Your electricity usage for a one month period is 500 kilowatt-hours. Assuming one month is equal to 30 days and that the electricity usage is uniform across the whole month, what would be the velocity of a 1 kg mass if it was given the kinetic energy equivalent to one day's electricity usage?

Divide 500 kW-h by 30 for the daily energy usage. Then convert that to Watt-seconds (joules). You should end up with 60 million.

Then set that equal to (1/2) M V^2 and solve for V.

To solve this problem, we need to find the kinetic energy and then calculate the velocity of the 1 kg mass.

Kinetic energy is given by the formula: KE = 0.5 * m * v^2, where KE is the kinetic energy, m is the mass, and v is the velocity.

First, let's calculate the kinetic energy equivalent to one day's electricity usage:

Electricity usage for one day = 500 kWh / 30 days = 16.67 kWh

To convert the kilowatt-hour (kWh) to joules (J), we use the conversion factor: 1 kWh = 3.6 * 10^6 J

Kinetic energy = 16.67 kWh * 3.6 * 10^6 J/kWh = 60 * 10^6 J

Now, let's find the velocity by rearranging the formula:

v^2 = (2 * KE) / m

v^2 = (2 * 60 * 10^6 J) / 1 kg

v = sqrt((2 * 60 * 10^6 J) / 1 kg)

v ≈ 10954 m/s

Therefore, the velocity of a 1 kg mass, given the kinetic energy equivalent to one day's electricity usage, would be approximately 10954 m/s.