All categories

4 years ago

What happens when a substance absorbs light?

1 answer:

7 0

<span>One of three things the photon passes through unscathed, the photon gets absorbed by the electron of an atom and stays put, or the photon gets absorbed and a second generation photon is generated by that electron</span>

You might be interested in

The Hubble Space Telescope has an aperture of 2.4 m and focuses visible light (400-700 nm). The Arecibo radio telescope in Puert

stealth61 [152]

Answer:

$y_{hubble} = 77\ \ m$

$y_{aceribo} = 1.1*10^6 \ \ m$

Explanation:

what is the smallest crater that each of these telescopes could resolve on our moon?

For moon ;

s = 3.8 × 10 ⁸ m

y = 1.22 λs/D

where;

λ = 400 nm = 400× 10 ⁻⁹

D = 2.4 m

The smallest crater for the hubble space is calculated as follows:

$y_{hubble} = 1.22*400*10^{-9}*3.8*10^8/2.4$

$y_{hubble} = 77\ \ m$

For Aceribo ;

y = 1.22 λs/D

where :

λ = 75 cm = 0.75 m

D = 305 m

$y_{acerbo} = 1.22*0.75 *3.8*10^8/305$

$y_{aceribo} = 1.1*10^6 \ \ m$

5 0

3 years ago

After you pick up a spare, your bowling ball rolls without slipping back toward the ball rack with a linear speed of 2.85 m/s. T

motikmotik

Answer:

V' = 0.84 m/s

Explanation:

given,

Linear speed of the ball, v = 2.85 m/s

rise of the ball, h = 0.53 m

Linear speed of the ball, v' = ?

rotation kinetic energy of the ball

$KE_r = \dfrac{1}{2}I\omega^2$

I of the moment of inertia of the sphere

$I = \dfrac{2}{5}MR^2$

v = R ω

using conservation of energy

$KE_r = \dfrac{1}{2}( \dfrac{2}{5}MR^2)(\dfrac{V}{R})2$

$KE_r = \dfrac{1}{5}MV^2$

Applying conservation of energy

Initial Linear KE + Initial roational KE = Final Linear KE + Final roational KE + Potential energy

$\dfrac{1}{2}MV^2 + \dfrac{1}{5}MV^2 = \dfrac{1}{2}MV'^2 + \dfrac{1}{5}MV'^2 + M g h$

$0.7 V^2 = 0.7 V'^2 + gh$

$0.7\times 2.85^2 = 0.7\times V'^2 +9.8\times 0.53$

V'² = 0.7025

V' = 0.84 m/s

the linear speed of the ball at the top of ramp is equal to 0.84 m/s

6 0

3 years ago

For a 50 kV anode voltage, what is the maximum photon energy of the x-ray radiation?

V125BC [204]

Answer:

The energy of photon, $E=8\times 10^{-15}\ J$

Explanation:

It is given that,

Voltage of anode, $V=50\ kV=50\times 10^3\ V=5\times 10^4\ V$

We need to find the maximum energy of the photon of the x- ray radiation. The energy required to raise an electron through one volt is called electron volt.

$E=eV$

e is charge of electron

$E=1.6\times 10^{-19}\times 5\times 10^4$

$E=8\times 10^{-15}\ J$

So, the maximum energy of the x- ray radiation is $8\times 10^{-15}\ J$ . Hence, this is the required solution.

6 0

3 years ago

You place a 3.0-m-long board symmetrically across a 0.5-m-wide chair to seat three physics students at a party at your house. If

kakasveta [241]

Answer:

A) for leftmost point the coordinate is -0.28m that means it should be 0.28m towards the right.

B) for rightmost case the coordinate is 0.28m which is where komila should sit.

Explanation:

Detailed calculation and explanation is shown in the image below

5 0

3 years ago

An electric field of 1.27 kV/m and a magnetic field of 0.490 T act on a moving electron to produce no net force. If the fields a

lesantik [10]

Answer:

v = 2591.83 m/s

Explanation:

Given that,

The electric field is 1.27 kV/m and the magnetic field is 0.49 T. We need to find the electron's speed if the fields are perpendicular to each other. The magnetic force is balanced by the electric force such that,

$qE=qvB\\\\v=\dfrac{E}{B}\\\\v=\dfrac{1.27\times 10^3}{0.49}\\\\v=2591.83\ m/s$

So, the speed of the electron is 2591.83 m/s.

8 0

3 years ago