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nadezda [96]
3 years ago
10

Which statement best describes why an island food web could be considered a closed system

Physics
1 answer:
iris [78.8K]3 years ago
6 0

B. Energy enters and exits the system matter does not

Explanation:

In thermodynamics, there are three types of system:

  • Open system: it is a system that exchanges both matter and energy with the surroundings - an example is an open bottle
  • Closed system: it is a system that exchanges only matter with the surroundings, while energy cannot enter and exit the system - an example is a closed bottle (air cannot enter/exit the bottle, but heat can be exchanged through the walls of the bottle)
  • Isolated system: it is a system that cannot exchange matter or energy with the surroundings - an example is a sealed thermic container (heat cannot pass through the walls)

In this problem, we are asked to identify which statement best describes the island food web as a closed system: based on the definitions above, we can say that the correct option is

B. Energy enters and exits the system matter does not

Learn more about thermodynamics:

brainly.com/question/4759369

brainly.com/question/3063912

brainly.com/question/3564634

#LearnwithBrainly

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You are riding on a carousel that is rotating at a constant 24 rpm. It has an inside radius of 4 ftand outside radius of 12 ft.
Nookie1986 [14]

Answer:

magnitude of the Coriolis acceleration is 44.235 ft/s² and the direction of the acceleration is along the axis of transmission

Explanation:

Given the data in the question;

Speed of carousel N = 24 rpm

From the diagram below, selected path direction defines the Axis of slip.

Hence, The Coriolis is acting along the axis of transmission

Now, we determine the angular speed ω of the carousel.

ω = 2πN / 60

we substitute in the value of N

ω = (2π × 24) / 60

ω = 2.5133 rad/s

Next, we convert the given velocity from mph to ft/s

we know that; 1 mph = 1.4667 ft/s

so

V_{slip = 6 mph = ( 6 × 1.4667 ) = 8.8002 ft/s

Now, we determine the magnitude of the Coriolis acceleration

a_c = 2( V_{slip × ω )

we substitute

a_c = 2( 8.8002 ft/s × 2.5133 rad/s )

a_c = 44.235 ft/s²

Hence, magnitude of the Coriolis acceleration is 44.235 ft/s² and the direction of the acceleration is along the axis of transmission

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Constants Modern wind turbines are larger than they appear, and despite their apparently lazy motion, the speed of the blades ti
nadezda [96]

It is given that the length of blade of the turbine is 58 m.

During the motion, the turbine will undergo rotational motion. Hence the radius of the circle traced by the turbine is equal to the length of the blade.

Hence radius r = 58 m.

The frequency of the turbine [f] =14 rpm.

Here rpm stands for rotation per minute.

Hence the frequency of the turbine in one second-

                                                      f=\frac{14}{60}\ s^-1

                                                      f=0.23333 Hz

Here Hz[ hertz] is the unit of frequency.

The angular velocity of the turbine \omega =2\pi f

                                                          \omega=2*3.14*0.2333

                                                          \omega=1.465124 radian/second

Now we have to calculate the centripetal  acceleration of the blade.

Let the linear velocity of the blade is v.

we know that  linear velocity v=ωr

The centripetal acceleration is calculated as-

                                                                      a_{c} =\frac{v^2}{r}

                                                                            =\frac{[\omega r]^2}{r}

                                                                            =\omega^2r

                                                                            =[1.465124]^2 *58

                                                                            =124.5021234 m/s^2      [ans]

8 0
3 years ago
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lisov135 [29]
1: Lost (?)
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7 0
4 years ago
A Pitot-static probe is used to measure the speed of an aircraft flying at 3000 m. If the differential pressure reading is 3300
adell [148]

Answer:

   v = 306.76 Km/h

Explanation:

given,

height of the aircraft = 3000 m

differential pressure reading = 3300 N/m²

density of air = 0.909 Kg/m³

speed of aircraft = ?

Assuming the air flowing above air craft is in-compressible, irrotational and steady so, we can use Bernoulli's equation to solve the problem.

using Bernoulli's equation

          \dfrac{v^2}{2} = \dfrac{\Delta P}{\rho}

where ρ is the density of the air at 3000 m

          v= \sqrt{\dfrac{2 \times \Delta P}{\rho}}

          v= \sqrt{\dfrac{2 \times 3300}{0.909}}

          v = \sqrt{7260.726}

                 v = 85.21 m/s

          v= 85.21 \times \dfrac{3600}{1000}

                 v = 306.76 Km/h

4 0
3 years ago
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