What unit is used for electromagnetic fields.

barxatty [35]

Answer:

The energies corresponding to each of the allowed orbitals are called energy levels.

Explanation:

A scientist known as Niels Bohr put forward that electrons in an atom covers some permitted orbitals with a specific energy. In other words, the energy of an electron in an atom is not continuous, but 'quantized.' The energies corresponding to each of the allowed orbitals are called energy levels.

$E = -\frac{E_0}{n^2} \\where \\E_0 = 13.6 eV (1 eV = 1.602\times 10^{-19}Joules)\\and\ n = 1,2,3...$

8 0

3 years ago

A 0.15-kg ball is thrown into the air and rises to a height of 20.0 m. How much kinetic energy did the ball initially have?

zzz [600]

IF the toss was straight upward, then the kinetic energy it got
from the toss is the gravitational potential energy it has at the top,
where it stops rising and starts falling.

Potential energy = (mass) x   (gravity) x (height)

   = (0.15 kg) x (9.8 m/s²) x (20 m)

   = 29.4 kg-m²/s² = 29.4 joules .

7 0

3 years ago

Read 2 more answers

A woman is walking at 4 m/s. She is accelerating at a rate of 1 m/s2. To find out what her velocity is after 3 seconds, what els

liq [111]

Answer:

distance

Explanation:

8 0

3 years ago

Read 2 more answers

In the Copper Plating Solution lab, you placed copper (Cu) pennies in vinegar. Vinegar is a weak acid with hydrogen ions (H+). W

PolarNik [594]

Bubbles came off from the copper pennies are hydrogen gas. There many free hydrogen ions in an acid like vinegar. And because of the chemical reaction between the copper and the vinegar, many hydrogen ions joined together to form hydrogen gas.

6 0

3 years ago

Suppose you want to determine the resistance of a resistor that is nominally 100 . You should be able to apply 10 V across the r

Butoxors [25]

Answer:

a) For y = 102 mA, R = 98.039 ohms

For y = 97 mA, R = 103.09 ohms

b) Check explanatios for b

Explanation:

Applied voltage, V = 10 V

For the first measurement, current $y_{1} = 102 mA = 0.102 A$

According to ohm's law, V = IR

R = V/I

Here, $I = y_{1}$

$R = \frac{V}{y_{1} } \\R = \frac{10}{0.102} \\R = 98.039 ohms$

For the second measurement, current $y_{2} = 97 mA = 0.097 A$

$R = \frac{V}{y_{2} }$

$R = \frac{10}{0.097} \\R = 103 .09 ohms$

b) $y = \left[\begin{array}{ccc}y_{1} &y_{2} \end{array}\right] ^{T}$

$y = \left[\begin{array}{ccc}y_{1} \\y_{2} \end{array}\right]$

$y = \left[\begin{array}{ccc}102*10^{-3} \\97*10^{-3} \end{array}\right]$

A linear equation is of the form y = Gx

The nominal value of the resistance = 100 ohms

$x = \left[\begin{array}{ccc}100\end{array}\right]$

$\left[\begin{array}{ccc}102*10^{-3} \\97*10^{-3} \end{array}\right] = \left[\begin{array}{ccc}G_{1} \\G_{2} \end{array}\right] \left[\begin{array}{ccc}100\end{array}\right]\\\left[\begin{array}{ccc}G_{1} \\G_{2} \end{array}\right] = \left[\begin{array}{ccc}102*10^{-5} \\97*10^{-5} \end{array}\right]$

3 0

3 years ago