A treasure chest lies 20.0 m below the surface of the ocean. What is the magnitu

Question

A treasure chest lies 20.0 m below the surface of the ocean. What is the magnitude of the force that acts on the rectangular top that is 0.750 m times 0.425 m? A) 980 N B) 2.39 times 10^3 N C) 9.48 times 10^4 N D) 4.71 times 10^5 N E) 2.00 times 10^6 N At a location where the acceleration due to gravity is 9.807 m/s^2, the atmospheric pressure is 9.891 times 10^4 Pa. A barometer at the same location is filled with an unknown liquid. What is the density of the unknown liquid if its height in the barometer is 1.163 m? A) 210 kg/m^3 B) 4336 kg/m^3 C) 5317 kg/m^3 D) 8672 kg/m^3

Explanation / Answer

1). We know that under the depth h of a fluid, it experiences a gauge pressure given by

P= Density* g* h

where Density= here it is density of water= 1000 kg/m3

g= acceleration due to gravity= 9.8m/s2

Then for a depth of 20.0m below ocean, Total pressure exerted on the rectangle will be

P= Patm+ (1000)*(9.8)*(20)= (101325+196000) Pa= 297325Pa.....................(1)

Now given area of the rectangle, A= 0.750m * 0.425m.............................(2)

Then Force acting on the rectangle will be

F= Pressure* Area= P* A

using equation 1 and 2 in above,

F= 297325* 0.750* 0.425= 9.477 *104 N

  Hence correct option: C). 9.48 *104 N

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.