Which of the following statements about the electromagnetic spectrum is true?

What do we call the days when the sun is directly overhead the equator?

NikAS [45]

Answer:

Equinox

Explanation:

It is an astronomical phenomenon that is repeated twice a year, in summer and winter, and that is caused because the axis on which our planet rotates is slightly inclined with respect to the plane of translation, this is its trajectory around the Sun.

For this reason there are two equinoxes a year, a phenomenon completely opposite to the solstice, but also marks the beginning of the spring and autumn seasons.

The equinoxes occur around March 20 or 21 and September 22 or 23, and it is the days when the Sun is exactly over the equator, which makes day and night more or less the same duration. The word ‘equinox’ comes from Latin and means precisely ‘same night’. In those two moments the part of the Earth closest to the Sun is the equator.

8 0

3 years ago

The magnetising current in a transformer is rich in ______________ harmonic?

sukhopar [10]

The magnetizing current in a transformer is rich in 3rd harmonic. This is because harmonics are AC voltages and currents with frequencies that are generally higher.

5 0

3 years ago

An electron is trapped between two large parallel charged plates of a capacitive system. The plates are separated by a distance

Lorico [155]

Answer:

The electron will get at about 0.388 cm (about 4 mm) from the negative plate before stopping.

Explanation:

Recall that the Electric field is constant inside the parallel plates, and therefore the acceleration the electron feels is constant everywhere inside the parallel plates, so we can examine its motion using kinematics of a constantly accelerated particle. This constant acceleration is (based on Newton's 2nd Law:

$F=m\,a\\q\,E=m\,a\\a=\frac{q\,E}{m}$

and since the electric field E in between parallel plates separated a distance d and under a potential difference $\Delta V$ , is given by:

$E=\frac{\Delta\,V}{d}$

then :

$a=\frac{q\,\Delta V}{m\,d}$

We want to find when the particle reaches velocity zero via kinematics:

$v=v_0-a\,t\\0=v_0-a\,t\\t=v_0/a$

We replace this time (t) in the kinematic equation for the particle displacement:

$\Delta y=v_0\,(t)-\frac{1}{2} a\,t^2\\\Delta y=v_0\,(\frac{v_0}{a} )-\frac{a}{2} (\frac{v_0}{a} )^2\\\Delta y=\frac{1}{2} \frac{v_0^2}{a}$

Replacing the values with the information given, converting the distance d into meters (0.01 m), using $\Delta V=100\,V$ , and the electron's kinetic energy:

$\frac{1}{2} \,m\,v_0^2= (11.2)\,\, 1.6\,\,10^{-19}\,\,J$

we get:

$\Delta\,y= \frac{1}{2} v_0^2\,\frac{m (0.01)}{q\,(100)} =11.2 (1.6\,\,10^{-19})\,\frac{0.01}{(1.6\,\,10^{-19})\,(100)}=\frac{11.2}{10000} \,meters=0.00112\,\,meters$ Therefore, since the electron was initially at 0.5 cm (0.005 m) from the negative plate, the closest it gets to this plate is: