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3 years ago

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What would happen if a person who is wearing a heavy winter jacket were to place a thermometer inside the jacket next to his or

her skin? What would happen if we took the same jacket, after it had been hanging in a closet, and placed a thermometer inside?

1 answer:

Scilla [17]3 years ago

7 0

Answer:

Explained.

Explanation:

Let me explain the two cases in by one.

(1) When Person wearing a jacket, heat from his or her body is going to get trapped inside his or her jacket to keep this person warm when a thermometer is placed inside, it would indicate a higher temperature on the scale.

(2) When Jacket is Hanging in a closet, its temperature would be roughly close to room temperature and when we would place a thermometer it would not indicate any change on its scale.

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A reconnaissance plane flies 404 km awayfrom its base at 730 m/s, then flies back to its base at 1095 m/s.What is it’s average s

elena-14-01-66 [18.8K]

The average speed of the plane is 875.999 m/s.

Average speed can be defined as the ratio of total distanced travelled by the object to that of total time taken to cover the distance.

Mathematically, Average speed = Av = $\frac{Total Distance}{Total Time}$

According to the question,

Speed of the plane away from its base V₁ = 730 m/s

Speed of the plane when it flies back V₂ = 1095 m/s

Plane flies the distance D = 404 km

Total Distance covered by the plane S = 404 * 2 km

(because the distance travelled by the plane when going away from the base and then flying back to the base is same)

Therefore S = 808 km = 808 ˣ 10³ m

Time taken by the plane while flying away from the base T₁ = $\frac{D}{V1}$

T₁ = $\frac{404000}{730}$ = 553.425 s

Time taken by the plane while flying back to the base T₂ = $\frac{D}{V2}$

T₂ = $\frac{404000}{1095}$ = 368.949s

Total Time T = T₁ + T₂ = 922.375 s

Therefore Av = $\frac{Total Distance}{Total Time}$

= $\frac{D}{T}$ m/s

= $\frac{808000}{922.375}$ m/s

= 875.999 m/s

The average speed of the plane will be 875.999 m/s.

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Wind erosion can be reduced by _____.

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Using land according to its capability. protect the soil surface with some form of cover. control runoff before it develops into an erosive force

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There is a 50 g sample of Ra-229. It has a half-life of 4 minutes.

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Via half-life equation we have:

$A_{final}=A_{initial}(\frac{1}{2})^{\frac{t}{h} }$

Where the initial amount is 50 grams, half-life is 4 minutes, and time elapsed is 12 minutes. By plugging those values in we get:

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4 years ago

What is the magnitude of the magnetic field at a point midway between them if the top one carries a current of 19.5 A and the bo

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Answer:

The magnetic field will be $\large{\dfrac{1.4 \times 10^{-4}}{d}} T$ , '2d' being the distance the wires.

Explanation:

From Biot-Savart's law, the magnetic field ( $\large{\overrightarrow{B}}$ ) at a distance ' $r$ ' due to a current carrying conductor carrying current ' $I$ ' is given by

$\large{\overrightarrow{B} = \dfrac{\mu_{0}I}{4 \pi}} \int \dfrac{\overrightarrow{dl} \times \hat{r}}{r^{2}}}$

where ' $\overrightarrow{dl}$ ' is an elemental length along the direction of the current flow through the conductor.

Using this law, the magnetic field due to straight current carrying conductor having current ' $I$ ', at a distance ' $d$ ' is given by

$\large{\overrightarrow{B}} = \dfrac{\mu_{0}I}{2 \pi d}$

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Given $\large{I_{t} = 19.5 A}$ and $\large{I_{B} = 12.5 A}$

Therefore, the magnetic field ( $\large{B_{t}}$ ) at 'P' due to the top wire

$B_{t} = \dfrac{\mu_{0}I_{t}}{2 \pi d}$

and the magnetic field ( $\large{B_{b}}$ ) at 'P' due to the bottom wire

$B_{b} = \dfrac{\mu_{0}I_{b}}{2 \pi d}$

Therefore taking the value of $\mu_{0} = 4\pi \times 10^{-7}$ the net magnetic field ( $\large{B_{M}}$ ) at the midway between the wires will be

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