Physics: Visit this website and locate the element mercury, whose chemical symbol is hg. click on the square to read about the history, properties, and uses of mercury. name three uses for mercury.

Visit this website and locate the element mercury,

Ответ:

197224

08.01.2022

We simply asked to name three uses for mercury.
The most common and well-known use of mercury is the production of thermometers. It's property to stay liquid at room temperature makes it ideal for a temperature indicator. However, the use of mercury is thermometers has been phased out due to health hazards.
It is also used to form an amalgam which is the result of its combination with silver or gold. Mercury has been used to mine gold and silver. This application has also been phased out.
Today's use of mercury includes mercury-vapor lamps which are the bright lamps used in high-ways.

0,0(0 оценок)

Ответ:

Rodriguezi9492

25.06.2022

Physics

$q_1=+0.375\ {10}^{-9}$

Explanation:

Electrostatic Forces

The force exerted between two point charges q_1 and q_2 separated a distance d is given by Coulomb's formula

$\displaystyle F=\frac{k\ q_1\ q_2}{d^2}$

The forces are attractive if the charges have different signs and repulsive if they have equal signs.

The problem described in the question locates three point charges in a straight line. The charges have the values shown below

$\displaystyle q_3=+5\ 10^{-9}\ c$

$\displaystyle q_2=-3\ 10^{-9}\ c$

The distance between q_3 and q_2 is

$\displaystyle d_2=4cm=0.04\ m$

The distance between q_3 and q_1 is

$\displaystyle d_1=2cm=0.02\ m$

We must find the value of q_1 such that

$\displaystyle |F_3|=0$

Applying Coulomb's formula for q_1 is

$\displaystyle F_{13}=\frac{k\ q_1\ q_3}{d_1^2}$

Now for q_2

$\displaystyle F_{23}=\frac{k\ q_2\ q_3}{d_2^2}$

If the total force on q_3 is zero, both forces must be equal. Note that being q2 negative, the force on q3 is to the right. The force exerted by q1 must go to the left, thus q1 must be positive. Equating the forces we have:

$\displaystyle F_{13}=F_{23}$