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stiks02 [169]
3 years ago
6

If you wish to observe features that are around the size of atoms, say 5.5 × 10^-10 m, with electromagnetic radiation, the radia

tion must have a wavelength of about the size of the atom itself.
Required:

a. What is its frequency?

b. What type of electromagnetic radiation might this be?
Physics
1 answer:
Tanya [424]3 years ago
3 0

Answer:

a) 5.5×10^17 Hz

b) visible light

Explanation:

Since the wavelength of the electromagnetic radiation must be about the size of the about itself, this implies that;

λ= 5.5 × 10^-10 m

Since;

c= λ f and c= 3×10^8 ms-1

f= c/λ

f= 3×10^8/5.5 × 10^-10

f= 5.5×10^17 Hz

The electromagnetic wave is visible light

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Why is pseudoscience bad?
USPshnik [31]

Answer:

It is quite difficult to picture a pseudoscientist—really picture him or her over the course of a day, a year, or a whole career. What kind or research does he or she actually do, what differentiates him or her from a carpenter, or a historian, or a working scientist? In short, what do such people think they are up to?

… it is a significant point for reflection that all individuals who have been called “pseudoscientists” have considered themselves to be “scientists”, with no prefix.

The answer might surprise you. When they find time after the obligation of supporting themselves, they read papers in specific areas, propose theories, gather data, write articles, and, maybe, publish them. What they imagine they are doing is, in a word, “science”. They might be wrong about that—many of us hold incorrect judgments about the true nature of our activities—but surely it is a significant point for reflection that all individuals who have been called “pseudoscientists” have considered themselves to be “scientists”, with no prefix.

What is pseudoscience?

“Pseudoscience” is a bad category for analysis. It exists entirely as a negative attribution that scientists and non‐scientists hurl at others but never apply to themselves. Not only do they apply the term exclusively as a discrediting slur, they do so inconsistently. Over the past two‐and‐a‐quarter centuries since the term popped into the Western European languages, a great number of disparate doctrines have been categorized as sharing a core quality—pseudoscientificity, if you will—when in fact they do not. It is based on this diversity that I refer to such beliefs and theories as “fringe” rather than as “pseudo”: Their defining characteristic is the distance from the center of the mainstream scientific consensus in whichever direction, not some essential property they share.

Scholars have by and large tended to ignore fringe science as regrettable sideshows to the main narrative of the history of science, but there is a good deal to be learned by applying the same tools of analysis that have been used to understand mainstream science. This is not, I stress, to imply that there is no difference between hollow‐Earth theories and geophysics; on the contrary, the differences are the point of the analysis. Focusing on the historical and conceptual relationship between the fringe and the core of the various sciences as that blurry border has fluctuated over the centuries provides powerful analytical leverage for understanding where contemporary anti‐science movements come from and how mainstream scientists might address them.

As soon as professionalization blossomed, tagging competing theories as pseudoscientific became an important tool for scientists to define what they understood science to be

The central claim of this essay is that the concept of “pseudoscience” was called into being as the shadow of professional science. Before science became a profession—with formalized training, credentialing, publishing venues, careers—the category of pseudoscience did not exist. As soon as professionalization blossomed, tagging competing theories as pseudoscientific became an important tool for scientists to define what they understood science to be. In fact, despite many decades of strenuous effort by philosophers and historians, a precise definition of “science” remains elusive. It should be noted however that the absence of such definitional clarity has not seriously inhibited the ability of scientists to deepen our understanding of nature tremendously.

Explanation:

8 0
2 years ago
I have a voltage source of 12V but a light that only burns at 5V. The lamp works on 18 mA. Calculate the resistance that you EXT
ratelena [41]

Answer:

The resistance that will provide this potential drop is 388.89 ohms.

Explanation:

Given;

Voltage source, E = 12 V

Voltage rating of the lamp, V = 5 V

Current through the lamp, I = 18 mA

Extra voltage or potential drop, IR =  E- V  

                                                    IR = 12 V - 5 V = 7 V

The resistance that will provide this potential drop (7 V) is calculated as follows:

IR = V

R = \frac{V}{I} = \frac{7 \ V}{18 \times 10^{-3} A} \ = 388.89 \ ohms

Therefore, the resistance that will provide this potential drop is 388.89 ohms.

7 0
3 years ago
Ryan is driving home from work and notices a deer leaping onto the road about 25 m in front of his car. He immediately applies t
Anvisha [2.4K]

Answer:

mu = 0.56

Explanation:

The friction force is calculated by taking into account the deceleration of the car in 25m. This can be calculated by using the following formula:

v^2=v_0^2+2ax\\

v: final speed = 0m/s (the car stops)

v_o: initial speed in the interval of interest = 60km/h

    = 60(1000m)/(3600s) = 16.66m/s

x: distance = 25m

BY doing a the subject of the formula and replace the values of v, v_o and x you obtain:

a=\frac{v^2-v_o^2}{2x}=\frac{0m^2/s^2-(16.66m/s)^2}{2(25m)}=-5.55\frac{m}{s^2}

with this value of a you calculate the friction force that makes this deceleration over the car. By using the Newton second's Law you obtain:

F_f=ma=(1490kg)(5.55m/s^2)=8271.15N

Furthermore, you use the relation between the friction force and the friction coefficient:

F_f= \mu N=\mu mg\\\\\mu=\frac{F_f}{mg}=\frac{8271.15N}{(1490kg)(9.8m/s^2)}=0.56

hence, the friction coefficient is 0.56

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2 years ago
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3 years ago
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slavikrds [6]

Answer:

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