simply need to multiply the rate of decrease per second by 20 hours or 7.2×10^4.

• 
  `\rightarrow \: (7.2×10^4 * 8.56 * 10 {}^( - 9) )psi`
• 
  `\rightarrow \:(61.632 \: * 10 {}^( - 5))psi`
• 
  `\rightarrow( \:6.2 \: * 10 {}^( - 4)) psi`

Thus, 6.2 × 10^-4 psi would be the approximate decrease in the tyre after 20 hours .

Answer 2

`\sf 6.16* 10^(-4) psi`

Explanation:

Given:

The air level in a moped's tire decreases by
`\sf 8.56* 10^(-9)`psi per second

To find:

Total decrease in air pressure in 20 hours or
`\sf 7.2* 10^4` seconds

Solution:

In order to calculate the total decrease in air pressure in 20 hours, we can multiply the rate of decrease by the total time:

`\sf \textsf{Total decrease = rate of decrease $* $ total time}`

`\textsf{Total decrease }\sf 8.56* 10^(-9) psi/second * 7.2* 10^4 seconds`

`\textsf{Total decrease} \sf = 6.16* 10^(-4) psi`

Therefore, the air pressure in the tire decreases by about:

`\sf 6.16* 10^(-4) psi` in 20 hours.