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2 years ago

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Sharks are generally negatively buoyant; the upward buoyant force is less than the weight force. This is one reason sharks tend

to swim continuously; water moving past their fins causes a lift force that keeps sharks from sinking. A 92 kgbull shark has a density of 1040 kg/m3. What lift force must the shark's fins provide if the shark is swimming in seawater? Bull sharks often swim into freshwater rivers. What lift force is required in a river?

1 answer:

e-lub [12.9K]2 years ago

4 0

Answer:

866.92 N

Explanation:

mass of shark, m = 92 kg

density, d = 1040 kg/m^3

Volume, V = mass / density

V = 92 / 1040 = 0.08846 kg/m^3

The lift force is the buoyant force acting on the shark is

Lift force = Volume x density of water x g

Lift force = 0.08846 x 1000 x 9.8

Lift force = 866.92 N

Thus, the lift force is 866.92 N.

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A 6.0-cm-diameter, 11-cm-long cylinder contains 100 mg of oxygen (O2) at a pressure less than 1 atm. The cap on one end of the c

g100num [7]

Explanation:

The given data is as follows.

Mass of oxygen present = 100 mg = $100 \times 10^{-3}$ g

So, moles of oxygen present are calculated as follows.

n = $\frac{100 \times 10^{-3}}{32}$

= $3.125 \times 10^{-3}$ moles

Diameter of cylinder = 6 cm = $6 \times 10^{-2}$ m

= 0.06 m

Now, we will calculate the cross sectional area (A) as follows.

A = $\pi \times \frac{(0.06)^{2}}{4}$

= $2.82 \times 10^{-3} m^{2}$

Length of tube = 11 cm = 0.11 m

Hence, volume (V) = $2.82 \times 10^{-3} \times 0.11$

= $3.11 \times 10^{-4} m^{3}$

Now, we assume that the inside pressure is P .

And, $P_{atm}$ = 100 kPa = 100000 Pa,

Pressure difference = 100000 - P

Hence, force required to open is as follows.

Force = Pressure difference × A

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We are given that force is 173 N.

Thus,

$(100000 - P) \times 2.82 \times 10^{-3}$ = 173

Solving we get,

P = $3.8650 \times 10^{4} Pa$

= 38.65 kPa

According to the ideal gas equation, PV = nRT

So, we will put the values into the above formula as follows.

PV = nRT

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6 0

1 year ago

A 96-mH solenoid inductor is wound on a form 0.80 m in length and 0.10 m in diameter. A coil is tightly wound around the solenoi

Sholpan [36]

Answer:

$i = 7.83 \mu A$

Explanation:

Induced EMF in the coil is given by the equation

$EMF = M\frac{di}{dt}$

so we have

$M = 31 \mu H$

also we know that rate of change in current in solenoid is given as

$\frac{di}{dt} = 2.5 A/s$

so induced EMF of coil is given as

$EMF = (31 \times 10^{-6})(2.5)$

$EMF = 77.5 \times 10^{-6} A/s$

now induced current in the coil will be given as

$i = \frac{EMF}{R}$

$i = \frac{77.5 \times 10^{-6}}{9.9}$

$i = 7.83 \mu A$

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1 year ago

A newly discovered planet has a mean radius of 7380 km. A vehicle on the planet\'s surface is moving in the same direction as th

Butoxors [25]

Answer:

292796435 seconds ≈ 300 million seconds

Explanation:

First of all, the speed of the car is 121km/h = 33.6111 m/s

The radius of the planet is given to be 7380 km = 7380000 m

From the relationship between linear velocity and angular velocity i.e., v=rw, the angular velocity of the car will be w=v/r = 33.6111/7380000 = 0.000000455 rad/s = 4.55 x 10⁻⁶ rad/sec

If the angular velocity of the vehicle about the planet's center is 9.78 times as large as the angular velocity of the planet then we have

w(vehicle) = 9.78 x w(planet)

w(planet) = w(vehicle)/9.78 = 4.55 x 10⁻⁶ / 9.78 = 4.66 x 10⁻⁷ rad/sec

To find the period of the planet's rotation; we use the equation

w(planet) = 2π÷T

Where w(planet) is the angular velocity of the planet and T is the period

From the equation T = 2π÷w = 2×(22/7) ÷ 4.66 x 10⁻⁷ = 292796435 seconds

Therefore the period of the planet's motion is 292796435 seconds which is approximately 300, 000, 000 (300 million) seconds

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2 years ago

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ololo11 [35]

Answer:

47.76°

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Uf - Ui = -0.458mJ

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rearranging the equation,

(μBcosθf) - (μBcosθi) = 0.458mJ

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μB * (cos 0 - cos θi) = 0.458mJ

but cos 0 = 1

(0.0243 * 0.0575) (1 - cos θi) = 0.458*10⁻³

1 - cos θi = 0.458*10⁻³ / 1.397*10⁻³

1 - cos θi = 0.3278

collect like terms

cosθi = 0.6722

θ = cos⁻ 0.6722

θ = 47.76°

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1 year ago

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