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4 years ago

Where is the steering nozzle located on a pwc?

2 answers:

3 0

At the rear.

PWC stands for personal watercraft, and it is a small powerboat. The main components of a PWC are the hull (body of the boat), deck (surface where people walk/stand), throttle (controls speed), steering nozzle and water intake.

Tanzania [10]4 years ago

3 0

Answer:

The steering nozzle is located at the back of the unit on the PWC

Explanation:

The detail of this question is given below:

PWC: PWC stands for Personal Watercraft which is operation on some certain and specific requirements set by the PWC operator. There are two things to be noted on PWC and these are steering wheel and steering nozzle.

Steering nozzle: PWC are moved by jet drive. Through jet drive water is pushed backwards into pump and then pump forcefully injected it out under full pressure by the use of steering nozzle which is located at the back of the unit. The steering control and steering nozzle work at the same direction. If the steering control move left than the nozzle automatically turns left. In order to operation PWC, the important thing to keep in mind is that you should be well strong enough to control it because it requires a lot of power. If the engine is switch off then you will definitely lose your control over it.

Answer Detail:

Level: High School

Subject: Physics

Keywords:

• PWC

• Steering control

• Steering nozzle

• engine

Learn more to evaluate:

PWC Control: brainly.com/question/4035935

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Anna71 [15]

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3 years ago

A convex mirror of focal length 34 cm forms an image of a soda bottle at a distance of 18 cm behind the mirror. The height of th

Mademuasel [1]

<h2>Answers:</h2>

In convex mirrors the focus is virtual and the focal distance is negative. This is how the reflected rays diverge and only their extensions are cut at a point on the main axis, resulting in a virtual image of the real object .

<h2 /><h2>a) Position of the soda bottle (the object) </h2><h2 />

The Mirror equation is:

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$ (1)

Where:

$f$ is the focal distance

$u$ is the distance between the object and the mirror

$v$ is the distance between the image and the mirror

We already know the values of $f$ and $v$ , let's find $u$ from (1):

$u=\frac{v.f}{v-f}$ (2)

Taking into account the explanation at the beginign of this asnswer:

$f=-34cm$ and $v=-18cm$

The <u>negative signs</u> indicate the <u>focal distance</u> and the <u>distance between the image and the mirror are virtual </u>

Then:

$u=\frac{(-18cm)(-34cm)}{-18cm-(-34cm)}$

$u=38.25cm$ (3) >>>>Position of the soda bottle

<h2>b) magnification of the image </h2><h2 />

The magnification $m$ of the image is given by:

$m=-\frac{v}{u}$ (4)

$m=-\frac{(-18cm)}{38.25cm}$

$m=0.47$ (5)>>>the image is 0.47 smaller than the object

<h2>The fact that this value is positive means the image is upright </h2>

<h2>c) Describe the image </h2><h2 />

According to the explanations and results obtained in the prior answers, the correct option is 9:

<h2>The image is <u>virtual, upright, smaller</u> than the object </h2><h2 /><h2>d) Height of the soda bottle (object) </h2>

Another way to find the magnification is by the following formula:

$m=\frac{h_{i}}{h_{o}}$ (6)

Where:

$h_{i}$ is the image height

$h_{o}$ is the object height

We already know the values of $m$ and $h_{i}$ , let's find $h_{o}$ :

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$h_{o}=\frac{6.8cm}{0.47}$

$h_{o}=14.46cm$ (8) >>>height of thesoda bottle

6 0

3 years ago

When the mass of the cylinder increased by a factor of 3, from 1.0 kg to 3.0 kg, what happened to the cylinder’s gravitational p

const2013 [10]

Answer: Fourth option. It increased by a factor of 3.

Solution:

m1=1.0 kg

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Ep1=(1.0 kg)*g*h

Ep1=g*h

m2=3.0 kg

Ep2=(3.0 kg)*g*h

Ep2=3*g*h

Replacing g*h by Ep1 in the equation above:

Ep2=3*Ep1

Then, the cylinder's gravitational potential energy increased by a factor of 3.

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4 years ago

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Nesterboy [21]

Answer:

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Explanation:

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Net force on the lower box

= T - mg μ_s = M a ( a is the acceleration created by net force in M )

Considering force on the upper box

mg μ_s = ma

a = g μ_s

Put this value of a in the equation above

T - m gμ_s = M g μ_s

T = mg μ_s + M g μ_s

= g μ_s ( M+m )

2 )

Largest tension required

T = 9.8 x .50 x ( 10+6 )

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4 years ago

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harkovskaia [24]

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