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Sloan [31]
3 years ago
11

If one's hands are being warmed by holding them to one side of a flame, the predominant form of heat transfer is what process?

Physics
1 answer:
avanturin [10]3 years ago
4 0

Answer:

b. radiation

Explanation:

The correct choice is

b. radiation

Radiation is the process of heat transfer where no medium is required. Air is present between hands and flame and thermal conductivity of air is very low.

The hands are not in direct contact with the flame, hence conduction process of heat transfer is not possible.

In convection process of heat transfer, we have the warmer part of the medium moving towards the colder part. hence the convection is also not responsible for transfer of heat in this scenario

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givi [52]

I would say the answer is B

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What’s is the movement of one object around another
allsm [11]

Answer:

revolution

Explanation:

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What is the resultant of the two vectors shown?
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Answer:

B is the right answer of the following statement

5 0
3 years ago
The density of ice is 0.93 g/cm3. what is the volume, in cm3, of a block of ice whose mass is 5.00 kg? remember to select an ans
serious [3.7K]

The Volume of the ice block is 5376.344 cm^3.

The density of a material is define as the mass per unit volume.

Here, the density of ice given is 0.93 g/cm^3

Mass of the ice block  given is 5 kg or 5000 g

Now calculate the volume of the ice block

density=mass/volume

0.93=5000/Volume

Volume =5376.344 cm^3

Therefore the volume of  ice block is 5376.344 cm^3

7 0
3 years ago
Assume a rectangular strip of a material with an electron density of n=5.8x1020 cm-3. The strip is 8 mm wide and 0.8 mm thick an
vampirchik [111]

Answer: I = 111.69 pA

Explanation: The hall effect is all about the fact that when a semiconductor is placed perpendicularly to a magnetic field, a voltage is generated which could be measured at right angle to the current path. This voltage is known as the hall voltage.

The hall voltage of a semiconductor sensor is given below as

V = I×B/qnd

Where V = hall voltage = 1.5mV =1.5/1000=0.0015V

I = current =?,

n= concentration of charge (electron density) = 5.8×10^20cm^-3 = 5.8×10^20/(100)³ = 5.8×10^14 m^-3

q = magnitude of an electronic charge=1.609×10^-19c

B = strength of magnetic field = 5T

d = thickness of sensor = 0.8mm = 0.0008m

By slotting in the parameters, we have that

0.0015 = I × 5/5.8×10^14 × 1.609×10^-19×0.0008

0.0015 = I×5/7.446×10^-8

I = (0.0015 × 7.446×10^-8)/5

I = 111.69*10^(-12)

I = 111.69 pA

3 0
3 years ago
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