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Alex
4 years ago
7

when you place your feet near the fireplace and they become warm what type of energy conversion occurs

Physics
2 answers:
JulsSmile [24]4 years ago
8 0
I think you forgot to give the options along with the question. I am answering the question based on my knowledge and research. "Radiant to thermal" is the type of energy conversion that occurs when <span>you place your feet near the fireplace and they become warm. I hope the answer has come to your great help.</span>
Valentin [98]4 years ago
8 0

Answer:

Radiation

Explanation:

When we place our feet near the fireplace then the heat transfer from the fire place to our feet by the mode of radiation.

In the process of radiation, the medium is not required.

The heat waves travel with the speed of light.

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Suppose that diameters of a new species of apple have a bell-shaped distribution with a mean of 7.2cm and a standard deviation o
Svet_ta [14]

Answer:

97.5%

Explanation:

By the empirical rule (68-95-99.7),

  1. 68% of data are within <em>μ </em>- <em>σ</em> and <em>μ </em>+ <em>σ</em>
  2. 95% of data are within <em>μ </em>- 2<em>σ</em> and <em>μ </em>+ 2<em>σ</em>
  3. 99.7% of data are within <em>μ </em>- 3<em>σ</em> and <em>μ </em>+ 2<em>σ</em>

<em>σ </em> and <em>μ</em> are the standard deviation and the mean respectively.

From the question,

<em>μ</em> = 7.2 cm

<em>σ</em> = 0.38 cm

7.96 = 7.2 + (<em>n</em> × 0.38)

<em>n</em> = 2

Hence, 7.96 represents <em>μ </em>+ 2<em>σ</em>.

P(X < <em>μ </em>+ 2<em>σ</em>) = P(X < <em>μ</em>) + P(<em>μ</em> < X < <em>μ </em>+ 2<em>σ</em>)

P(X < <em>μ</em>) is the percentage less than the mean = 50%.

P(<em>μ</em> < X < <em>μ </em>+ 2<em>σ</em>) is half of P(<em>μ </em>- 2<em>σ</em> < X < <em>μ </em>+ 2<em>σ</em>) = 95% ÷ 2 = 47.5%.

Considering this, for apples that are no more than 7.96 cm,

P(X < 7.96) = P(X < 7.2) + P(7.2 < X < 7.96) = 50% + 47.5% = 97.5%

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A mole of water molecules would just about fill a
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B. tablespoon

Explanation:

Considering the question given, a mole of water is small compare to the options given from the question aside the table spoon.

A mole of water is just 18g and that's equivalent to 18 mL.

18 mL of water will fill a table spoon but not a cup which is about 237mL. The wheel barrow and gallon bucket have larger volumes of which 18 mL of water will never fill them.

So, a mole of water can successfully fill a table spoon.

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3 years ago
What happens to the body’s immune system when attacked by HIV ?
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Answer:

HIV attacks a specific type of immune system cell in the body. It's known as the CD4 helper cell or T cell. When HIV destroys this cell, it becomes harder for the body to fight off other infections.

Explanation:

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3 years ago
High-altitude mountain climbers do not eat snow, but always melt it first with a stove. To see why, calculate the energy absorbe
gavmur [86]

Answer:

The conditions are not given in the question. Here is the complete question.

High-altitude mountain climbers do not eat snow, but always melt it first with a stove. To see why, calculate the energy absorbed from a climber's body under the following conditions. The specific heat of ice is 2100 J/kg⋅C∘, the latent heat of fusion is 333 kJ/kg, the specific heat of water is 4186 J/kg⋅C∘

a) eat 1.0kg of -15 degrees Celsius snow which your body warms to body temperature of 37 degrees Celsius ?

b) melt 1.0kg of -15 degrees Celsius snow using a stove and drink the resulting 1.0kg of water at 2 degrees Celsius, which your body has to warm to 37 degrees Celsius ?

Explanation:

Let's calculate the heat required to convert ice from -15°C to 0°C and then the heat required to convert it into the water.

a)

Heat required to convert -15°C ice to 0°C.

  ms_{ice}ΔT = (1.0)(2.100×10³)(15) = 3.150×10^{4}J

Heat required to convert 1.0 kg ice to water.

    mL_{ice} = 1×3.33×10^{5} = 3.33×10^{5}J

Heat required to convert 1.0 kg water at 0°C to 37°C.

   ms_{water}ΔT = 1×4.186×10^{3}×37 = 1.548×10^{5} J

Total heat required = 3.150×10^{4} + 3.33×10^{5} + 1.548×10^{5}

                              = 5.19×10^{5} J

b)

Heat required to warm 1.0 kg water at 2°C to water at 37°C.

ms_{water}ΔT = 1×4.186×10^{3}×35

                = 1.465×10^{5}J

   

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3 years ago
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