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3 years ago

6

A spherical submersible 1.84 m in radius, armed with multiple cameras, descends under water in a region of the Atlantic Ocean kn

own for shipwrecks and finds its first shipwreck at a depth of 2.37 ✕ 103 m. Seawater has density 1.025 ✕ 103 kg/m3, and the air pressure at the ocean's surface is 1.013 ✕ 105 Pa. What is the absolute pressure at the depth of the shipwreck?

1 answer:

Drupady [299]3 years ago

5 0

Answer:

23932242.5 Pa

Explanation:

$P_a$ = Atmospheric pressure = $1.013\times 10^5\ Pa$

$P_w$ = Pressure of seawater

$\rho$ = Density of sea water = $1.025\times 10^3\ kg/m^3$

h = Depth of shipwreck = $2.37\times 10^3\ m$

g = Acceleration due to gravity = 9.81 m/s²

The absolute pressure is given by

$P_{ab}=P_a+P_w\\\Rightarrow P_{ab}=1.013\times 10^5+1.025\times 10^3\times 9.81\times 2.37\times 10^3\\\Rightarrow P_{ab}=23932242.5\ Pa$

The absolute pressure at the depth of the shipwreck is 23932242.5 Pa

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A flywheel is a mechanical device used to store rotational kinetic energy for later use. Consider a flywheel in the form of a un

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Answer:

<em>a) 6738.27 J</em>

<em>b) 61.908 J</em>

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Explanation:

The complete question is

A flywheel is a mechanical device used to store rotational kinetic energy for later use. Consider a flywheel in the form of a uniform solid cylinder rotating around its axis, with moment of inertia I = 1/2 mr2.

Part (a) If such a flywheel of radius r1 = 1.1 m and mass m1 = 11 kg can spin at a maximum speed of v = 35 m/s at its rim, calculate the maximum amount of energy, in joules, that this flywheel can store?

Part (b) Consider a scenario in which the flywheel described in part (a) (r1 = 1.1 m, mass m1 = 11 kg, v = 35 m/s at the rim) is spinning freely at its maximum speed, when a second flywheel of radius r2 = 2.8 m and mass m2 = 16 kg is coaxially dropped from rest onto it and sticks to it, so that they then rotate together as a single body. Calculate the energy, in joules, that is now stored in the wheel?

Part (c) Return now to the flywheel of part (a), with mass m1, radius r1, and speed v at its rim. Imagine the flywheel delivers one third of its stored kinetic energy to car, initially at rest, leaving it with a speed vcar. Enter an expression for the mass of the car, in terms of the quantities defined here.

moment of inertia is given as

$I$ = $\frac{1}{2}$ $mr^{2}$

where m is the mass of the flywheel,

and r is the radius of the flywheel

for the flywheel with radius 1.1 m

and mass 11 kg

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$I$ = $\frac{1}{2}$ $*11*1.1^{2}$ = 6.655 kg-m^2

The maximum speed of the flywheel = 35 m/s

we know that v = ωr

where v is the linear speed = 35 m/s

ω = angular speed

r = radius

therefore,

ω = v/r = 35/1.1 = 31.82 rad/s

maximum rotational energy of the flywheel will be

E = $Iw^{2}$ = 6.655 x $31.82^{2}$ = <em>6738.27 J</em>

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b) second flywheel has

radius = 2.8 m

mass = 16 kg

moment of inertia is

$I$ = $\frac{1}{2}$ $mr^{2}$ = $\frac{1}{2}$ $*16*2.8^{2}$ = 62.72 kg-m^2

According to conservation of angular momentum, the total initial angular momentum of the first flywheel, must be equal to the total final angular momentum of the combination two flywheels

for the first flywheel, rotational momentum = $Iw$ = 6.655 x 31.82 = 211.76 kg-m^2-rad/s

for their combination, the rotational momentum is

$(I_{1} +I_{2} )w$

where the subscripts 1 and 2 indicates the values first and second flywheels

$(I_{1} +I_{2} )w$ = (6.655 + 62.72)ω

where ω here is their final angular momentum together

==> 69.375ω

Equating the two rotational momenta, we have

211.76 = 69.375ω

ω = 211.76/69.375 = 3.05 rad/s

Therefore, the energy stored in the first flywheel in this situation is

E = $Iw^{2}$ = 6.655 x $3.05^{2}$ = <em>61.908 J</em>

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c) one third of the initial energy of the flywheel is

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where m is the mass of the car

$v_{car}$ is the velocity of the car

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2246.09 = $\frac{1}{2}mv_{car} ^{2}$

making m the subject of the formula

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In a string of length L, the wavelength of the n-th harmonic of the standing wave produced in the string is given by:

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$\lambda=\frac{2}{1} \cdot 3.5 m=7.0 m$

The wavelength of the 2nd harmonic is:

$\lambda=\frac{2}{2} \cdot 3.5 m=3.5 m$

The wavelength of the 4th harmonic is:

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It is not possible to find any integer n such that $\lambda=5 m$ , therefore the correct options are A, B and D.

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3 years ago

Read 2 more answers

Consider three force vectors Fi with magni- tude 53 N and direction 116º, F2 with mag- nitude 57 N and direction 217°, and F3 wi

Flauer [41]

Answer:

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c. Option 2

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Resolving the vectors into x and y components.

Fnet= -2.04i+37.62j

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Then the direction becomes 180-43.37

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3 years ago