Which of the following represents the greatest energy transition from a higher energy level to a lower one?

1 answer:

8 0

You need to use Planck's law:
E = h·υ = (h·c)/λ

Without making all the calculations, a fraction is bigger than another when the denominator is smaller. Therefore you need to find the smallest wavelength (λ) which is 450nm.

You could also be helped by colors: in order of decreasing energy, you have blue - green - yellow - red.

In any case, the correct answer is a).

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A cyclist is travelling eastwards at a velocity of 40 ms and rain is falling vertically at a speed of 10 ms. Find the velocity o

Genrish500 [490]

41.23

v(B,E)=40 GOING EAST

V(R,E)=10 COMING DOWN

then 41.23

6 0

3 years ago

Is kevin Showing work help him please

7nadin3 [17]

Answer:

I believe it is not a chemical change for the components are not changing.

This excerpt from thoughtco.com states that for it to be a chemical change, the atoms would be rearranged.

<em>In other words, a chemical change is a chemical reaction involving the rearrangement of atoms. While a physical change can often be reversed, a chemically change typically cannot be, except through more chemical reactions. When a chemical change occurs, there is also a change in the energy of the system.</em>

So, in this case, they are not being rearranged, just combined to form one big fruit salad.

4 0

3 years ago

Kevin has a mass of 87 kg and is skating with in-line skates. He sees his 22-kg younger brother up ahead standing on the sidewal

neonofarm [45]

Answer:

$V_ k = 3 m/s$

Explanation:

Using the conservation of the linear momentum P:

$P_i = P_f$

so:

$M_kV_k = M_sV_s$

where $M_k$ is the mass of kevin, $V_k$ is the velocity of kevin before he grabs her brother, $M_s$ is the velocity of her brother and him after grabs her brother and $V_s$ is the velocity of both after the collition.

So, replacing the values and solving for Vk, we get:

$(87 kg)V_k = (87 kg+22 kg)(2.4m/s)$

$V_ k = 3 m/s$

8 0

3 years ago

Scientists experimenting with two charged objects of 2.56 x 10-6 C and 3.34 x 10-7 C displayed a repulsion of 2.26 x 10-3 N. How

Pachacha [2.7K]

Answer:

The objects were 1.8m apart.

Explanation:

We will start stating the Coulomb's Law. It says that:

$F_e=\frac{Kq_1q_2}{r^{2}}$

Where F_e is the electric force between the objects, q_1 and q_2 are the magnitude of the charge of the objects, r is the distance between them and K is the Coulomb's constant ( $K=8.9*10^{9} \frac{Nm^{2} }{C^{2} }$ in vacuum). Solving for the distance r we have: