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4 years ago

11

Which air molecules within our atmosphere are important in absorbing infrared radiation, and therefore keeping our atmosphere wa

rmer than it would be without them?

1 answer:

kozerog [31]4 years ago

6 0

Answer:

Ozone

Explanation:

The Ozone or green house gases prevents this Ultraviolent radiation from the sun.

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Can a moving object have gravitational energy?

Evgesh-ka [11]

Answer:

It is worth noting that the higher the gravitational energy of an object moving downwards, the lower the kinetic energy, and the lower the kinetic energy of an object moving upwards, the higher its gravitational energy.

Gravitational potential energy is acquired by an object when it has been moved against a gravitational field. For example, an object raised above the surface of the Earth will gain energy, which is released if the object is allowed to fall back to the ground.

5 0

4 years ago

Does a battery produce electric currents?

Leokris [45]

Answer:

A battery is a device that stores chemical energy and converts it to electrical energy. The chemical reactions in a battery involve the flow of electrons from one material (electrode) to another, through an external circuit. The flow of electrons provides an electric current that can be used to do work.

6 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

Two gliders move toward each other on a linear air track, which we assume is frictionless. Glider A has a mass of 0.50 kg, and g

Nataly_w [17]

Answer:

-0.4 m/s

-3.552 m/s

Explanation:

$m_1$ = Mass of first glider = 0.5 kg

$m_2$ = Mass of second glider = 0.3 kg

$u_1$ = Initial Velocity of first glider = 2 m/s

$u_2$ = Initial Velocity of second glider = -2 m/s

$v_1$ = Final Velocity of first glider

$v_2$ = Final Velocity of second glider = 2 m/s

As the linear momentum of the system is conserved we have

$m_1u_1+m_2u_2=m_1v_1+m_2v_2\\\Rightarrow v_1=\dfrac{m_1u_1+m_2u_2-m_2v_2}{m_1}\\\Rightarrow v_1=\dfrac{0.5\times 2+0.3\times (-2)-0.3\times 2}{0.5}\\\Rightarrow v_1=-0.4\ m/s$

The velocity of glider A is -0.4 m/s

$u_1$ = 0

$u_2$ = -5 m/s

$v_2$ = 0.92 m/s

$m_1u_1+m_2u_2=m_1v_1+m_2v_2\\\Rightarrow v_1=\dfrac{m_1u_1+m_2u_2-m_2v_2}{m_1}\\\Rightarrow v_1=\dfrac{0.5\times 0+0.3\times (-5)-0.3\times 0.92}{0.5}\\\Rightarrow v_1=-3.552\ m/s$

The velocity of glider A is -3.552 m/s

8 0

3 years ago

mm UHF . 1 i . MM p r gm. w Emmi 1 . mi u m w a if 1 H I . V P V .. V . 1 . r n . I I a.

allsm [11]

C.labratory
I hope this helps

7 0

4 years ago