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Novosadov [1.4K]
2 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]2 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]2 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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You and a friend each carry identical boxes from the first floor of a building to a room located on the second floor, farther do
melamori03 [73]

Answer:

Work done in both the cases will be same

Explanation:

As we know that the work done against gravity is given as

W = F_g .d

here we know that gravitational force is a conservative force and the work done against gravitational force is independent of the path

So here the work done by person to move the object between two different heights will be independent of the path they choose

So for the first person and second person will be same in both the cases because the height through which the boxes are transferred will be same in both the cases

8 0
3 years ago
Diagram of an atom with labels
lina2011 [118]

See this. I hope you find your answer

6 0
3 years ago
Please help me .... ​
miss Akunina [59]

Explanation:

Orbital speed= 2pi x radius / time period

=2pi x 1.5x10^11 / 365.25

=2.58x10^9m/day

8 0
3 years ago
The accepted value for the density of iron is 7.87 g/cm3. a student records the mass of a 20.00 cm3 block of iron as 153.8 grams
dangina [55]
The density of an object can be calculated using the formula Density = Mass/Volume.

Experimental Density:

Density = 153.8g / 20.00 cm^3
Density = 7.69g/cm^3

Percent error equation:

% Error = | Theoretical Value - Experimental Value|/Theoretical Value * 100
% Error = | 7.87g/cm^3 - 7.69g/cm^3|/7.87g/cm^3 * 100
% Error = 2.29%

Therefore a is the correct answer.
6 0
2 years ago
A compact disc (CD) stores music in a coded pattern of tiny pits 10−7m deep. The pits are arranged in a track that spirals outwa
andreev551 [17]

(a) 50 rad/s

The angular speed of the CD is related to the linear speed by:

\omega=\frac{v}{r}

where

\omega is the angular speed

v is the linear speed

r is the distance from the centre of the CD

When scanning the innermost part of the track, we have

v = 1.25 m/s

r = 25.0 mm = 0.025 m

Therefore, the angular speed is

\omega=\frac{1.25 m/s}{0.025 m}=50 rad/s

(b) 21.6 rad/s

As in part a, the angular speed of the CD is given by

\omega=\frac{v}{r}

When scanning the outermost part of the track, we have

v = 1.25 m/s

r = 58.0 mm = 0.058 m

Therefore, the angular speed is

\omega=\frac{1.25 m/s}{0.058 m}=21.6 rad/s

(c) 5550 m

The maximum playing time of the CD is

t =74.0 min \cdot 60 s/min = 4,440 s

And we know that the linear speed of the track is

v = 1.25 m/s

If the track were stretched out in a straight line, then we would have a uniform motion, therefore the total length of the track would be:

d=vt=(1.25 m/s)(4,440 s)=5,550 m

(d) -6.4\cdot 10^{-3} rad/s^2

The angular acceleration of the CD is given by

\alpha = \frac{\omega_f - \omega_i}{t}

where

\omega_f = 21.6 rad/s is the final angular speed (when the CD is scanned at the outermost part)

\omega_i = 50.0 rad/s is the initial angular speed (when the CD is scanned at the innermost part)

t=4440 s is the time elapsed

Substituting into the equation, we find

\alpha=\frac{21.6 rad/s-50.0 rad/s}{4440 s}=-6.4\cdot 10^{-3} rad/s^2

5 0
3 years ago
Read 2 more answers
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