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Gala2k [10]
3 years ago
8

Jack was sitting, reading a book when all of a sudden he felt he was "losing his mind." His heart beat faster and he began to sw

eat and tremble. His behavior mostly resembles __________ disorder
Physics
1 answer:
Tatiana [17]3 years ago
7 0

Answer:

panic anxiety

Explanation:

Panic anxiety occur unexpectedly, sometimes even when waking up from sleep characterized by its unexpectedness and debilitating, immobilizing intensity it is caused by Severe stress, such as the death of a loved one, divorce, or job loss can also trigger panic anxiety. Panic attacks can also be caused by medical conditions and other physical cause e.g sitting in a position for hours under an unfavorable condition doing a recurring activity.  Panic anxiety may occur as part of another disorder, such as panic disorder, social phobia, or depression. Regardless of the cause, panic attacks are treatable.

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Classify the chemical equations as being balanced or not balanced.
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A spring is originally 0.4m. When a 23N force is applied it stretches to a total of 0.7m. How far is the spring been extended? W
sergejj [24]

Recall the force applied to a spring and its extension/compression:

F = kΔx

Δx is the change in the spring's length (measured from its original length).

x = 0.7 m

x₀ = 0.4 m

Δx = x-x₀ = 0.7 m - 0.4 m = 0.3 m

F = 23 N


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8 0
3 years ago
Two equal charges with magnitude Q and Q experience a force of 12.3442 when held at a distance r. What is the force between two
andre [41]

Answer:

197.5072.

Explanation:

According to the Coulomb's law, the magnitude of the electrostatic force of interaction between two charges \rm q_1 and \rm q_2 which are separated by the distance \rm d is given by

\rm F = \dfrac{kq_1q_2}{d^2}.

<em>where,</em> k is the Coulomb's constant.

For the case, when,

  • \rm q_1 = Q.
  • \rm q_2 = Q.
  • \rm d=r.
  • \rm F=12.3442.

Then, using Coulomb's law,

\rm 12.3442 = \dfrac{kQQ}{r^2}=\dfrac{kQ^2}{r^2}\ \ \ \ .......\ (1).

For the case, when,

  • \rm q_1 = 2Q.
  • \rm q_2 = 2Q.
  • \rm d=\dfrac r2.

Then, using Coulomb's law, the new electric force between the charges is given by,

\rm F' = \dfrac{k(2Q)(2Q)}{\left (\dfrac r2\right )^2}\\=\dfrac{k\ 4Q^2}{\dfrac{r^2}{4}}\\=4\times 4 \times \dfrac{kQ^2}{r^2}\\=16\ \dfrac{kQ^2}{r^2}\\=16\times 12.3442\ \ \ \ \ \ \ \ (Using\ (1))\\=197.5072.

8 0
3 years ago
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