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3 years ago

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A smoke detector is an electronic fire-protection device that automatically senses the presence of smoke as a key indicator of f

ire and creates a loud sound as a warning. In 3-5 sentences, describe what is happening inside a photoelectric smoke detector as soon as smoke enters it

1 answer:

Colt1911 [192]3 years ago

7 0

Answer:

Photoelectric-type alarms aim a light source into a sensing chamber at an angle away from the sensor. Smoke enters the chamber, reflecting light onto the light sensor; triggering the alarm.

Explanation:

nfpa.org is the website with theanswer

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A photon of wavelength 7.33 pm scatters at an angle of 157° from an initially stationary, unbound electron. What is the de Brogl

Ann [662]

Answer:

4.63 p.m.

Explanation:

The problem given here can be solved by the Compton effect which is expressed as

$\lambda^{'}-\lambda=\frac{h}{m_e c}(1-cos\theta)$

here, $\lambda$ is the initial photon wavelength, $\lambda^{'}$ is the scattered photon wavelength, h is he Planck's constant, $m_e$ is the free electron mass, c is the velocity of light, $\theta$ is the angle of scattering.

Given that, the scattering angle is, $\theta=157^{\circ}$

Putting the respective values, we get

$\lambda^{'}-\lambda=\frac{6.626\times 10^{-34} }{9.11\times 10^{-31}\times 3\times 10^{8} } (1-cos157^\circ ) m\\\lambda^{'}-\lambda=2.42\times 10^{-12} (1-cos157^\circ ) m\\\lambda^{'}-\lambda=2.42(1-cos157^\circ ) p.m.$

Therfore,

$\lambda^{'}-\lambda=4.64 p.m.$

Here, the photon's incident wavelength is $\lamda=7.33pm$

So,

$\lambda^{'}=7.33+4.64=11.97 p.m$

From the conservation of momentum,

$\vec{P_\lambda}=\vec{P_{\lambda^{'}}}+\vec{P_e}$

here, $\vec{P_\lambda}$ is the initial photon momentum, $\vec{P_{\lambda^{'}}}$ is the final photon momentum and $\vec{P_e}$ is the scattered electron momentum.

Expanding the vector sum, we get

$P^2_{e}=P^2_{\lambda}+P^2_{\lambda^{'}}-2P_\lambda P_{\lambda^{'}}cos\theta$

Now expressing the momentum in terms of De-Broglie wavelength

$P=h/\lambda$ and putting it in the above equation we get,

$\lambda_{e}=\frac{\lambda \lambda^{'}}{\sqrt{\lambda^{2}+\lambda^{2}_{'}-2\lambda \lambda^{'} cos\theta}}$

Therfore,

$\lambda_{e}=\frac{7.33\times 11.97}{\sqrt{7.33^{2}+11.97^{2}-2\times 7.33\times 11.97\times cos157^\circ }} p.m.\\\lambda_{e}=\frac{87.7401}{18.935} = 4.63 p.m.$

This is the de Broglie wavelength of the electron after scattering.

8 0

3 years ago

A tank is filled with an ideal gas at 400 K and pressure of 1.00 atm.

bekas [8.4K]

To find the temperature it is necessary to use the expression and concepts related to the ideal gas law.

Mathematically it can be defined as

$PV=nRT$

Where

P = Pressure

V = Volume

n = Number of moles

R = Gas constant

T = Temperature

When the number of moles and volume is constant then the expression can be written as

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$

Or in practical terms for this exercise depending on the final temperature:

$T_2 = \frac{P_2T_1}{P_1}$

Our values are given as

$T_1 = 400K\\P_1 = 1atm\\P_2 = 2atm$

Replacing

$T_2 = \frac{(2)(400)}{1}\\T_2 = 800K$

Therefore the final temperature of the gas is 800K

6 0

3 years ago

Work is the product of force and an object's

Neporo4naja [7]

Answer:

C. displacement

Explanation:

7 0

3 years ago

Read 2 more answers

A bowling (mass = 7.2 kg, radius = 0.11 m) and a billiard ball (mass = 0.38 kg, radius = 0.028 m) may each be treated as uniform

cestrela7 [59]

Hope this helps you!

7 0

3 years ago

A comet is in an elliptical orbit around the sun. its closest approach to the sun is a distance of 4.5 1010 m (inside the orbit

barxatty [35]

r1 = 5*10^10 m , r2 = 6*10^12 m

v1 = 9*10^4 m/s

From conservation of energy

K1 +U1 = K2 +U2

0.5mv1^2 - GMm/r1 = 0.5mv2^2 - GMm/r2

0.5v1^2 - GM/r1 = 0.5v2^2 - GM/r2

M is mass of sun = 1.98*10^30 kg

G = 6.67*10^-11 N.m^2/kg^2

0.5*(9*10^4)^2 - (6.67*10^-11*1.98*10^30/(5*10^10)) = 0.5v2^2 - (6.67*10^-11*1.98*10^30/(6*10^12))

v2 = 5.35*10^4 m/s

4 0

4 years ago