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3 years ago

What is needed for steering control on a PWC??

2 answers:

7 0

The most basic requirement for the steering control of a PWC is the jet propulsion of water to make it move forward and control is sideways motion by controlling the direction of the jet propulsion.

Further Explanation:

The PWC is the personal water craft that is used to run on the water surface. The personal water craft makes use of the high pressure water jet for applying a forward reaction force on the water craft.

The high speed water jet applies a reaction force on the craft in the forward reaction. This force follows the Newton’s third law of motion. According to it, every action has equal and opposite reaction.

Therefore, the jet propulsion applies the sufficient force and it is also moved sideways to control the steering of the water craft.

Learn More:

A 30.0-kg box is being pulled across a carpeted floor by a horizontal force of 230 n brainly.com/question/7031524
The amount of kinetic energy an object has depends on its brainly.com/question/137098
A 50-kg meteorite moving at 1000 m/s strikes earth. Assume the velocity is along the line brainly.com/question/6536722

Answer Details:

Grade: High School

Chapter: Newton’s law of Motion

Subject: Physics

Keywords: PWC, Personal water Craft, newton’s law of motion, action, reaction, steering control, jet propulsion, force, forward direction.

posledela3 years ago

6 0

In steering in control on a PWC, the most needed requirement is to have power in which will allows the one using it to control the engine for it to stop or for it to move. Without power, the person using it could be out of balance or fail to start the engine. This is the basic requirement because even though it is shut off or the engine is off, the PWC is still on and it could still move without proper knowledge of the engine.

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Answer:

$5.8\; {\rm kg\cdot m \cdot s^{-1}}$ .

Explanation:

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Thus, the change in the velocity of the mass would be:

$\begin{aligned}& (\text{Change in Velocity}) \\ =\; & (\text{Final Velocity}) - (\text{Initial Velocity}) \\ =\; & (-4.0\; {\rm m\cdot s^{-1}}) - (6.0\; {\rm m\cdot s^{-1}}) \\ =\; & (-10\; {\rm m\cdot s^{-1})\end{aligned}$ .

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$\begin{aligned} & \text{change in momentum} \\ =\; & (\text{mass}) \times (\text{change in velocity}) \\ =\; & 0.58\; {\rm kg} \times (-10\; {\rm m\cdot s^{-1}}) \\ =\; & (-5.8\; {\rm kg \cdot m \cdot s^{-1}})\end{aligned}$ .

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3 years ago