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slavikrds [6]
3 years ago
13

PLEASE    A point P is placed between two charges, A and B. If the electric field experienced by point P due to charge A is 8.7

× 106 newtons/coulomb and due to charge B is 5.5 × 106 newtons/coulomb, calculate the net electric field at point P.
A>3.2 × 106 newtons/coulomb
B>5.5 × 106 newtons/coulomb
C>8.7 × 106 newtons/coulomb
D>1.4 × 107 newtons/coulomb
Physics
2 answers:
Alik [6]3 years ago
4 0
In my opinion or what I get from your question:A>3.2 × 106 newtons/coulomb.
gavmur [86]3 years ago
4 0

its A , your welcome :)

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g initial angular velocity of 39.1 rad/s. It starts to slow down uniformly and comes to rest, making 76.8 revolutions during the
MrRa [10]

Answer:

Approximately -1.58\; \rm rad \cdot s^{-2}.

Explanation:

This question suggests that the rotation of this object slows down "uniformly". Therefore, the angular acceleration of this object should be constant and smaller than zero.

This question does not provide any information about the time required for the rotation of this object to come to a stop. In linear motions with a constant acceleration, there's an SUVAT equation that does not involve time:

v^2 - u^2 = 2\, a\, x,

where

  • v is the final velocity of the moving object,
  • u is the initial velocity of the moving object,
  • a is the (linear) acceleration of the moving object, and
  • x is the (linear) displacement of the object while its velocity changed from u to v.

The angular analogue of that equation will be:

(\omega(\text{final}))^2 - (\omega(\text{initial}))^2 = 2\, \alpha\, \theta, where

  • \omega(\text{final}) and \omega(\text{initial}) are the initial and final angular velocity of the rotating object,
  • \alpha is the angular acceleration of the moving object, and
  • \theta is the angular displacement of the object while its angular velocity changed from \omega(\text{initial}) to \omega(\text{final}).

For this object:

  • \omega(\text{final}) = 0\; \rm rad\cdot s^{-1}, whereas
  • \omega(\text{initial}) = 39.1\; \rm rad\cdot s^{-1}.

The question is asking for an angular acceleration with the unit \rm rad \cdot s^{-1}. However, the angular displacement from the question is described with the number of revolutions. Convert that to radians:

\begin{aligned}\theta &= 76.8\; \rm \text{revolution} \\ &= 76.8\;\text{revolution} \times 2\pi\; \rm rad \cdot \text{revolution}^{-1} \\ &= 153.6\pi\; \rm rad\end{aligned}.

Rearrange the equation (\omega(\text{final}))^2 - (\omega(\text{initial}))^2 = 2\, \alpha\, \theta and solve for \alpha:

\begin{aligned}\alpha &= \frac{(\omega(\text{final}))^2 - (\omega(\text{initial}))^2}{2\, \theta} \\ &= \frac{-\left(39.1\; \rm rad \cdot s^{-1}\right)^2}{2\times 153.6\pi\; \rm rad} \approx -1.58\; \rm rad \cdot s^{-1}\end{aligned}.

7 0
3 years ago
You can analogize the photoelectric energy to
Anon25 [30]

Answer:

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5 0
4 years ago
How many leptons are in ⁷₃Li?
shutvik [7]

-- Electrons are leptons.  There are <em>three</em> electrons in each neutral Lithium atom.

The last two parts of the question are absurd.

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5 0
3 years ago
A person is standing on a level floor. His head, upper torso, arms, and hands together weigh 458 N and have a center of gravity
Step2247 [10]

Answer:

the location of the center of gravity for the entire body is 1.08 m

Explanation:

Given the data in the question;

w1 = 458 N, y1 = 1.34 m

w2 = 120 N, y2 = 0.766 m

w3 = 89.8 N, y2 = 0.204 m

The location arrangement of the body part is vertical, locate the overall centre of gravity by simply replacing the horizontal position x by the vertical position y as measured relative to the floor.

so,

Y_{centre of gravity} = (w1y1 + w2y2 + w3y3 ) / ( w1 + w2 + w3 )

so we substitute in our values

Y_{centre of gravity} = (458×1.34 + 120×0.766 + 89.8×0.204 ) / ( 458 + 120 + 89.8 )

Y_{centre of gravity} = 723.9592 / 667.8

Y_{centre of gravity} = 1.08 m

Therefore, the location of the center of gravity for the entire body is 1.08 m

5 0
3 years ago
The following equation is an example of decay?<br><br> 232/90 TH---4/2 HE +228/88 RA?
DaniilM [7]

Answer:

Alpha decay

Explanation:

  • Alpha decay is one of the three major types of decays, others being, beta decay and gamma decay.
  • <em><u>When a radioactive isotope undergoes alpha decay it emits alpha particles. An alpha particle is equivalent to the nucleus of Helium atom.</u></em>
  • <em><u>Therefore, an atom undergoing decay, its atomic mass is decreased by 4 and its atomic number is decreased by 2. </u></em>
  • Thus, since 232/90 Th, has undergone alpha decay its mass number is reduced by 4 to 228 and its atomic number by 2 to 88, and becomes 228/88 Ra.
5 0
3 years ago
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