Polarizing windows, filters, etc. are often used to reduce the amount of light that enters the lens of a camera or into a room o

The correct choice is D!

Clarification:

Answer:

0.018 J

Explanation:

The work required to bring the charge from infinity to the point P is equal to the change in its electric potential energy. This can be expressed as

where

represents the charge's magnitude

and signifies the potential difference between point P and infinity.

After substituting into the formula, we arrive at

Answer:

The molar mass of the metal in grams per mole is calculated to be 8.87.

Explanation:

Initially, we can consider a sample of the compound weighing 100 g. This results in:

• 52.92% metal: 52.92 g M
• 47.80% oxygen: 47.80 g O

 By utilizing the molar mass of oxygen, which is 16 g / mol, we can determine the quantity of moles of oxygen in the sample via the rule of three:

moles of oxygen=2.9875

The formula for the metal oxide indicates that:

2 M⁺³ + 3 O²⁻ ⇒ M₂O₃

From the previous equation, it is evident that 3 oxygen ions are necessary to react with 2 metal ions. Hence:

Given 52.92 g of metal in the sample, the molar mass of the metal is:

molar mass≅ 8.87 g/mol

The molar mass of the metal in grams per mole is 8.87.

The value that most closely corresponds to this is Beryllium (Be), which has an atomic mass of 9.0122 g / mol.

Answer:

The current flowing through the tube is 0.601 A

Explanation:

Given data;

the diameter of the fluorescent tube is d = 3 cm

the incoming negative charge in the tube is -e = 3 x 10¹⁸ electrons/second

the outgoing positive charge equals +e = 0.75 x 10¹⁸ electrons/ second

The current within the fluorescent tube results from both positive and negative charges which contribute to maintaining electrical neutrality in the conductor (fluorescent tube).

Q = It

I = Q/t

where;

I signifies current in Amperes (A)

Q represents charge measured in Coulombs (C)

t denotes time which is in seconds (s)

1 electron (e) accounts for 1.602 x 10⁻¹⁹ C

Thus, 3 x 10¹⁸ e/s can be computed as

= (3 x 10¹⁸ e/s  x 1.602 x 10⁻¹⁹ C) / 1e

= 0.4806 C/s

This reflects the negative charge per second (Q/t) = 0.4806 C/s

Meanwhile, the positive charge per second amounts to

(0.75 x 10¹⁸ e/s  x 1.602 x 10⁻¹⁹ C) / 1e

Thus, the positive charge per second is equal to 0.12015 C/s

The total charge per second in the tube is then obtained by summing both charges: Q / t = (0.4806 C/s + 0.12015 C/s)

                                                                I = 0.601 A

Therefore, the current within the tube computes to 0.601 A