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gizmo_the_mogwai [7]
3 years ago
15

good morning to you all! can someone please answer this, ill give you brainliest and your earning 50 points.

Physics
2 answers:
elena55 [62]3 years ago
7 0

Despite the risks, Human populations can be found near volcanoes. This can be associated with the availability of which natural resource?

  • <em>Geothermal</em><em> </em><em>energy</em>

<u>Geothermal</u><u> </u><u>energy</u><u> </u><u>is</u><u> </u><u>the</u><u> </u><u>f</u><u>o</u><u>r</u><u>m</u><u> </u><u>of</u><u> </u><u>energy</u><u> </u><u>obtained</u><u> </u><u>from</u><u> </u><u>the</u><u> </u><u>heat</u><u> </u><u>of</u><u> </u><u>the</u><u> </u><u>earth</u><u>.</u><u> </u><u>Near</u><u> </u><u>volcanoes</u><u> </u><u>the</u><u> </u><u>land</u><u> </u><u>is</u><u> </u><u>more</u><u> </u><u>hot</u><u> </u><u>compared</u><u> </u><u>to</u><u> </u><u>the</u><u> </u><u>normal</u><u> </u><u>areas</u><u>.</u><u> </u><u>So</u><u>,</u><u> </u><u>the</u><u> </u><u>heat</u><u> </u><u>of</u><u> </u><u>the</u><u> </u><u>lava</u><u> </u><u>will</u><u> </u><u>help</u><u> </u><u>in</u><u> </u><u>producing</u><u> </u><u>Geothermal</u><u> </u><u>energy</u><u>.</u><u>.</u><u>.</u><u>~</u>

otez555 [7]3 years ago
5 0

Answer:

Geothermal energy

Explanation:

  • The volcanoes erupt the hot magma or lava
  • This is the main source of geothermal energy.
  • As energy is most important thing for survival so many humans establish community near volcanoes
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A 3.5 kW drill transfers 5 000 kJ of kinetic energy during 15 seconds of use. What is the percentage efficiency of the drill?
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Answer:

1.04%

Explanation:

Given that,

The power of drill = 3.5 kW = 3500 W

Transferred kinetic energy = 5000 kJ during 15 seconds of use.

We need to find the percentage efficiency of the drill. It can be given by :

\eta=\dfrac{P_o}{P_i}\times 100

Where

Po and Pi are output and input powers.

P_o=\dfrac{5000\times 10^3}{15}\\\\=3.34\times 10^5\ W

So,

\eta=\dfrac{3500}{3.34\times10^{5}}\times100\\\\=1.04\%

So, the percentage efficiency of the drill is 1.04%.

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What temperature is required to transfer waste heat to the environment for a heat engine to be 100 percent efficient?
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This can be seen as a trick question because heat engines can typically never be 100 percent efficient. This is due to the presence of inefficiencies such as friction and heat loss to the environment. Even the best heat engines can only go up to around 50% efficiency.
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A potter's wheel is spinning with an initial angular velocity of 11 rad/s . It rotates through an angle of 80.0 rad in the proce
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  • The angular acceleration of the wheel approximately <u>-0.76 rad/s² or proportionally as deceleration approximately 0.76 rad/s</u>.
  • It need approximately <u>14.474 s</u> to come to rest.

<h2>Introduction</h2>

Hi ! I will help you to discuss about Proportionally Changes in Circular Motion. The analogy of proportionally changes in circular motion is same as the analogy of proportionally changes in direct motion. Here you will hear again the terms acceleration and change in speed, only expressed in the form of a certain angle coverage. Before that, in circular motion, it is necessary to know the following conditions:

  • 1 rotation = 2π rad
  • 1 rps = 2π rad/s
  • 1 rpm = \sf{\frac{1}{60} \: rps} = \sf{\frac{1}{30}\pi \: rad/s}

<h2>Formula Used</h2>

The following equations apply to proportionally changes circular motion:

<h3>Relationship between Angular Acceleration and Change of Angular Velocity </h3>

\boxed{\sf{\bold{\omega_t = \omega_0 + \alpha \times t}}}

With the following conditions:

  • \sf{\omega_t} = final angular velocity  (rad/s)
  • \sf{\omega_0} =  initial angular velocity (rad/s)
  • \sf{\alpha} = angular acceleration (rad/s²)
  • t = interval of the time (s)
<h3>Relationship between Angular Acceleration and Change of \sf{\theta} (Angle of Rotation) </h3>

\boxed{\sf{\bold{\theta = \omega_0 \times t + \frac{1}{2} \times \alpha \times t^2}}}

Or

\boxed{\sf{\bold{(\omega_t)^2= (\omega_0)^2 + 2 \times \alpha \times \theta}}}

With the following condition :

  • \sf{\theta} = change of the sudut (rad)
  • \sf{\alpha} = angular acceleration (rad/s²)
  • t = interval of the time (s)
  • \sf{\omega_t} = final angular velocity  (rad/s)
  • \sf{\omega_0} =  initial angular velocity (rad/s)

<h2>Problem Solving</h2>

We know that :

  • \sf{\omega_t} = final angular velocity  = 0 rad/s >> see in the sentence "in the process of coming to rest."
  • \sf{\omega_0} =  initial angular velocity = 11 rad/s
  • \sf{\theta} = change of the sudut = 80.0 rad

What was asked :

  • \sf{\alpha} = angular acceleration = ... rad/s²
  • t = interval of the time = ... s

Step by step :

  • \sf{\alpha} = ... rad/s²

\sf{(\omega_t)^2= (\omega_0)^2 + 2 \times \alpha \times \theta}

\sf{0^2= (11)^2 + 2 \times \alpha \times 80}

\sf{0 = 121 + 160 \alpha}

\sf{-160 \alpha = 121}

\sf{\alpha = \frac{121}{-160}}

\sf{\alpha = -0.75625 \: rad/s^2 \approx \boxed{-0.76 \: rad/s^2}}

  • t = ... s

\sf{\alpha = \frac{\omega_0 - \omega_t}{t}}

\sf{-0.76 = \frac{0 - 11}{t}}

\sf{-0.76t = -11}

\sf{t = \frac{- 11}{-0.76}}

\boxed{\sf{t \approx 14.474 \: s}}

<h3>Conclusion</h3>

So :

  • The angular acceleration of the wheel approximately -0.76 rad/s² or proportionally as deceleration approximately 0.76 rad/s.
  • It need approximately 14.474 s to come to rest.
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