D. convergent plate boundary involving an oceanic plate
Expression to calculate energy from voltage: E= V*Q where E= energy, V= voltage, and Q= charge
Additional help:
-To find the Voltage ( V )
[ V = I x R ] V (volts) = I (amps) x R (Ω)
-To find the Current ( I )
[ I = V ÷ R ] I (amps) = V (volts) ÷ R (Ω)
-To find the Resistance ( R )
[ R = V ÷ I ] R (Ω) = V (volts) ÷ I (amps)
I hope that helps to some extent-
The correct answer is
Air resistance
In fact, when a ball is in free fall, there are two forces acting on it:
- its weight (force of gravity), acting downward
- the air resistance, acting upward
The effect of the weight is to accelerate the ball, because its direction is the same as the direction of motion of the ball, while the effect of the air resistance is to slow down the ball, because its direction is opposite to that of the motion.
A watering can is used to hold a water that we will use to water the plants. The water has both mass and volume. Two watering cans are most often different by the volume they contain.
Many various units for volume are used but most often used unit is liter. In a metric system basic units are those such as meter, kilogram and liter while in imperial system units used are those such as foote, inch, pound and gallon.
Unit for volume in metric system is cubic meter. It is equal to a volume of a cube whose all sides measure 1m. This is equal to 1000L. For watering cans that contain several liters units used is decimeter cubed. 1dm^3 = 1L
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
