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solniwko [45]
3 years ago
9

The first confirmed detections of extrasolar planets occurred in ____________. The first confirmed detections of extrasolar plan

ets occurred in ____________. the mid-17th century the mid-20th century the 1990s 2009
Physics
1 answer:
nirvana33 [79]3 years ago
4 0

Answer:

1992 (Early 1990s)

Explanation:

First of all, i would like to define an extrasolar planet as a planet that orbits a start that is not our own.

The first confirmed detections of extrasolar planets occured in the early 1990s (specifically 1992, some say 1995). The name of the first extrasolar planet is widely believed to be called Dimidium or 51 Pegasi b.  

Extrasolar were searched by monitoring stars for slight dimming that might occur as unseen planets pass in front of them.

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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}.

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