3energy storage consider the capacitor as a device for storing energy. the practical limitation on the potential difference between the plates of a capacitor is the "dielectric strength" of the dielectric material between the plates. this is the maximum field strength that can exist without causing a spark. it is typically on the order of 107 v/m for a good liquid dielectric with k 2.3 and ρ ~: 1 g/cm3 (a) neglecting the weight of the electrodes and case, what is the energy density of the capacitor in j/kg? le., how many joules of energy can be stored per kilogram of dielectric? physics 222 winter 2018 january 21, 2018 (b) the energy density of gasoline is about 46 mj/kg, and gasoline has a density of about 0.75 g/cm3. assume a gas tank holds about 12 gallons when full. what mass of dielectric would be required to provide the same energy as a tank of gasoline?

Is this a question of an answer

    m₁v₁ + m₂v₂ = (m₁ + m₂)vmomentum is conserved.the left-hand side of the equation represents the momentum in the system before the collision.the right-hand side represents the momentum in the system after the collisionm₁ and v₁ represents the mass and the velocity of the physics blockm₂ and v₂ represents the mass and velocity of the ball of puttyv represents the final velocity of the combined mass.the question states the two masses stick together. they combined mass has its own velocity with the body mass the sum of the two objects' massestheir combined momentum is (m₁ + m₂)vto find the magitude of (m₁ + m₂)v, we need to add the vectors on the left-hand side of the equation: m₁v₁ + m₂v₂.note the diagram. i have positioned the vectors in accordance with their directions.the magnitude of the resultant can be calculated using the pythagorean theorem since this is a right triangle situation (north and west are perpendicular)|(m₁ + m₂)v| =  √[ (m₁v₁)² + (m₂v₂)² ]|(m₁ + m₂)v| =  √[ (1.5kg·8.0m/s)² + (0.40kg·20m/s)² ]|(m₁ + m₂)v| =  14.4 kg m/sfor two significant figures (assuming that 20m/s is two), this rounds to 14 kg m/s

Chope i i hope i did

Following the american gauge wire standards ,the larger gauge number, the slimmer the wire and thus the smaller current carrying capacity it can have.then both statements would be incorrect, as they are saying the opposite of what it is actually is.(smaller gauge numbers have thicker wires and can carry more current)