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klio [65]
3 years ago
5

Two light bulbs are equally bright, but one has a power of 6 W while the other has a power of 10 W. In one hour the higher power

bulb uses J more energy than the lower power bulb.
Physics
1 answer:
joja [24]3 years ago
3 0
If this is a t/f This is true.
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Sound cannot travel through a vacuum. Explain why.
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3 0
3 years ago
Calculate the energy of a photon having a wavelength in thefollowing ranges.(a) microwave, with λ = 50.00 cmeV(b) visible, with
IgorLugansk [536]

(a) 2.5\cdot 10^{-6}eV

The energy of a photon is given by:

E=\frac{hc}{\lambda}

where

h=6.63\cdot 10^{-34}Js is the Planck constant

c=3\cdot 10^8 m/s is the speed of light

\lambda is the wavelength

For the microwave photon,

\lambda=50.00 cm = 0.50 m

So the energy is

E=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{0.50 m}=4.0\cdot 10^{-25} J

And converting into electronvolts,

E=\frac{4.0\cdot 10^{-25}J}{1.6\cdot 10^{-19} J/eV}=2.5\cdot 10^{-6}eV

(b) 2.5 eV

For the energy of the photon, we can use the same formula:

E=\frac{hc}{\lambda}

For the visible light photon,

\lambda=500 nm = 5 \cdot 10^{-7}m

So the energy is

E=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{5\cdot 10^{-7} m}=4.0\cdot 10^{-19} J

And converting into electronvolts,

E=\frac{4.0\cdot 10^{-19}J}{1.6\cdot 10^{-19} J/eV}=2.5 eV

(c) 2500 eV

For the energy of the photon, we can use the same formula:

E=\frac{hc}{\lambda}

For the x-ray photon,

\lambda=0.5 nm = 5 \cdot 10^{-10}m

So the energy is

E=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{5\cdot 10^{-10} m}=4.0\cdot 10^{-16} J

And converting into electronvolts,

E=\frac{4.0\cdot 10^{-16}J}{1.6\cdot 10^{-19} J/eV}=2500 eV

6 0
3 years ago
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