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madreJ [45]
2 years ago
14

At a depth of 10.9 km, the Challenger Deep in the Marianas Trench of the Pacific Ocean is the deepest site in any ocean. Yet, in

1960, Donald Walsh and Jacques Piccard reached the Challenger Deep in the bathyscaph Trieste. Assuming that seawater has a uniform density of 1024 kg/m3, approximate the hydrostatic pressure (in atmospheres) that the Trieste had to withstand. (Even a slight defect in the Trieste structure would have been disastrous.)
Physics
2 answers:
gavmur [86]2 years ago
8 0

Answer:7434.47 atmosphere

Explanation:

hydrostatic pressure (in atmospheres) = atmospheric pressure + pressure at depth

specific gravity = density of substance / density of water = 1024/1000= 1.024

specific weight = 1.024 * 9.79 kN/m^3 = 10.02496 kN/m^3

= 14.7 psi + 10.02496 kN/m^3* 10.9 km * 1000m = 109286.764 psi (1km = 1000m)

to atmosphere = 109286.764 psi/14.7 psi * 1 atmosphere = 7434.47 atmosphere

KATRIN_1 [288]2 years ago
7 0

Answer:

P = 1081.304\,atm

Explanation:

The hydrostic pressure is equal to:

P = P_{atm} + P_{man}

P = 1\,atm + \left(\frac{1\,atm}{101,325\,Pa} \right)\cdot \left(1,024\,\frac{kg}{m^{3}} \right)\cdot (9.807\,\frac{m}{s^{2}} )\cdot (10,900\,m)

P = 1081.304\,atm

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2 years ago
Describe the relationship between the length and period of a pendulum in the language of direct proportions
Wittaler [7]

The period of the pendulum is directly proportional to the square root of the length of the pendulum

Explanation:

The period of a simple pendulum is given by the equation

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From the equation, we see that when the length of the pendulum increases, the period of the pendulum increases as the square root of L, T\propto \sqrt{L}. This means that

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2 years ago
while working out a man performed 2525j of work in 19seconds . what was his power A:132.9w. B:241.5w C 47.975w. D100.5w
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7 0
3 years ago
Read 2 more answers
Chloe wanted to find out if the color of a food would affect how quickly kindergarten children would eat it for lunch. She belie
Harrizon [31]
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6 0
3 years ago
A 1350 kg uniform boom is supported by a cable. The length of the boom is l. The cable is connected 1/4 the
olchik [2.2K]

Answer:

Tension= 21,900N

Components of Normal force

Fnx= 17900N

Fny= 22700N

FN= 28900N

Explanation:

Tension in the cable is calculated by:

Etorque= -FBcostheta(1/2L)+FT(3/4L)-FWcostheta(L)= I&=0 static equilibrium

FTorque(3/4L)= FBcostheta(1/2L)+ FWcostheta(L)

Ftorque=(Fcostheta(1/2L)+FWcosL)/(3/4L)

Ftorque= 2/3FBcostheta+ 4/3FWcostheta

Ftorque=2/3(1350)(9.81)cos55° + 2/3(2250)(9.81)cos 55°

Ftorque= 21900N

b) components of Normal force

Efx=FNx-FTcos(90-theta)=0 static equilibrium

Fnx=21900cos(90-55)=17900N

Fy=FNy+ FTsin(90-theta)-FB-FW=0

FNy= -FTsin(90-55)+FB+FW

FNy= -21900sin(35)+(1350+2250)×9.81=22700N

The Normal force

FN=sqrt(17900^2+22700^2)

FN= 28.900N

4 0
3 years ago
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