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Lostsunrise [7]
3 years ago
11

What is self-esteem? An individual's desire to grow An individual's sense of self worth An individual's drive to fulfill biologi

cal needs An individual's knowledge about him or herself
Physics
2 answers:
mariarad [96]3 years ago
6 0
An individuals sense of self worth.
Eddi Din [679]3 years ago
3 0

Answer:

An individual's sense of self worth

Explanation:

Self-esteem is the quality that belongs to the individual who is satisfied with his or her identity, that is, a person who has confidence and values himself. In psychology, self-esteem is a subjective assessment that a given individual makes of himself. In this case, characteristics such as dignity, respect and trust are present in that person's personality.

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What do you know about tides?
Neporo4naja [7]

Tides are the rise and fall of sea levels caused by the combined effects of the gravitational forces exerted by the Moon and the Sun, and the rotation of the Earth. Tide tables can be used to find the predicted times and amplitude (or "tidal range") of tides at any given locale.

For further expiation please contact me at 678-987-2411. Disclaimer(This is not a real number)

8 0
3 years ago
Which of the following statements does not represent a benefit of peer relationships? A. Jason and Sean argued over whose turn i
Wewaii [24]
C does not represent a benefit, so that seems most logical.

A) they comprised. 
B) at least ones listening
D) nothings wring with that
3 0
3 years ago
Read 2 more answers
Three equal charge 1.8*10^-8 each are located at the corner of an equilateral triangle ABC side 10cm.calculate the electric pote
Arlecino [84]

Answer:

If all these three charges are positive with a magnitude of 1.8 \times 10^{-8}\; \rm C each, the electric potential at the midpoint of segment \rm AB would be approximately 8.3 \times 10^{3}\; \rm V.

Explanation:

Convert the unit of the length of each side of this triangle to meters: 10\; \rm cm = 0.10\; \rm m.

Distance between the midpoint of \rm AB and each of the three charges:

  • d({\rm A}) = 0.050\; \rm m.
  • d({\rm B}) = 0.050\; \rm m.
  • d({\rm C}) = \sqrt{3} \times (0.050\; \rm m).

Let k denote Coulomb's constant (k \approx 8.99 \times 10^{9}\; \rm N \cdot m^{2} \cdot C^{-2}.)

Electric potential due to the charge at \rm A: \displaystyle \frac{k\, q}{d({\rm A})}.

Electric potential due to the charge at \rm B: \displaystyle \frac{k\, q}{d({\rm B})}.

Electric potential due to the charge at \rm A: \displaystyle \frac{k\, q}{d({\rm C})}.

While forces are vectors, electric potentials are scalars. When more than one electric fields are superposed over one another, the resultant electric potential at some point would be the scalar sum of the electric potential at that position due to each of these fields.

Hence, the electric field at the midpoint of \rm AB due to all these three charges  would be:

\begin{aligned}& \frac{k\, q}{d({\rm A})} + \frac{k\, q}{d({\rm B})} + \frac{k\, q}{d({\rm C})} \\ &= k\, \left(\frac{q}{d({\rm A})} + \frac{q}{d({\rm B})} + \frac{q}{d({\rm C})}\right) \\ &\approx 8.99 \times 10^{9}\; \rm N \cdot m^{2} \cdot C^{-2} \\ & \quad \quad \times \left(\frac{1.8 \times 10^{-8} \; \rm C}{0.050\; \rm m} + \frac{1.8 \times 10^{-8} \; \rm C}{0.050\; \rm m} + \frac{1.8 \times 10^{-8} \; \rm C}{\sqrt{3} \times (0.050\; \rm m)}\right) \\ &\approx 8.3 \times 10^{3}\; \rm V\end{aligned}.

4 0
3 years ago
Please can someone give a clear explanantion, <br><br> no extra links thanks
Tems11 [23]

Answer:

the extension recorded by the student would be smaller than the actual extension of the spring

3 0
2 years ago
4.5 billion km, the average separation between the sun and Neptune (report answer in hours). How long does it take light to trav
Liula [17]

Answer:

t = 4.17 hours

Explanation:

given,

The distance between Sun and Neptune, d = 4.5 billion Km

                                                                         = 4.5 x 10⁹ Km

                                                                          = 4.5 x 10¹¹ m

The velocity of light, c = 3 x 10⁸ m/s

The velocity is always equal to displacement by the time.

                                           <em>V = d / t    m/s</em>

∴                                           t = d / V

                                               = 4.5 x 10¹¹ m / 3 x 10⁸ m/s

                                               = 15,000 s

                                               = 4.17 h

Hence, the time taken by the light rays to reach the Neptune is, t = 4.17 h

4 0
3 years ago
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