Answer: 0.5 seconds or 2.625 seconds
Explanation:
At t = 0, The ball is 4 ft above the ground.
The height of the football varies with time in the following way:
s(t) = -16 t² + 50 t + 4
we need to find the time in which the height would of the football would be 25 ft:
⇒25 = -16 t² + 50 t + 4
we need to solve the quadratic equation:
⇒ 16 t² - 50 t + 21 = 0
⇒ t = 0.5 s or 2.625 s
Therefore, at t = 0.5 s or 2.625 s, the football would be 25 ft above the ground.
Answer: option A. strong nuclear force.
Explanation:
The diagram shows the subatomic particles inside the nucelous: protons and neutrons.
As you know, the protons are positively charged partilces inside the nucleous.
Being those particles charged with the same kind of charge they experiment electrostatic repulsion. So, how do you explain that they can stand together in such small space as it is the nucleous?
The responsible of keeping the subatomic particles together is the so called strong nuclear force.
Strong nuclear force or simply strong force is one of the four fundamental interactions or forces: i) gravitational, ii) electromagnetic, iii) weak nuclear force, and iv) strong nuclear force.
Strong nuclear force is the strongest force of nature and acts only in short distances as those inside the nucleous and is responsible for both the atraction among quarks and the atraction among protons to bind them together inside the atomic nucleous.
I think the answer is photosynthis, when plants turn light into food and energy.
Answer:
There are two different types of crust: thin oceanic crust that underlies the ocean basins, and thicker continental crust that underlies the continents. These two different types of crust are made up of different types of rock.
Explanation:
There ya go !
:) hoping this helped ya out
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
