Answer:
The theory of relativity usually encompasses two interrelated theories by Albert Einstein: special relativity and general relativity, proposed and published in 1905 and 1915, respectively. Special relativity applies to all physical phenomena in the absence of gravity. General relativity explains the law of gravitation and its relation to other forces of nature.It applies to the cosmological and astrophysical realm, including astronomy.
The theory transformed theoretical physics and astronomy during the 20th century, superseding a 200-year-old theory of mechanics created primarily by Isaac Newton. It introduced concepts including spacetime as a unified entity of space and time, relativity of simultaneity, kinematic and gravitational time dilation, and length contraction. In the field of physics, relativity improved the science of elementary particles and their fundamental interactions, along with ushering in the nuclear age. With relativity, cosmology and astrophysics predicted extraordinary astronomical phenomena such as neutron stars, black holes, and gravitational waves
Answer:
A fire hose must be able to shoot water to the top of a building 35.0 m tall ... Water enters this hose at a steady rate of 0.500 m3/s and shoots out of a round nozzle. ... I know that Flow rate=0.500 m3/s=A*V. I know the pressure needed to ... The first equation has no potential while the second has no kinetic.
Explanation:
Answer:
62.8 μC
Explanation:
Here is the complete question
The volume electric charge density of a solid sphere is given by the following equation: ρ = (0.2 mC/m⁵)r²The variable r denotes the distance from the center of the sphere, in spherical coordinates. What is the net electric charge (in μC) of the sphere if the radius of the sphere is 0.5 m?
Solution
The total charge on the sphere Q = ∫∫∫ρdV where ρ = volume charge density = 0.2r² and dV = volume element in spherical coordinates = r²sinθdθdrdΦ
So, Q = ∫∫∫ρdV
Q = ∫∫∫ρr²sinθdθdrdΦ
Q = ∫∫∫(0.2r²)r²sinθdθdrdΦ
Q = ∫∫∫0.2r⁴sinθdθdrdΦ
We integrate from r = 0 to r = 0.5 m, θ = 0 to π and Φ = 0 to 2π
So, Q = ∫∫∫0.2r⁴sinθdθdrdΦ
Q = ∫∫∫0.2r⁴[∫sinθdθ]drdΦ
Q = ∫∫0.2r⁴[-cosθ]drdΦ
Q = ∫∫0.2r⁴-[cosπ - cos0]drdΦ
Q = ∫∫∫0.2r⁴-[-1 - 1]drdΦ
Q = ∫∫0.2r⁴-[- 2]drdΦ
Q = ∫∫0.2r⁴(2)drdΦ
Q = ∫∫0.4r⁴drdΦ
Q = ∫0.4r⁴dr∫dΦ
Q = ∫0.4r⁴dr[Φ]
Q = ∫0.4r⁴dr[2π - 0]
Q = ∫0.4r⁴dr[2π]
Q = ∫0.8πr⁴dr
Q = 0.8π∫r⁴dr
Q = 0.8π[r⁵/5]
Q = 0.8π[(0.5 m)⁵/5 - (0 m)⁵/5]
Q = 0.8π[0.125 m⁵/5 - 0 m⁵/5]
Q = 0.8π[0.025 m⁵ - 0 m⁵]
Q = 0.8π[0.025 m⁵]
Q = (0.02π mC/m⁵) m⁵
Q = 0.0628 mC
Q = 0.0628 × 10⁻³ C
Q = 62.8 × 10⁻³ × 10⁻³ C
Q = 62.8 × 10⁻⁶ C
Q = 62.8 μC
Solve the following word problems.
1. The ratio of red marbles and blue marbles that Carlo has is 8: 3. When he
exchanged 35 red marbles for 20 blue marbles from his brother, he was left with
equal number of red and blue marbles.
How many red and blue marbles did he have at the beginning
How many red and blue marbles did he have now
5Newtons or 5N
Ten newtons minus five Newton’s is 5 Newton’s