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3 years ago

A PWC is overtaking another vessel. Which vessel must give way?

1 answer:

8 0

The PWC must give way. The boat that is being overtaken needs to just keep going the same way keeping the same speed, while the PWC gives way and increases the speed in order to overtake the other vessel. Very similar to how driving a normal vehicle works when driving on ground.

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A traditional set of cycling rollers has two identical, parallel cylinders in the rear of the device that the rear tire of the b

mars1129 [50]

Answer:

ω2 = 216.47 rad/s

Explanation:

given data

radius r1 = 460 mm

radius r2 = 46 mm

ω = 32k rad/s

solution

we know here that power generated by roller that is

power = T. ω ..............1

power = F × r × ω

and this force of roller on cylinder is equal and opposite force apply by roller

so power transfer equal in every cylinder so

( F × r1 × ω1) ÷ 2 = ( F × r2 × ω2 ) ÷ 2 ................2

ω2 = $\frac{460\times 32}{34\times 2}$

ω2 = 216.47

8 0

3 years ago

A subway train is traveling at 22.2 m/s when it approaches a slower train 50m ahead traveling in the same direction at 6.94 m/s.

Amiraneli [1.4K]

Answer:

Time that they collide = 4.99s

Relative speed of the trains when they collide: The relative speed of The first train relative to the second, slower train at collision = 4.781 m/s

Explanation:

We will use the equations of motion to obtain the solution required

At time t = 0

speed of first train = 22.2 m/s

Initial space between the two trains = 50 m

Speed of second train = 6.94 m/s

For the first car, distance covered by the first train = y

y = distance covered between the beginning of the deceleration and the point where the the two trains hit one another.

u = initial velocity = 22.2 m/s

t = time taken for all this to happen

a = deceleration = - 2.1 m/s²

y = ut + (1/2)at²

y = 22.2t - 1.05t² (eqn 1)

For the second train,

At t = 0, y = 50 m

Let the new distance moved by the second train before collision = (y - 50)

u = initial velocity = 6.94 m/s

t = time taken = t

a = acceleration of the second train = 0 m/s² (constant velocity)

(y - 50) = ut + (1/2)at²

y - 50 = 6.94t

y = 6.94t + 50 (eqn 2)

substituting for y in eqn 2 using the expression obtained in eqn 1

y = 22.2t - 1.05t²

y = 6.94t + 50

22.2t - 1.05t² = 6.94t + 50

1.05t² - 15.26t + 50 = 0

Solving this quadratic equation

t = 4.99 s or 9.54 s

The position of the two trains are the same at those two times, but the first time is when they hit each other.

t = 4.99 s

At 4.99 s, the the velocity of the first train

v = u + at

v = 22.2 + (-2.1×4.99) = 11.721 m/s in the same direction as the second train.

Relative velocity at this point will be

= 11.721 - 6.94 = 4.781 m/s

Relative speed of the trains when they collide: The relative speed of The first train relative to the second, slower train at collision = 4.781 m/s

Hope this Helps!!!

4 0

2 years ago

4. Define resistance and describe what would happen to a light bulb if the voltage increased but the resistance stayed the same.

Arlecino [84]

Current would increase proportionally to voltage. Power dissipation (heating) would increase with the square of the voltage. And resistance means, "the refusal to accept or comply with something"

4 0

2 years ago

PLS HELP WILL MARK BRAINLIEST IF RIGHT NEED IMMEDIATELY PLEASEEEEE

timofeeve [1]

Answer:

Explanation:

Prescribe medications.

Prescribe treatments or therapies.

Treat patients using psychological therapies.

Collect medical information from patients, family members, or other medical professionals.

Record patient medical histories.

Develop medical treatment plans

3 0

2 years ago

Read 2 more answers

A penny is dropped from the top of a tower. It hits the ground below after 2.5 s. How tall was the tower?

pychu [463]

Answer:

C. 30.6m

Explanation:

To find the height of the tower, we are to use Newtons law of motion to solve this problem. Since the penny is falling from the top of the tower, it is acted by the acceleration due to gravity. The formula to be used is:

$H=ut+\frac{1}{2}gt^2$

Where H is the height of the tower, t is the time taken to hit the ground, u is the initial velocity and g is the acceleration due to gravity.

Given that, t = 2.5 s, g =9.8 m/s², u = 0 m/s (at the top of tower)

$H=ut+\frac{1}{2}gt^2\\\\H=0(2.5)+ \frac{1}{2}(9.8)(2.5)^2\\\\H=30.6\ m$

3 0

3 years ago