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3 years ago

A cathedral hull allows a boat to do what?

Physics

1 answer:

nirvana33 [79]3 years ago

7 0

Answer: A cathedral hull allows a boat to

(Get up on a plane quicker)

Explanation:

A boat is a device that is used for water transportation and other purposes. It is made up of various parts which includes:

- The HULL: this is the base( bottom) of a boat with the pointy forward part called the BOW and the flat back part called the STERN.

There are different types of hulls which can be constructed by boat manufacturers and it depends on the use of the boat. They include:

1.)displacement hull,

2.)planing hull,

3.) flat hull,

4.) round hull,

5.)v-shaped hull and

6.) multi-hulled boats.

Cathedral hull is a planing hull that allows a boat to get on a plane quicker. A boat is said to be "on a plane" when more power (and speed) is applied, lift increases, and the boat, in effect, rides over its bow wave, reducing wetted area and thus reducing drag. I hope this helps, thanks.

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3 years ago

1. Plot the following graphs:

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Answer:

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3 years ago

A third point charge q3 is now positioned halfway between q1 and q2. The net force on q2 now has a magnitude of F2,net = 14.413

natima [27]

Answer:

The value of charge q₃ is 40.46 μC.

Explanation:

Given that.

Magnitude of net force $F=14.413\ N$

Suppose a point charge q₁ = -3 μC is located at the origin of a co-ordinate system. Another point charge q₂ = 7.7 μC is located along the x-axis at a distance x₂ = 8.2 cm from q₁. Charge q₂ is displaced a distance y₂ = 3.1 cm in the positive y-direction.

We need to calculate the distance

Using Pythagorean theorem

$r=\sqrt{x_{2}^2+y_{2}^2}$

Put the value into the formula

$r=\sqrt{(8.2\times10^{-2})^2+(3.1\times10^{-2})^2}$

$r=0.0876\ m$

We need to calculate the magnitude of the charge q₃

Using formula of net force

$F_{12}=kq_{2}(\dfrac{q_{3}}{r_{3}^2}+\dfrac{q_{1}}{r_{1}^2})$

Put the value into the formula

$14.413=9\times10^{9}\times7.7\times10^{-6}(\dfrac{q_{3}}{(0.0438)^2}+\dfrac{-3\times10^{-6}}{(0.0876)^2})$

$(\dfrac{q_{3}}{(4.38\times10^{-2})^2}+\dfrac{-3\times10^{-6}}{(0.0876)^2})=\dfrac{14.413}{9\times10^{9}\times7.7\times10^{-6}}$

$\dfrac{q_{3}}{(0.0438)^2}=207\times10^{-4}+3.909\times10^{-4}$

$q_{3}=0.0210909\times(0.0438)^2$

$q_{3}=40.46\times10^{-6}\ C$

$q_{3}=40.46\ \mu C$

Hence, The value of charge q₃ is 40.46 μC.

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3 years ago

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4 years ago

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