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ss7ja [257]
3 years ago
5

What happens as photons of visible light are absorbed by dye molecules?

Physics
1 answer:
rjkz [21]3 years ago
3 0
Certain electrons in the dye molecule move to a higher energy level, with the difference in energy between the lower and higher energy levels being the same as the energy of the absorbed photons.
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Multiply 0.00032 cm by 4.02 cm and express the answer in scientific notation
FromTheMoon [43]
0.00032cm*4.02=1.2864 × 10^-3 in scientific notation.
4 0
3 years ago
Read 2 more answers
Cynthia pushes a ball with a horizontal speed of 6.50 m/s off a bench with a height of
Igoryamba
The answer is a because if you look really close
7 0
3 years ago
The volumes of the block is 10,500 mL and density is 8.52 g/mL. Find the block's mass Don't forget units and a space
Schach [20]

Answer:

<h2>89,460 g</h2>

Explanation:

The mass of a substance when given the density and volume can be found by using the formula

mass = Density × volume

From the question we have

mass = 8.52 × 10,500

We have the final answer as

<h3>89,460 g</h3>

Hope this helps you

3 0
3 years ago
Determine the kinetic of 600 kg roller coaster car that is moving with the speed of 35.2 m/s.
antiseptic1488 [7]
The answer is 10,560 Joules or 1.1*10^4


Explanation:

Step 1: Calculate
The equation for Kinetic Energy is

Kinetic energy=.5 times Mass times Velocity²
KE=.5*m*v²

so we plug in our numbers
KE=.5*600*35.2²

This works out to be 10,560 Joules or 1.1*10^4
5 0
3 years ago
An object is located 50 cm from a converging lens having a focal length of 15 cm. Which of the following is true regarding the i
crimeas [40]

Answer:

It is real, inverted, and smaller than the object.

Explanation:

Let's start by using the lens equation to find the location of the image:

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}

where we have:

q = ? is the distance of the image from the lens

f = 15 cm is the focal length (positive for a converging lens)

p = 50 cm is the distance of the object from the lens

Solving the equation for q, we find

\frac{1}{q}=\frac{1}{15 cm}-\frac{1}{50 cm}=0.047 cm^{-1}

q=\frac{1}{0.047 cm^{-1}}=+21.3 cm

The sign of q is positive, so the image is real.

Now let's also write the magnification equation:

h_i = - h_o \frac{q}{p}

where  

h_i, h_o are the size of the image and of the object

By substituting p = 50 cm and q = 21.3 cm, we find

h_i = - h_o \frac{21.3 cm}{50 cm}=-0.43 h_o

So we notice that:

|h_i| < |h_o| : this means that the image is smaller than the object

h_i < 0 : this means that the image is inverted

so, the correct option is:

It is real, inverted, and smaller than the object.

7 0
3 years ago
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