Two lightbulbs have powers P1 and P2 when separately connected across an ideal battery with potential difference ΔV. The bulbs a
1 answer:
Answer:
Explanation:
Lets take
Resistance of bulb 1 =R₁
Resistance of bulb 2 =R₂
As we know that power P
P= ΔV²/R
Given that voltage difference is same for both bulbs
So
P₁R₁= ΔV² --------1
P₂R₂= ΔV² -----------2
When these resistance are connected in series then equivalent resistance R
R=R₁+ R₂
The new power P'
P'=ΔV²/R
P'R=ΔV² ------3
From equation 1 ,2 and 3
P'(R₁+ R₂) = ΔV²
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Answer:
q = 3.6 10⁵ C
Explanation:
To solve this exercise, let's use one of the consequences of Gauss's law, that all the charge on a body can be considered at its center, therefore we calculate the electric field on the surface of a sphere with the radius of the Earth
r = 6 , 37 106 m
E = k q / r²
q = E r² / k
q =
q = 4.5 10⁵ C
Now let's calculate the charge on the planet with E = 222 N / c and radius
r = 0.6 r_ Earth
r = 0.6 6.37 10⁶ = 3.822 10⁶ m
E = k q / r²
q = E r² / k
q =
q = 3.6 10⁵ C
Answer:
2.59 m
Explanation:
Coordinates of origin = (0, 0)
Coordinates of Point p where the fly reach = (2.3 m, 1.2 m)
Use the distance formula of coordinates to find the distance between the origin and the point P.
d = 2.59 m
Thus, the distance between the origin and the point P is 2.59 m.