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postnew [5]
1 year ago
14

Water in a metal pot is heated over a campfire. Primarily, by wat means of energy transfer is thermal energy moved from the fire

to the bottom of the pot?
Physics
1 answer:
Art [367]1 year ago
7 0

when water in a metal pot is heated over a campfire. Primarily, by means of convection energy transfer is thermal energy moved from the fire to the bottom of the pot

<h3 /><h3>What is thermal energy?</h3>

It can be defined as the form of the energy in which heat is transferred from one body to another body due to their molecular movements, thermal energy is also known as heat energy.

Thermal energy can flow by different means such as conduction, convection, and radiation.

As given in the problem when water in a metal pot is heated over a campfire. The energy we receive around the campfire is mostly due to the effect of the radiation of the heat but the Primary means of energy transfer is thermal energy moved from the fire to the bottom of the pot because of the convection.

Thus, when water is heated in a metal pot over a campfire. Thermal energy is primarily transferred from the fire to the bottom of the pot through convection.

Learn more about thermal energy from here

brainly.com/question/3022807

#SPJ1

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A high-speed flywheel in a motor is spinning at 450 rpm when a power failure suddenly occurs. The flywheel has mass 40.0 kg and
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Answer:

A) \omega_f=17.503\ rad.s^{-1}

B) t=55.6822\ s

C) \theta=1312\ rad

Explanation:

Given:

  • mass of flywheel, m=40\ kg
  • diameter of flywheel, d=0.72\ m
  • rotational speed of flywheel, N_i=450\ rpm \Rightarrow \omega_i=\frac{450\times 2\pi}{60} =15\pi\ rad.s^{-1}
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<u>Using equation of motion:</u>

\theta=\omega_i.t+\frac{1}{2} \alpha.t^2

360\pi=15\pi\times 35+\frac{1}{2} \times \alpha\times35^2

\alpha=-0.8463\ rad.s^{-2}

Negative sign denotes deceleration.

A)

Now using the equation:

\omega_f=\omega_i+\alpha.t

\omega_f=15\pi-0.8463\times 35

\omega_f=17.503\ rad.s^{-1} is the angular velocity of the flywheel when the power comes back.

B)

Here:

\omega_f=0\ rad.s^{-1}

Now using the equation:

\omega_f=\omega_i+\alpha.t

0=15\pi-0.8463\times t

t=55.6822\ s is the time after which the flywheel stops.

C)

Using the equation of motion:

\theta=\omega_i.t+\frac{1}{2} \alpha.t^2

\theta=15\pi\times 55.68225-0.5\times 0.8463\times 55.68225^2

\theta=1312\ rad revolutions are made before stopping.

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P = (19355 + 55742.4) / 3 = 75097.4/3

P = 25032.47 W

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