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Pani-rosa [81]
3 years ago
7

While walking along the shore of a lake Travis felt a cold breeze. What type of heat energy transfer is this an example of?

Physics
1 answer:
AfilCa [17]3 years ago
3 0
Out of the 3 types of heat transfer, this scenario would be most likely to be an example of convection.

Convection is where the transferring of heat is resulted through the movements of fluid, but in this case it is air. What happens is that when a part of the whole mass of air is heated, the hotter air rises and the cooler air descends and takes place of the hotter air before it was heated. Then, the cooler air becomes hotter and the hotter air before becomes the cooler air of both, which then results to the repeat of the exchange of places. This creates a motion until the whole mass has achieved mutual temperature, the heat source has stopped or extinguished, or there is a shift of temperature.
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All living creatures contain carbon
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One hypothesis states that plate movement results from convection currents in the
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I believe is A) Inner core
8 0
3 years ago
The earth's radius is 6.37×106m; it rotates once every 24 hours. What is the earth's angular speed? What is the speed of a point
Zarrin [17]

Answer:

a) w = 7.27 * 10^-5 rad/s

b) v1 = 463.1 m/s

c) v1 = 440.433 m/s

Explanation:

Given:-

- The radius of the earth,  R = 6.37 * 10 ^6 m

- The time period for 1 revolution T = 24 hrs

Find:

What is the earth's angular speed?

What is the speed of a point on the equator?

What is the speed of a point on the earth's surface located at 1/5 of the length of the arc between the equator and the pole, measured from equator?

Solution:

- The angular speed w of the earth can be related with the Time period T of the earth revolution by:

                                  w = 2π / T

                                  w = 2π / 24*3600

                                  w = 7.27 * 10^-5 rad/s

- The speed of the point on the equator v1 can be determined from the linear and rotational motion kinematic relation.

                                 v1 = R*w

                                 v1 = (6.37 * 10 ^6)*(7.27 * 10^-5)

                                 v1 = 463.1 m/s

- The angle θ subtended by a point on earth's surface 1/5 th between the equator and the pole wrt equator is.

                                 π/2  ........... s

                                 x     ............ 1/5 s

                                 x = π/2*5 = 18°    

- The radius of the earth R' at point where θ = 18° from the equator is:

                                R' = R*cos(18)

                                R' = (6.37 * 10 ^6)*cos(18)

                                R' = 6058230.0088 m

- The speed of the point where θ = 18° from the equator v2 can be determined from the linear and rotational motion kinematic relation.

                              v2 = R'*w

                              v2 = (6058230.0088)*(7.27 * 10^-5)

                              v2 = 440.433 m/s

5 0
3 years ago
What is the work required for a penguin to push a box 2 meters with a force of 8 newtons?
choli [55]

Work done is given by product of force and displacement due to that force

So here we will have

Work = Force \times displacement

here we know that

Force = 8 N

displacement = 2 m

Now work done is given as

W = 8\times 2

W = 16 J

so it will do 16 J work to move the box

3 0
3 years ago
A mass of .1539 kg moves down a 5 meter ramp in 2 seconds. What
enot [183]

Answer:

Velocity=2.5m/s

KE=4809.375J

Explanation:

Velocity=5m/2s=2.5m/s

KE=½×1539kg×(2.5m/s)²

KE=4809.375J

4 0
3 years ago
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