It uses electromagnetic radiation waves to enable military communications, navigation, radar, nonintrusive inspection of aircraft, and other equipment. Hope this helps.
In 16 times
KE= o.5 m times V squared
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
Answer: 20cm
Explanation:
Given the following data:
Magnification of image = - 1.50
Distance of image from the lens = 30.0cm
The Magnification of an image is the ratio of the image distance from the lens and the object distance from the lens.
Mathematically,
Magnification(M) = - (image distance (di) / object distance (do))
M = - (di/do)
-1.50 = - (30/do)
do = 30 / 1.5
do = 20 cm
Object is placed at a distance of 20cm from the lens.
An upright base / 6ft long
