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4 years ago

Which is an example of conduction?

Physics

2 answers:

cupoosta [38]4 years ago

8 0

A hand is burned when touching a pot handle

tatuchka [14]4 years ago

7 0

Answer: The correct answer is a hand is burned when touching a pot handle.

Explanation: There are 3 modes of heat transfer:

Conduction: This type of heat transfer happens when there is direct contact between the two object.

Convection: This type of heat transfer happens when there is a movement of fluid (liquid or gas).

Radiation: This type of heat transfer happens when there is direct transfer of energy through space.

From the given options,

A hand is burned when touching a pot handle is an example of conduction because there is direct contact between the hand and pot handle.

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A motorcycle travels 90.0 km/h. How many seconds will it take the motorcycle to cover 2.10 x 103 m?

mina [271]

Answer:

The time taken is 84 seconds, please give the brainliest award would be much appreciated.

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3 years ago

Which state of matter can be compressed?

MA_775_DIABLO [31]

Gases............In gases, the atoms are much more spread out than in solids or liquids, and the atoms collide randomly with one another. A gas will fill any container, but if the container is not sealed, the gas will escape. Gas can be compressed much more easily than a liquid or solid

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3 years ago

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Which describes one event that causes an eclipse?

Burka [1]

Answer:

Earths shadow covering the moon would create a lunar eclipse.

Explanation:

because i just know

3 0

3 years ago

An AM radio station broadcasts isotropically (equally in all directions) with an average power of 3.40 kW. A receiving antenna 6

lara [203]

To solve the problem we will apply the concepts related to the Intensity as a function of the power and the area, as well as the electric field as a function of the current, the speed of light and the permeability in free space, as shown below.

The intensity of the wave at the receiver is

$I = \frac{P_{avg}}{A}$

$I = \frac{P_{avg}}{4\pi r^2}$

$I = \frac{3.4*10^3}{4\pi(4*1609.34)^2} \rightarrow 1mile = 1609.3m$

$I = 6.529*10^{-6}W/m^2$

The amplitude of electric field at the receiver is

$I = \frac{E_{max}^2}{2\mu_0 c}$

$E_{max}= \sqrt{2I\mu_0 c}$

The amplitude of induced emf by this signal between the ends of the receiving antenna is

$\epsilon_{max} = E_{max} d$

$\epsilon_{max} = \sqrt{2I \mu_0 cd}$

Here,

I = Current

$\mu_0$ = Permeability at free space

c = Light speed

d = Distance

Replacing,

$\epsilon_{max} = \sqrt{2(6.529*10^{-6})(4\pi*10^{-7})(3*10^{8})(60.0*10^{-2})}$

$\epsilon_{max} = 0.05434V$

Thus, the amplitude of induced emf by this signal between the ends of the receiving antenna is 0.0543V

6 0

4 years ago

How do your results from ray tracing compare to your results from using the thin-lens equation?

EastWind [94]

Answer:

20cm

Explanation:

A convex lens has a positive focal length and the object placed in front of it produce both virtual and real image (image distance can be negative or positive depending on the nature of the image).

According to the lens equation

$\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$ where;

f is the focal length of the lens

u is the object distance

v is the image distance

If the magnification is - 0.6

mag = v/u = -0.5

v = -0.5u

since v = 10cm

10 = -0.5u

u = -10/0.5

u =-20 cm

Substitute u = -20cm ( due to negative magnification)and v = 10cm into the lens formula to get the focal length f

$\frac{1}{f} = \frac{1}{u} + \frac{1}{v}\\\frac{1}{f} = \frac{1}{-20} + \frac{1}{10}\\\frac{1}{f} = \frac{1}{-20} + \frac{1}{10}\\\frac{1}{f} = \frac{-1+2}{20} \\\frac{1}{f} = \frac{1}{20} \\cross \ multiply\\f = 20\\f = 20 cm$

Hence the focal length of the convex lens is 20cm

7 0

3 years ago