A distance of 2.00 mm separates two objects of equal mass. If the gravitational force between them is 0.0104 N, find the mass of

Answer:

a) 16675.75 Kg/m³ b) 77.6%

Explanation:

The weight of the crown is 60 N, with gold's density at 19300 Kg/m³, lead's at 11340 kg/m³, water's at 1000kg/m³, and gravitational acceleration at 9.8 m/s².

The upthrust acting on the crown equals the weight in air minus its weight in water: 60 - 56.4 results in 3.6 N.

This leads to the mass of water displaced being 3.6 / 9.8, as weight equals mass times gravity.

The mass of displaced water is 0.367 Kg.

The density of water relates mass to volume as: 1000 = 0.367 / volume.

Cross-multiplying helps us determine the volume:

The crown’s volume becomes 0.367 / 1000 = 0.000367 m³ since it displaces an equal volume of water per Archimedes' principle.

Let V1 denote the gold volume and V2 the lead volume.

Total volume for the crown becomes V1 + V2.

Likewise, using the relationship of densities:

Density of gold translates to mass of gold over V1 and lead density translates to mass of lead over V2.

Thus, 19300 = mass of gold in the crown / V1 and 11340 = mass of lead in the crown / V2.

Combining them gives us: 19300 V1 = mass of gold and 11340 V2 = mass of lead.

Adding together leads to: 19300 V1 + 11340 V2 = weight of the crown / 9.8.

So, 19300 V1 + 11340 V2 = 6.12.

From V1 + V2 = 0.000367, we can express V1 in relation to V2.

V1 = 0.000367 - V2.

Substituting this into the mass equation results in:

19300 (0.000367 - V2) + 11340 V2 = 6.12.

Expanding gives:

7.083 - 19300 V2 + 11340 V2 = 6.12.

Reorganizing yields:

-7960 V2 = 6.12 - 7.083.

So, -7960 V2 = -0.963.

This leads to V2 = -0.963 / -7960 = 0.000121 (the lead volume in the crown).

Substituting V2 back into the total volume equation gives:

V1 + 0.000121 = 0.000367 m³

Thus, V1 = 0.000367 - 0.000121 = 0.000246 m³ (the gold volume in the crown).

Which leads to the mass of gold in the crown = 19300 × 0.000246 = 4.748 kg.

The mass of lead equals 11340 × 0.000121 = 1.372 kg.

The average density for the crown calculates as (mass of gold + mass of lead) / total volume = 6.12 / 0.000367 = 16675.75 kg/m³.

b) The percentage of gold by weight computes to mass of gold / total mass × 100 = approximately 77.6%.

Response:

Magnitude of the electrostatic force acting on the +32 µC charge,

Clarification:

Let q₁ = +32 µC, located at x₁ = 0

q₂ = +20 µC, positioned at x₂ = 40 cm = 0.4 m

q₃ = -60 µC, placed at x₃ = 60 cm = 0.6 m

Define the force magnitude on the +32 µC charge from the +20 µC charge as F₁ (the force on q₁ due to q₂).

Define the force magnitude on the +32 µC charge from the -60 µC charge as F₂ (the force on q₁ due to q₃).

The resultant electrostatic force on the 32 µC charge is

Answer:

10000 V

0.00225988700565 m²

Explanation:

E = Electric field =

d = Distance = 2.5 mm

Q = Charge = 80 nC

= Permittivity of free space =

The potential difference is calculated as

The potential difference across the plates amounts to 10000 V

Area is determined by

The area of each plate measures 0.00225988700565 m²

Capacitance is determined by

The capacitance is