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LuckyWell [14K]
3 years ago
9

Transverse thrusters are used to make large ships fully maneuverable at low speeds without tugboat assistance. A transverse thru

ster consists of a propeller mounted in a duct; the unit is then mounted below the waterline in the bow or stern of the ship. The duct runs completely across the ship. Calculate the thrust developed by a 1900 kW unit supplied to the propeller if the duct is 2.6 m in diameter and the ship is stationary.
Physics
1 answer:
suter [353]3 years ago
8 0

Answer:

Thrust developed = 212.3373 kN

Explanation:

Assuming the ship is stationary

<u>Determine the Thrust developed</u>

power supplied to the propeller ( Punit ) = 1900 KW

Duct distance ( diameter ; D  ) = 2.6 m

first step : <em>calculate the area of the duct </em>

A = π/4 * D^2

   =  π/4 * ( 2.6)^2  = 5.3092 m^2

<em>next : calculate the velocity of propeller</em>

Punit = (A*v*β ) / 2  * V^2     ( assuming β = 999 kg/m^3 ) also given V1 = 0

∴V^3 = Punit * 2 / A*β

         = ( 1900 * 10^3 * 2 ) / ( 5.3092 * 999 )

hence V2 = 8.9480 m/s

<em>Finally determine the thrust developed </em>

F = Punit / V2

  = (1900 * 10^3) / ( 8.9480)

  = 212.3373 kN

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Answer:

12.0 meters

Explanation:

Given:

v₀ = 0 m/s

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a₂ = 1.43 m/s²

t₂ = 2.42 s

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First, find the velocity reached at the end of the first acceleration.

v = at + v₀

v = (0.281 m/s²) (5.44 s) + 0 m/s

v = 1.53 m/s

Next, find the position reached at the end of the first acceleration.

x = x₀ + v₀ t + ½ at²

x = 0 m + (0 m/s) (5.44 s) + ½ (0.281 m/s²) (5.44 s)²

x = 4.16 m

Finally, find the position reached at the end of the second acceleration.

x = x₀ + v₀ t + ½ at²

x = 4.16 m + (1.53 m/s) (2.42 s) + ½ (1.43 m/s²) (2.42 s)²

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The shanghai maglev train is capable of reaching speeds of up to 350km h-1 what is the speed in m/h
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A point charge q1 is held stationary at the origin. A second charge q2 is placed at point a, and the electric potential energy o
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Answer:

U_{b}=+7.3*10^{-8}J

Explanation:

Given data

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q₂ moves to point b where a negative work done on it  W_{a-b}=-1.9*10^{-8}J

Required

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Solution

When a particle moves from a point where the potential is Ua to a point where it is Ub the change in potential energy is equal to work done where the force exerted on the charge is conservative and work done is given by:

W_{a-b}=U_{a}-U_{b}\\U_{b}=U_{a}-W_{a-b}

Now  substitute the given values

So

U_{b}=5.4*10^{-8}J-(-1.9*10^{-8}J)\\U_{b}=+7.3*10^{-8}J

3 0
3 years ago
During a neighborhood baseball game in a vacant lot, a particularly wild hit sends a 0.146 kg baseball crashing through the pane
Nonamiya [84]

Answer:

Impulse, |J| = 0.6716 kg-m/s

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Explanation:

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To find,

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Solution,

(a) Impulse of an object is equal to the change in its momentum. It is given by :

J=m(v-u)

J=0.146\ kg(10.7-15.3)\ m/s

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(b) Another definition of impulse is given by the product of force and time of contact.

t = 0.0106 s

J=F\times \Delta t

F=\dfrac{J}{\Delta t}

F=\dfrac{0.6716\ kg-m/s}{0.0106\ s}

F = 63.35 N

Hence, this is the required solution.

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2 years ago
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