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andre [41]
3 years ago
15

Determine the force exerted by two particles that have a charge of +0.35 Coulombs each and are 1 meter apart

Physics
2 answers:
konstantin123 [22]3 years ago
6 0

The electric force between two charges is:

F = (9 x 10⁹) Q₁ Q₂ / D²

F is the force, in Newtons

Q₁ and Q₂ are the two charges, in Coulombs

D is the distance between them, in meters

For these two particles:

F = (9 x 10⁹) (0.35) (0.35) / (1)²

F = (9 x 0.35 x 0.35 x 10⁹) / (1)

<em>F = 1.10 x 10⁹  Newtons</em>

Thatsa lotta force . . . like <em>124 thousand tons</em> !

The reason it's so big is because the charges in this question are so big ... 0.35 Coulombs each.  1 Coulomb is a huge amount of charge.

Each of the particles feels the same force, pushing it away from the other particle.  (The electric force between two charges is always the same in both directions, just like the gravitational force between two masses.)

Viktor [21]3 years ago
3 0

Acc. to columb's law;

  • F = k × Q1Q2/r²

Clearly, from the question you have:

  • Charge on both particles = +0.35 columbs
  • Distance between them = 1 meter = 1 × 10-¹⁰ Å

Now just put them up in the formula:

F = k × Q1Q2/r²

F = 9 × 10⁹ × 0.35 × 0.35/1 × 10-¹⁰

F = 9 × 10⁹ × 35 × 35/ 10-¹⁰ × 100 × 100

F = 9 × 10⁹ × 35 × 35/ 10-¹⁰ × 10⁴

F = 9 × 10⁹ × 35 × 35/ 10-⁶

F = 9 × 10⁹+⁶ × 1225

F = 11025 × 10¹⁵

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Answer:

Explanation:

Remark

This is a second class lever. It is much more efficient than the fishing pole problem. All distances are measured from the pivot in these kinds of questions.

Givens

d1 = 1.5

d2 = ?

m1 = 50 kg

m2 = 30 kg

The lighter child will have to sit further away from the pivot to make the two conditions equal.

Formula

d1*m1 = d2*m2

1.5*50 = d2 * 30

75 = 30 * d2

75/30 = d2

d2 = 2.5

Remark

Notice that the distance is longer for the lighter child. The fact that these are masses and not forces does not matter, but you should take note of it. There is a difference between masses and forces. See the fishing pole problem.

Answer to the multiple Choice question. No motion on this kind of problem means equal moments. The answer is D

Problem 2

1) The wheels are further apart making B more stable. The wider the distance the wheels are apart, the harder it would be to tip the concrete mixer over

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3 years ago
1. A listener stands 20.0 m from a speaker that pumps out music with a power output of 100.0 W.
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(1.a) The surface area being vibrated by the time the sound reaches the listener is 5,026.55 m².

(1.b) The intensity of the sound wave as it reaches the person listening is 0.02 W/m².

(1.c) The relative intensity of the sound as heard by the listener is 103 dB.

(2.a) The speed of sound if the air temperature is 15⁰C is 340.3 m/s.

(2.b) The frequency of the sound heard by the suspect is 614.3 Hz.

<h3>Surface area being vibrated</h3>

The surface area being vibrated by the time the sound reaches the listener is calculated as follows;

A = 4πr²

A = 4π x (20)²

A = 5,026.55 m²

<h3>Intensity of the sound</h3>

The intensity of the sound is calculated as follows;

I = P/A

I = (100) / (5,026.55)

I = 0.02 W/m²

<h3>Relative intensity of the sound</h3>

B = 10log(\frac{I}{I_0} )\\\\B = 10 \times log(\frac{0.02}{10^{-12}} )\\\\B = 103 \ dB

<h3>Speed of sound at the given temperature</h3>

v= 331.3\sqrt{1 + \frac{T}{273} } \\\\v = 331.3\sqrt{1 + \frac{15}{273} } \\\\v = 340.3 \ m/s

<h3>Frequency of the sound</h3>

The frequency of the sound heard is determined by applying Doppler effect.

f_o = f_s(\frac{v \pm v_0}{v \pm v_s} )

where;

  • -v₀ is velocity of the observer moving away from the source
  • -vs is the velocity of the source moving towards the observer
  • fs is the source frequency
  • fo is the observed frequency
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f_0 = f_s(\frac{v-v_0}{v- v_s} )

f_0 = 512(\frac{340.3 - 10}{340.3 - 65} )\\\\f_0 = 614.3 \ Hz

Learn more about intensity of sound here: brainly.com/question/17062836

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Answer;

C. The brightness of each bulb would remain the same even though the total resistance of the circuit would decrease.

Explanation;

-If light bulbs are connected in parallel to a voltage source, the brightness of the individual bulbs remains more-or-less constant as more and more bulbs are added to the circuit.

-The current increases as more bulbs are added to the circuit and the overall resistance decreases. In addition, if one bulb is removed from the circuit the other bulbs do not go out. Each bulb is independently linked to the voltage source

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Answer:

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… 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.

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Explanation:

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