Centripital acceleration. When an object moves in a circle, centripital acceleration acts to accelerate the object towards the center of that circle
Answer:
For a body moving at a uniform velocity you can calculate the speed by dividing the distance traveled by the amount of time it took, for example one mile in 1/2 hour would give you 2 miles per hour. If the velocity is non-uniform all you can say is what the average speed is.
Newton's law of universal gravitation, says that every particle attracts every other particle in the universe with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.
Or in simple words, every particle in the world attracts each other to themselves, but the particle with most mass would attract with more force compared to a particle with less mass.
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Answer:
a) ΔV₁ = 21.9 V, b) U₀ = 99.2 10⁻¹² J, c) U_f = 249.9 10⁻¹² J, d) W = 150 10⁻¹² J
Explanation:
Let's find the capacitance of the capacitor
C = 
C = 8.85 10⁻¹² (8.00 10⁻⁴) /2.70 10⁻³
C = 2.62 10⁻¹² F
for the initial data let's look for the accumulated charge on the plates
C = 
Q₀ = C ΔV
Q₀ = 2.62 10⁻¹² 8.70
Q₀ = 22.8 10⁻¹² C
a) we look for the capacity for the new distance
C₁ = 8.85 10⁻¹² (8.00 10⁻⁴) /6⁴.80 10⁻³
C₁ = 1.04 10⁻¹² F
C₁ = Q₀ / ΔV₁
ΔV₁ = Q₀ / C₁
ΔV₁ = 22.8 10⁻¹² /1.04 10⁻¹²
ΔV₁ = 21.9 V
b) initial stored energy
U₀ = 
U₀ = (22.8 10⁻¹²)²/(2 2.62 10⁻¹²)
U₀ = 99.2 10⁻¹² J
c) final stored energy
U_f = (22.8 10⁻¹²) ² /(2 1.04 10⁻⁻¹²)
U_f = 249.9 10⁻¹² J
d) the work of separating the plates
as energy is conserved work must be equal to energy change
W = U_f - U₀
W = (249.2 - 99.2) 10⁻¹²
W = 150 10⁻¹² J
note that as the energy increases the work must be supplied to the system
Well depending on what current the heater pulls im going to assume about 13, and 13A for the hair dryer, thats 26A on the 40A circuit.
I dont see how a lightbulb could overload the circuit.
Anyway, assuming the circuit is overloaded by some really big heater- the circuit would trip, the fuse would go and remain off. Most houses are fitted with seperate circuits for lights and sockets, so the light may remain on depending on the breaker board. - the reason for them all being able to run with the sudden overload may be due to a surge.
One solution to this is not to put such a large heater on the circuit with other appliances.
Another may be to dry your hair in the dark
