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PilotLPTM [1.2K]
1 year ago
10

True or FalseIf an object is submerged in water, on earth, at a depth of 3km, the Pressure on the object is2.94 x 10^7 Pa

Physics
1 answer:
zhenek [66]1 year ago
6 0

The pressure of a submerged object in a fluid is given by:

P=\rho gh+P_{atm}

where ρ is the density of the fluis, g is the acceleration of gravity, h is the depth of the object and Patm is the pressure of the atmosphere. In this case we know that:

• The density of water is 1000 kg/m^3

,

• The acceleration of gravity is 9.8 m/s^2

,

• The depth of the object is 3 km, that is, 3000 m.

,

• The atmospheric pressure is 101325 pascals.

Plugging these values in the equation given above we have:

\begin{gathered} P=(1000)(9.8)(3000)+101325 \\ P=2.95\times10^7 \end{gathered}

Therefore, the pressure at this depth is 2.95x10^7 Pa.

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A person has legs of length L = 1.10 m. (a) If the maximum angle between the legs during walking is ϕ = 50o , what is the person
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a). Maximum Length L=0.929m

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Explanation:

Length legs=L=1.1m

angle=50

the step that give the person forms a triangle whose two sides are known and the angle that forms between them, then using trigonometry as the image

Divide the original triangle in two and form a right triangle so the angle is 25 and the L is hypotenuse and the opposite is the step length

a).

sin(\alpha) =\frac{op}{h}

op=h*sin(\frac{\alpha }{2})\\ op=1.1m*sin(\frac{50}{2})\\op=0.464m

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L=0.464m*2

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period=T

T=\frac{1}{time}=\frac{1}{t}\\ T=\frac{1}{1.2s}\\T=0.83 s^{-1}\\ T=0.83Hz

c).

T1=2\pi *\sqrt{\frac{L}{g}} \\T1=2\pi *\sqrt{\frac{1.1m}{9.8\frac{m}{s^{2}}}}\\ T1=2.1 Hz

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t=\frac{1}{T1}=t= \frac{1}{2.1}=0.47s

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The speed is the relation between the distance with time so:

Vt=\frac{0.928m}{1.2s} \\Vt=0.773 \frac{m}{s} \\Vt=\frac{0.928m}{0.475s} \\Vt=1.953 \frac{m}{s}

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