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jenyasd209 [6]
3 years ago
12

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.
Physics
1 answer:
bearhunter [10]3 years ago
8 0

Answer:

P = 1.09 \times 10^8 Pa

Explanation:

As we know that the pressure inside the liquid level is given as

P = \rho g h + P_o

here we have

\rho = 1024 kg/m^3

h = 10.9 km

also we know that

P_o = 1.01 \times 10^5 Pa

now we have

P = (1.01 \times 10^5) + (1024)(9.81)(10.9 \times 10^3)

P = 1.09 \times 10^8 Pa

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scientists have determined the composition of the lithosphere and asthenosphere but have struggled to discover much about the lo
dalvyx [7]

Explanation:

The lithosphere is composed of the crust and a  portion of the upper mantle. the athenosphere is partially molten upper mantle material that behaves plasticly & can flow.

It is harder to work on the lower layers because its gets extremely hot & sometimes it can get so hot that it melts tools used  to research these layers.  

7 0
3 years ago
Assume air resistance is negligible unless otherwise stated. Calculate the displacement in m and velocity in m/s at the followin
tankabanditka [31]

Answer: 1) t=0.5 s; S=6.225 m and v=14.9 m/s

2) t=1 s; S=14.9 m and v=19.8 m/s

3) t=1.5 s; S=26.05 m and v=24.7 m/s

Explanation:

The displacement S is given by

S=ut+\frac{1}{2} at^{2}

and  final velocity v is given by

v=u+at

where u is the initial velocity

a is acceleration

t is time taken

Case 1: when time is 0.5 s

The displacement is

S=ut+\frac{1}{2} at^{2}\\S=10\times  0.5 +\frac{1}{2}\times 9.8\times 0.5^{2}\\\\S=6.225 m

the velocity is

v=u+at\\v=10+9.8\times 0.5\\v=14.9 m/s

Case 2: when t=1 sec

The displacement is

S=ut+\frac{1}{2} at^{2}\\S=10\times  1 +\frac{1}{2}\times 9.8\times 1^{2}\\\\S=14.9 m

the velocity is

v=u+at\\v=10+9.8\times 1\\v=19.8 m/s

Case 3: t=1.5 s

The displacement is

S=ut+\frac{1}{2} at^{2}\\S=10\times  1.5 +\frac{1}{2}\times 9.8\times 1.5^{2}\\\\S=26.05 m

the velocity is

v=u+at\\v=10+9.8\times 1.5\\v=24.7 m/s

4 0
3 years ago
A fidget spinner experiences a constant torque of 1.4 N.m. If the spinner is initially at rest, what is its angular momentum 2.0
motikmotik

Answer:

2.8N-msec

Explanation:

We have given torque \tau =1.4N-m

Initial time t_1=0\ sec

Final time t_2=2\ sec

There is relation between angular momentum and torque

that is \frac{dL}{dt}=\tau

dL=\tau dt

\int dL=\tau \oint_{t_1}^{t_2}dt

L=\tau (t_1-t_2)=1.4\times (2-0)=2.8N-msec

4 0
4 years ago
The applied force of 3 washers will increase the applied force on the car to n.
Aleks04 [339]

Answer: 0.147

Explanation:

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3 0
3 years ago
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This problem uses the same concepts as Multiple-Concept, except that kinetic, rather than static, friction is involved. A crate
Bingel [31]

Answer:

1.96 m/s^2

Explanation:

angle of inclination, θ = 30.6°

coefficient of friction, μ = 0.36

Acceleration of the crate

a = g Sin θ - μ g Cos θ

a = 9.8 ( Sin 30.6 - 0.36 x Cos 30.6)

a = 9.8 ( 0.509 - 0.309)

a = 1.96 m/s^2

Thus, the acceleration f the crate is 1.96 m/s^2.

3 0
3 years ago
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