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Novosadov [1.4K]
3 years ago
7

Olivia wants to find out whether a substance will fluoresce. She says she should put it in a microwave oven. Do you agree with h

er? Why or why not?
Physics
2 answers:
andreyandreev [35.5K]3 years ago
4 0
Disagree.
Fluoresce objects will only glow when put under actual Ultraviolet light. This is due to the molecules becoming excited by the ultraviolet radiation.


Microwaves give micro-waves that are present in another spectrum of wave length and will not be able to fluoresce the molecules. If it’s not “ultra violet “.... it’s not going to glow.
konstantin123 [22]3 years ago
4 0

Answer:

Sample Response: I disagree with her because many substances show fluorescence under ultraviolet light, not microwaves.

Explanation:

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Sloan [31]
Is there more information ?
5 0
3 years ago
A police car parked on the side of the highway emits a 1200 Hz sound that bounces off a vehicle farther down the highway and ret
emmasim [6.3K]

f' = frequency observed by the police car after sound reflected from the vehicle and comes back to police car = 1250 Hz

f = frequency emitted by the police car  = 1200 Hz

V = speed of sound = 340 m/s

v = speed of vehicle = ?

frequency observed by the police car is given as

f' = f (V + v)/(V - v)

inserting the values in the above equation

1250 = 1200 (340 + v)/(340 - v)

v = 6.9 m/s

4 0
4 years ago
Read 2 more answers
What property do the following elements have in common? Li, C, and F A) They are poor conductors of electricity. B) Each element
Aloiza [94]

Answer: Option (D) is the correct answer.

Explanation:

The given elements Li, C and F are all second period elements. So, when we move from left to right across a period then there occurs increase in number of valence electrons as there occurs increase in total number of electrons.

So, it means more electrons are added to the same energy level.

Thus, we can conclude that a property of valence electrons for each element is located in the same energy level is common in the given elements.

7 0
3 years ago
Read 2 more answers
You are given two unknown point charges, Q1, and Q2. At a point on the line joining them, one third of the way from Q1 to Q2 the
podryga [215]
The formula that is applicable here is E = kQ/r^2 in which the energy of attraction is proportional to the charges and inversely proportional to the square of the distance. In this case, 
kQ1/(r1)^2 = kQ2/(r2)^2  r1=l/3, r2=2l/3solve Q1/Q2
kQ1/(l/3)^2 = kQ2/(2l/3)^2  kQ1/(l^2/9) = kQ2/(4l^2/9)Q1/Q2 =   1/4
3 0
3 years ago
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A light, rigid rod is 55.8 cm long. Its top end is pivoted on a frictionless horizontal axle. The rod hangs straight down at res
VMariaS [17]

To solve this problem we will apply the principle of conservation of energy. For this purpose, potential energy is equivalent to kinetic energy, and this clearly depends on the position of the body. In turn, we also note that the height traveled is twice that of the rigid rod, therefore applying these concepts we will have

KE = PE

\frac{1}{2} mv^2 = mgh

v = \sqrt{2gh}

v = \sqrt{2(9.8)(2(55.8*10^{-2}))}

v = 4.67m/s

Therefore the minimum speed at the bottom is required to make the ball go over the top of the circle is 4.67m/s

4 0
3 years ago
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