Province. Is what is located away from the capital.
Answer:
a) 0.568 kg
b) 474 kg/m³
Explanation:
Given:
Inner radius = 8.82 cm = 0.0882 m
Outer radius = 9.91 cm = 0.0991 m
Density of the liquid = 948.00 Kg/m³
a) The volume of the sphere = 
or
volume of sphere = 0.0012 m³
also, volume of half sphere = 
or
volume of half sphere = 
or
Volume of half sphere =0.0006 m³
Now, from the Archimedes principle
Mass of the sphere = Weight of the volume of object submerged
or
Mass of the sphere = 0.0006× 948.00 = 0.568 kg
b) Now, density = 
or
Density = 
or
Density = 474 kg/m³
Lose electrons because they have a negative charge.
Answer:
In the clarification portion elsewhere here, the definition of the concern is mentioned.
Explanation:
So like optical telescopes capture light waves, introduce it to concentrate, enhance it, as well as make it usable through different instruments via study, so radio telescopes accumulate weak signal light waves, introduce that one to focus, enhance it, as well as make this information available during research. To research naturally produced radio illumination from stars, galaxies, dark matter, as well as other natural phenomena, we utilize telescopes.
Optical telescopes detect space-borne visible light. There are some drawbacks of optical telescopes mostly on the surface:
- Mostly at night would they have been seen.
- Unless the weather gets cloudy, bad, or gloomy, they shouldn't be seen.
Although radio telescopes monitor space-coming radio waves. Those other telescopes, when they are already typically very massive as well as costly, have such an improvement surrounded by optical telescopes. They should be included in poor weather and, when they travel through the surrounding air, the radio waves aren't obscured by clouds. Throughout the afternoon and also some at night, radio telescopes are sometimes used.
Given Information:
Current = I = 20 A
Diameter = d = 0.205 cm = 0.00205 m
Length of wire = L = 1 m
Required Information:
Energy produced = P = ?
Answer:
P = 2.03 J/s
Explanation:
We know that power required in a wire is
P = I²R
and R = ρL/A
Where ρ is the resistivity of the copper wire 1.68x10⁻⁸ Ω.m
L is the length of the wire and A is the area of the cross-section and is given by
A = πr²
A = π(d/2)²
A = π(0.00205/2)²
A = 3.3x10⁻⁶ m²
R = ρL/A
R = 1.68x10⁻⁸*(1)/3.3x10⁻⁶
R = 5.09x10⁻³ Ω
P = I²R
P = (20)²*5.09x10⁻³
P = 2.03 Watts or P = 2.03 J/s
Therefore, 2.03 J/s of energy is produced in 1.00 m of 12-gauge copper wire carrying a current of 20 A
