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3 years ago

Which wavelength produces fluorescence? Why do you think this wavelength produces fluorescence while the other does not?

Physics

1 answer:

Maurinko [17]3 years ago

4 0

Answer:

Long wavelength

Explanation:

Wavelengths that corresponds to the bands of blue and red are strongly absorbed whereas the wavelengths that lie in the mid-range corresponds to green light that are absorbed weakly.

Fluorescence produced is always directed towards longer wavelengths of the spectra as compared to the corresponding spectra for absorption.

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Most irregular galaxies are thought to have formed _____

Musya8 [376]

Most irregular galaxies were have thought to have been formed from collisions from other galaxies <span />

6 0

4 years ago

Read 2 more answers

Four fixed point charges are at the corners of a square with sides of length L. Q1 is positive and at (OL) Q2 is positive and at

Ne4ueva [31]

Answer:

A) See Annex

B) Fq₁₂ = K * Q₁*Q₂ /16 [N] (repulsion force)

C) Fq₃₂ = K * Q₃*Q₂ /16 [N] (repulsion force)

D) Fq₄₂ = K * Q₄*Q₂ /32 [N] (attraction force)

E) Net force (its components)

Fnx = (2,59/64 )* K*Q² [N] in direction of original Fq₃₂

Fny =(2,59/64 )* K*Q² [N] in direction of original Fq₁₂

Explanation:

For calculation of d (diagonal of the square, we apply Pythagoras Theorem)

d² = L² + L² ⇒ d² = 2*L² ⇒ d = √2*L² ⇒ d= (√2 )*L

d = 4√2 units of length (we will assume meters, to work with MKS system of units)

B) Force of Q₁ exerts on charge Q₂

Fq₁₂ = K * Q₁*Q₂ /(L)² Fq₁₂ = K * Q₁*Q₂ /16 (repulsion force in the direction indicated in annex)

C) Force of Q₃ exerts on charge Q₂

Fq₃₂ = K * Q₃*Q₂ /(L)² Fq₃₂ = K * Q₃*Q₂ /16 (repulsion force in the direction indicated in annex)

D) Force of -Q₄ exerts on charge Q₂

Fq₄₂ = K * Q₄*Q₂ / (d)² Fq₄₂ = K * Q₄*Q₂ /32 (Attraction force in the direction indicated in annex)

E) Net force in the case all charges have the same magnitude Q (keeping the negative sign in Q₄)

Let´s take the force that Q₄ exerts on Q₂ and Q₂ = Q ( magnitude) and

Q₄ = -Q

Then the force is:

F₄₂ = K * Q*Q / 32 F₄₂ = K* Q²/32 [N]

We should get its components

F₄₂(x) = [K*Q²/32 ]* √2/2 and so is F₄₂(y) = [K*Q²/32 ]* √2/2

Note that this components have opposite direction than forces Fq₁₂ and

Fq₃₂ respectively, and that Fq₁₂ and Fq₃₂ are bigger than F₄₂(x) and F₄₂(y) respectively

In new conditions

Fq₁₂ = K * Q₁*Q₂ /16 becomes Fq₁₂ = K * Q²/ 16 [N] and

Fq₃₂ = K* Q₃*Q₂ /16 becomes Fq₃₂ = K* Q² /16 [N]

Note that Fq₁₂ and Fq₃₂ are bigger than F₄₂(x) and F₄₂(y) respectively

Then over x-axis we subtract Fq₃₂ - F₄₂(x) = Fnx

and over y-axis, we subtract Fq₁₂ - F₄₂(y) = Fny

And we get:

Fnx = K* Q² /16 - [K*Q²/32 ]* √2/2 ⇒ Fnx = K*Q² [1/16 - √2/64]

Fnx = (2,59/64 )* K*Q²

Fny has the same magnitude then

Fny =(2,59/64 )* K*Q²

The fact that Fq₁₂ and Fq₃₂ are bigger than F₄₂(x) and F₄₂(y) respectively, means that Fnx and Fny remains as repulsion forces

5 0

3 years ago

Is the net force on an object moving with uniform velocity zero or not zero?

NikAS [45]

Answer:

Zero

Explanation:

The net force acting on an object moving with uniform velocity is zero. This is an accordance with the Newton's first law of motion.

Newton's first law of motion states that a body will continue in its state of rest or uniform motion unless it is acted upon by an external force.

Since this body is moving with uniform motion, it is not accelerating. When a body is acceleration, the net force on it is not zero.

But this one is moving with uniform motion. The net force on the body is balanced and zero.

3 0

3 years ago

A small object with a 5.0-mC charge is accelerating horizontally on a friction-free surface at 0.0050 m/s2 due only to an electr

kolbaska11 [484]

Answer:

$0.002 N/C$

Explanation:

Parameters given:

Charge of object, q = 5 mC = $5 * 10^{-3} C$

Acceleration of object, a = $0.005 m/s^2$

Mass of object, m = 2.0 g

The Electric field exerts a particular force on the object, causing it to accelerate (Electrostatic force).

We know that Electrostatic force, F, is given in terms of Electric field, E, as:

F = qE

This means that the object exerts a force of -qE on the Electric force (Action with equal and opposite reaction).

The object also has a force, F, due to its acceleration a. This force is the product of its mass and acceleration. Mathematically:

F = ma

Equating the two forces of the object, we get:

-qE = ma

=> $E = \frac{-ma}{q}$

Solving for E, we have:

$E = \frac{-2 * 10^{-3} * 0.005}{5 * 10^{-3}} \\\\\\E = -0.002 N/C$

The magnitude will be:

$|E| = |-0.002| N/C = 0.002 N/C$

The electric field has a magnitude of 0.002 N/C.

4 0

4 years ago

When a spring is compressed 2.50 × 10^–2 meter

kompoz [17]

The working of a spring is given by:

$T_{el}= \frac{k\Delta x^2}{2}$

Entering the unknowns, we have:

$T_{el}=\frac{k\Delta x^2}{2} \\ 1.25*10^{-2}*2*J=k(2.5*10^{-2}m)^2 \\ k= \frac{2.5*10^{-2}*J}{(2.5*10^{-2}m)^2} \\ k=40* \frac{J}{m^2} \\ k=40 \frac{N*m}{m^2} \\ \boxed {k=40* \frac{N}{m} }$

If you notice any mistake in my english, please let me know, because i am not native.

4 0

3 years ago