I would say B. A stable home and varied activities
Answer:
27.95[kW*min]
Explanation:
We must remember that the power can be determined by the product of the current by the voltage.

where:
P = power [W]
V = voltage [volt]
I = amperage [Amp]
Now replacing:
![P=110*8.47\\P=931.7[W]](https://tex.z-dn.net/?f=P%3D110%2A8.47%5C%5CP%3D931.7%5BW%5D)
Now the energy consumed can be obtained mediate the multiplication of the power by the amount of time in operation, we must obtain an amount in Kw per hour [kW-min]
![Energy = 931.7[kW]*30[days]*10[\frac{min}{1day} ]=279510[W*min]or 27.95[kW*min]](https://tex.z-dn.net/?f=Energy%20%3D%20931.7%5BkW%5D%2A30%5Bdays%5D%2A10%5B%5Cfrac%7Bmin%7D%7B1day%7D%20%5D%3D279510%5BW%2Amin%5Dor%2027.95%5BkW%2Amin%5D)
Answer:
the answer will be 5.6 j..
hope you like the answer.....
Coulomb's Law
Given:
F = 3.0 x 10^-3 Newton
d = 6.0 x 10^2 meters
Q1 = 3.3x 10^-8 Coulombs
k = 9.0 x 10^9 Newton*m^2/Coulombs^2
Required:
Q2 =?
Formula:
F = k • Q1 • Q2 / d²
Solution:
So, to solve for Q2
Q2 = F • d²/ k • Q1
Q2 = (3.0 x 10^-3 Newton) • (6.0 x 10^2 m)² / (9.0 x 10^9
Newton*m²/Coulombs²) • (3.3x 10^-8 Coulombs)
Q2 = (3.0 x 10^-3 Newton) • (360 000 m²) / (297 Newton*m²/Coulombs)
Q2 = 1080 Newton*m²/ (297 Newton*m²/Coulombs)
Then, take the reciprocal of the denominator and start
multiplying
Q2 = 1080 • 1 Coulombs/297
Q2 = 1080 Coulombs / 297
Q2 = 3.63636363636 Coulombs
Q2 = 3.64 Coulumbs
"Constant velocity" is another way of saying "zero acceleration".