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Bas_tet [7]
3 years ago
5

Consider the 65 N light fixture supported as in the figure. Find the tension in the supporting wires.

Physics
1 answer:
ASHA 777 [7]3 years ago
3 0

By using Lami's theorem formula, the tension in the supporting wires is 48.6 Newtons

TENSION

  • Tension is also a force having Newton as S.I unit.
  • The tension in the wire will be the same.

This question can be solved by using either vector diagram or by using  Lami's theorem.

The sum of two given angles  = 42 + 42 = 84 degrees

The third angle = 180 - 84 = 96 degrees.

Below is the Lami's theorem formula

\frac{T}{sin\alpha } = \frac{T}{sin\beta } = \frac{W}{sinY}

Where

\alpha  = \beta = 42 + 90 = 132 degrees

Y = 96 degrees

W = 65 N

By using the formula, we have

\frac{T}{sin\alpha } =  \frac{W}{sinY}

T/sin 132 = 65/sin96

Cross multiply

T = 0.743 x 65.57

T = 48.56 N

Therefore, the tension in the supporting wires is 48.6 Newtons approximately.

Learn more about Tension here: brainly.com/question/24994188

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If a 15g irregular shaped object is submerged in a graduated cylinder of water, the level rises from 10 ml to 25 ml, what is the
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Answer:

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Explanation:

6 0
3 years ago
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
3 years ago
What is the kinetic energy of a bicycle with a mass of 14 kg traveling at a<br> velocity of 3.0 m/s?
Sav [38]

46.6666 that is the mass number

Explanation:

14 divided 3.0

5 0
3 years ago
A mixture of helium and oxygen is used in scuba diving tanks to help prevent ""the bends"". 46 L helium and 12 L oxygen are comb
marin [14]

Answer

given,

For helium

Volume,V = 46 L

Pressure,P = 1 atm

Temperature,T = 25°C  =  273 +25 = 298 K

R=0.0821 L . atm /mole.K

n₁ = ?

number of moles

we know

P V = n R T

n_1 =\dfrac{46 \times 1}{0.0821\times 298}

  n₁ = 1.89 moles

For oxygen

Volume,V = 12 L

Pressure,P = 1 atm

Temperature,T = 25°C  =  273 +25 = 298 K

R=0.0821 L . atm /mole.K

n₂ = ?

number of moles

we know

P V = n R T

n_2 =\dfrac{12 \times 1}{0.0821\times 298}

  n₂ = 0.49 moles

Total volume of tank = 5 L

temperature of tank = 298 K

Partial pressure of helium

P_1=\dfrac{n_1 R T}{V}

P_1=\dfrac{1.89\times 0.0821\times 298}{5}

     P₁ = 9.25 atm

Partial pressure of oxygen

P_2=\dfrac{n_2 R T}{V}

P_2=\dfrac{0.49\times 0.0821\times 298}{5}

    P₂ = 2.39 atm

total pressure

    P = P₁ + P₂

    P = 9.25 + 2.39

    P = 11.64 atm

8 0
3 years ago
The intensity of the radiation from the Sun measured on Earth is 1360 W/m2 and frequency is f = 60 MHz. The distance between the
Mama L [17]

Answer: (a) power output = 3.85×10²⁶W

(b). There is no relative change in power as it is independent from frequency

(c). 590 W/m²

Explanation:

given Radius between earth and sun to be = 1.50 × 10¹¹m

Intensity of the radiation from the sun measured on earth to be = 1360 W/m²

Frequency = 60 MHz

(a). surface area A of the sun on earth is = 4πR²

substituting value of R;

A = 4π(.50 × 10¹¹)² = 2.863 10²³×m²

A = 2.863 10²³×m²

now to get the power output of the sun we have;

<em>P </em>sun = <em>I </em><em>sun-earth </em><em>A </em><em>sun-earth</em>

where A = 2.863 10²³×m², and <em>I </em> is 1360 W/m²

<em>P </em>sun =  2.863 10²³ × 1360

<em>P </em>sun = 3.85×10²⁶W

(c). surface area A of the sun on mars is = 4πR²

now we substitute value of 2.28 ×10¹¹ for R sun-mars, we have

A sun-mars = 4π(2.28× 10¹¹)²

A sun-mars = 6.53 × 10²³m²

now to calculate the intensity of the sun;

<em>I </em><em>sun-mars = </em><em>P </em>sun / A sun-mars

where <em>P </em>sun = 3.85×10²⁶W and A sun-mars = 6.53 × 10²³m²

<em>I </em><em>sun-mars =  </em>3.85×10²⁶W / 6.53 × 10²³m²

<em>I </em><em>sun-mars = </em>589.6 ≈ 590 W/m²

<em>I </em><em>sun-mars = </em>590 W/m²

6 0
3 years ago
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