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GenaCL600 [577]
3 years ago
10

Please Help!!!!! Quickly!!!! brainiest to correct answer !!!!!!

Physics
1 answer:
iragen [17]3 years ago
5 0
Just show a picture so I can help ur information is misleading...
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the ocean floor is, on average, 4267 m below sea level. What is the pressure in the atmosphere at this depth?
hichkok12 [17]
The pressure at the depth h in the ocean is given by (Stevin's law)
p= p_0 + \rho g h
where
p_0 = 1.0 \cdot 10^5 Pa is the atmospheric pressure
and \rho g h is the pressure exerted by the column of water of height h=4267 m, with \rho = 1000 kg/m^3 being the water density and g=9.81 m/s^2.
Substituting, we find
p=1.0 \cdot 10^5 Pa + (1000 kg/m^3)(9.81 m/s^2)(4267 m)=4.20 \cdot 10^7 Pa
We want to convert this into atmospheres: we know that 1 atm corresponds to the atmospheric pressure at sea level, so 1 atm=1.0\cdot 10^5 Pa, therefore we just need to divide by this number:
p= \frac{4.20 \cdot 10^7 Pa}{1.0 \cdot 10^5 Pa/atm} =420 atm
7 0
3 years ago
Un resorte se alarga 5 cm bajo la acción de una fuerza de 39,2 N. ¿Cuál es la constante del resorte? Si ahora la fuerza es 68,6
Lorico [155]

Answer:

k=784 N/m

\Delta x=8,8 cm

Explanation:

Usando la ley de Hook tenemos:

F=k\Delta x

Solving it for k we have:

k=\frac{F}{\Delta x}

k=\frac{39,2}{0,05}

k=784 N/m

Usando la misma ecuación y sabiendo k tenemos:

\Delta x=\frac{F}{k}

\Delta x=\frac{68,6}{784}

\Delta x=8,8 cm

Espero esto te ayude!

6 0
3 years ago
What part of this transverse wave is the green arrow identifying?<br><br> Help plz
cupoosta [38]

It is the 'crest' part that the green arrow is identifying.

3 0
3 years ago
What occurs when an object travels in a curved path
KengaRu [80]
There must be a centripetal force to move the object move in a curve path.
5 0
3 years ago
Read 2 more answers
A 1.0-cm-diameter pipe widens to 2.0, then narrows to 0.5. Liquid flows through the first segment at a speed of 4.0. What is the
uysha [10]

Answer:

3.14 ×  10⁻⁴  m³  /s

Explanation:

The flow rate (Q) of a fluid is passing through different cross-sections remains of pipe always remains the same.

Q = Area x velocity

Given:

Diameters of 3 sections of the pipe are given as  

d1  =  1.0  cm,  d2  =  2.0  cm  and  d3  =  0.5  cm.

Speed in the first segment of the pipe is  

v1  =  4  m/s.

From the equation of continuity the flow rate through different cross-sections remains the same.

Flow  rate  =  Q  =  A1  v1  =  A2  v2  =  A3  v3.

Q = A1v1

   =π/4  d²1  v1  =  π/4  * 0.01² ×4.0 m³/s  =  3.14 ×  10⁻⁴  m³  /s

3 0
3 years ago
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