In 6 days, the temperature dropped 18 degrees. How many degrees per day did the temperature drop?

Q1 is located at the origin, Q2 is located at x = 2.50 cm and Q3 is located at x = 3.50 cm. Q1 has a charge of +4.92μC and Q3 ha

Inessa05 [86]

Answer:

$+1.11\mu C$

Explanation:

A charge located at a point will experience a zero electrostatic force if the resultant electric field on it due to any other charge(s) is zero.

$Q_1$ is located at the origin. The net force on it will only be zero if the resultant electric field intensity due to $Q_2$ and $Q_3$ at the origin is equal to zero. Therefore we can perform this solution without necessarily needing the value of $Q_1$ .

Let the electric field intensity due to $Q_2$ be + $E_2$ and that due to $Q_3$ be - $E_3$ since the charge is negative. Hence at the origin;

$+E_2-E_3=0..................(1)$

From equation (1) above, we obtain the following;

$E_2=E_3.................(2)$

From Coulomb's law the following relationship holds;

$+E_2=\frac{kQ_2}{r_2^2}\\$

$-E_3=\frac{kQ_3}{r_3^2}$

where $r_2$ is the distance of $Q_2$ from the origin, $r_3$ is the distance of $Q_3$ from the origin and k is the electrostatic constant.

It therefore means that from equation (2) we can write the following;

$\frac{kQ_2}{r_2^2}=\frac{kQ_3}{r_3^2}.................(3)$

k can cancel out from both side of equation (3), so that we finally obtain the following;

$\frac{Q_2}{r_2^2}=\frac{Q_3}{r_3^2}................(4)$

Given;

$Q_2=?\\r_2=2.5cm=0.025m\\Q_3=-2.18\mu C=-2.18* 10^{-6}C\\r_3=3.5cm=0.035m$

Substituting these values into equation (4); we obtain the following;

$\frac{Q_2}{0.025^2}=\frac{2.18*10^{-6}}{0.035^2}\\\\hence;\\\\Q_2=\frac{0.025^2*2.18*10^{-6}}{0.035^2}\\$

$Q_2=\frac{0.00136*10^{-6}}{0.00123}=1.11*10^{-6}C\\\\Q_3=+1.11\mu C$

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Easy chemical formula question (9th grade)

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Silver and Silver (II) are different because silver (II) has more than 1 electric charge. It has 2! lol

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Explanation:

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If stellar parallax can be measured to a precision of about 0.01 arcsec using telescopes on the Earth to observe stars, to what

marin [14]

Answer:

It corresponds to a distance of 100 parsecs away from Earth.

Explanation:

The angle due to the change in position of a nearby object against the background stars it is known as parallax.

It is defined in a analytic way as it follows:

$\tan{p} = \frac{1AU}{d}$

Where d is the distance to the star.

$p('') = \frac{1}{d}$ (1)

Equation (1) can be rewritten in terms of d:

$d(pc) = \frac{1}{p('')}$ (2)

Equation (2) represents the distance in a unit known as parsec (pc).

The parallax angle can be used to find out the distance by means of triangulation. Making a triangle between the nearby star, the Sun and the Earth (as is shown in the image below), knowing that the distance between the Earth and the Sun (150000000 Km), is defined as 1 astronomical unit (1AU).

For the case of ( $p('') = 0.01$ ):

$d(pc) = \frac{1}{0.01}$

$d(pc) = 100$

Hence, it corresponds to a distance of 100 parsecs away from Earth.

<em>Summary:</em>

Notice how a small parallax angle means that the object is farther away.

Key terms:

Parsec: Parallax of arc second

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