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quester [9]
3 years ago
9

A pressure cooker contains water and steam in equilibrium at a pressure greater than atmo-spheric pressure. How does this greate

r pressure increase cooking speed
Physics
1 answer:
Amanda [17]3 years ago
6 0

Answer:

As the pressure inside the pressure cooker increases as it is heated, the temperature also increases.

Explanation:

A pressure cooker contains water and steam in equilibrium at a pressure greater than atmospheric pressure.

According to the ideal gas law, as the pressure increases, the temperature also increases.

As the pressure inside the pressure cooker increases as it is heated, the temperature also increases.

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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

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Answer:

false

Explanation:

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