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3 years ago

Which description best models the energy transfer that occurs in the radiative zone?

Physics

2 answers:

Zinaida [17]3 years ago

5 0

Answer : Option D) A dropped wallet is kicked around the floor of a busy train station.

Explanation : The description that best suits the model for the energy transfer that occurs in the radiative zone is -

<h3>A dropped wallet is kicked around the floor of a busy train station.</h3>

As in the radiation zone, which is also called as radiative zone or radiative region in the layer of a star's interior where energy is primarily transported toward the exterior by means of radiative diffusion and thermal conduction. This can be correlated with the example of a dropped wallet in a busy train station as the wallet would be kicked by many people who are in a hurry to travel, this represents the radiative diffusion. So, every time the wallet is being kicked by someone the energy is getting transformed.

hodyreva [135]3 years ago

4 0

the answer is D) a dropped wallet is kicked around the floor of a busy train station

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lora16 [44]

Answer:

Explanation:

Circle motion

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12=√Vo²+2²

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Ha1=Hb2

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r2=6×3/11.83

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4 years ago

395,000 meters in 9000

marta [7]

Answer:

43.88 meters per second

Explanation:

The computation of the speed is shown below:

As we know that

$Speed = \frac{Distance}{time}$

where,

Distance is 395,000 meters

Time is 9,000 seconds

Now placing these values to the formula

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$= \frac{395,000}{9,000}$

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As speed shows the relation between the distance and time and the same is to be considered i.e by applying the formula

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4 years ago

A jet aircraft with a mass of 4,475 kg has an engine that exerts a force (thrust) equal to 60,800 N. (a) What is the jet's accel

BartSMP [9]

Answer:

a) 13.59 m/s²

b) 67.95 m/s

c) 169.875 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

m = Mass

Force

$F=ma\\\Rightarrow a=\frac{F}{m}\\\Rightarrow a=\frac{60800}{4475}\\\Rightarrow a=13.59\ m/s^2$

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$v=u+at\\\Rightarrow v=0+13.59\times 5\\\Rightarrow v=67.95\ m/s$

Velocity attained at 5 seconds is 67.95 m/s

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{67.95^2-0^2}{2\times 13.59}\\\Rightarrow s=169.875\ m$

Distance traveled in the 5 seconds is 169.875 m

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Answer:

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