COLLEGE PHYSICS
COLLEGE PHYSICS
2nd Edition
ISBN: 9781464196393
Author: Freedman
Publisher: MAC HIGHER
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Chapter 11, Problem 50QAP
To determine

(a)

Convert 1500kPa to N/m2

Expert Solution
Check Mark

Answer to Problem 50QAP

  1500kPa is equal to 15×105N/m2

Explanation of Solution

Given info:

Pressure, P=1500kPa

Calculation:

The value of 1500kPa in N/m2 is calculated as

  1500kPa=1500×103Pa(1kPa=1×103Pa)=15×105N/m2

Conclusion:

  1500kPa is equal to 15×105N/m2

To determine

(b)

Convert 35psi to N/m2

Expert Solution
Check Mark

Answer to Problem 50QAP

  35psi is equal to 241317N/m2

Explanation of Solution

Given info:

Pressure, 35psi

Calculation:

The value of 35psi in N/m2 is calculated as

  35psi=35×6894.76N/m2(1psi=6894.76N/m2)=241317N/m2

Conclusion:

  35psi is equal to 241317N/m2

To determine

(c)

Convert 2.85atm to N/m2

Expert Solution
Check Mark

Answer to Problem 50QAP

  2.85atm is equal to 2.87×105N/m2

Explanation of Solution

Given info:

Pressure, 2.85atm

Calculation:

The value of 2.85atm in N/m2 is calculated as

  2.85atm=2.85×1.01×105Pa(1atm=1.01×105Pa)=2.87×105N/m2

Conclusion:

  2.85atm is equal to 2.87×105N/m2

To determine

(d)

Convert 883torr to N/m2

Expert Solution
Check Mark

Answer to Problem 50QAP

  883torr is equal to 117721.52N/m2.

Explanation of Solution

Given info:

Pressure, 883torr

Calculation:

The value of 883torr in N/m2 is calculated as

  883torr=883×133.32Pa(1torr=133.32Pa)=117721.52N/m2

Conclusion:

  883torr is equal to 117721.52N/m2.

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What is the pressure of the gas in the cylinder, in kPa (kiloPascal)? h = 0.700 m, ρmercury = 13,600 kg/m3, 1.0 atm = 1.00 × 105 Pa = 100 kPa, and g = 10.0 m/s2. Your answer needs to have 3 significant figures, including the negative sign in your answer if needed. Do not include the positive sign if the answer is positive. No unit is needed in your answer, it is already given in the question statement.
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Chapter 11 Solutions

COLLEGE PHYSICS

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