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3 years ago

How can electric energy be converted to heat energy?

1 answer:

5 0

To do that, you must pass electric current through a substance
that electrons have to spend energy to pass through.
The substance will be one that gets warm and dissipates heat
when electric current flows through it.
We'll say that the substance has "resistance", which we can measure.
The amount of heat that appears when current flows through it
will be (current²)·(resistance).

A few examples of things used for that purpose:

-- resistors
-- burners on electric stoves
-- coils of resistor-wire in a toaster
-- aquarium heater
-- electric clothes iron
-- electric coffee pot
-- blow-dryer
-- electric hair-curling iron
-- skinny tungsten wire in a light-bulb .

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PLZ HELP FAST!!!!!!!!!Which characteristics describe a maritime polar air mass? A.cold and humid air B.cold and dry air C.warm a

svet-max [94.6K]

"Maritime" air . . . . . always high relative humidity

"Polar" air . . . . . always cold

"Maritime polar" air . . . . . choice-A

6 0

3 years ago

As microwave light travels through a liquid, it moves at a speed of 2.2 x 108 m/s. If the frequency of this light wave is 1.1 x

Alexandra [31]

λ = 2 m.

The easiest way to solve this problem is using the equation of frecuency of a wave f = v/λ, where v is the velocity of the wave, and λ is the wavelength.

To calculate the wavelength of a microwave light travels through a liquid, it moves at a speed of 2.2 x 10⁸ m/s. If the frecuency of the light wave is 1.1 x 10⁸ Hz, we have to clear λ from the equation f = v/λ:

f = v/λ -------> λ = v/f

λ = 2.2 x 10⁸ m/s / 1.1 x 10⁸ Hz

λ = 2 m (wavelength of the microwave)

7 0

3 years ago

Mr. Phillips' car iso parked on a steep hill

Wewaii [24]

? There is no question

7 0

3 years ago

Name two objects that have a high density.

Sonbull [250]

Answer:

Iron and stone

Explanation:

4 0

3 years ago

X rays of wavelength 0.0169 nm are directed in the positive direction of an x axis onto a target containing loosely bound electr

mamaluj [8]

Answer:

a) 4.04*10^-12m

b) 0.0209nm

c) 0.253MeV

Explanation:

The formula for Compton's scattering is given by:

$\Delta \lambda=\lambda_f-\lambda_i=\frac{h}{m_oc}(1-cos\theta)$

where h is the Planck's constant, m is the mass of the electron and c is the speed of light.

a) by replacing in the formula you obtain the Compton shift:

$\Delta \lambda=\frac{6.62*10^{-34}Js}{(9.1*10^{-31}kg)(3*10^8m/s)}(1-cos132\°)=4.04*10^{-12}m$

b) The change in photon energy is given by:

$\Delta E=E_f-E_i=h\frac{c}{\lambda_f}-h\frac{c}{\lambda_i}=hc(\frac{1}{\lambda_f}-\frac{1}{\lambda_i})\\\\\lambda_f=4.04*10^{-12}m +\lambda_i=4.04*10^{-12}m+(0.0169*10^{-9}m)=2.09*10^{-11}m=0.0209nm$

c) The electron Compton wavelength is 2.43 × 10-12 m. Hence you can use the Broglie's relation to compute the momentum of the electron and then the kinetic energy.

$P=\frac{h}{\lambda_e}=\frac{6.62*10^{-34}Js}{2.43*10^{-12}m}=2.72*10^{-22}kgm\\$

$E_e=\frac{p^2}{2m_e}=\frac{(2.72*10^{-22}kgm)^2}{2(9.1*10^{-31}kg)}=4.06*10^{-14}J\\\\1J=6.242*10^{18}eV\\\\E_e=4.06*10^{-14}(6.242*10^{18}eV)=0.253MeV$

5 0

3 years ago