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swat32
4 years ago
13

The electric potential difference between the ground and a cloud in a particular thunderstorm is 3.0 ✕ 109 V. What is the magnit

ude of the change in the electric potential energy of an electron that moves between the ground and the cloud?
Engineering
1 answer:
Naily [24]4 years ago
4 0

Answer:

The question is incomplete, below is the complete question "The electric potential difference between the ground and a cloud in a particular thunderstorm is 3.0*10^{9}V. What is the magnitude of the change in the electric potential energy of an electron that moves between the ground and the cloud?"

Answer:

U=3.0*10^{9}eV

Explanation:

data given,

Potential difference,V=3.0*10^9V

charge on an electron, q=1e.

Recall that the relationship between potential difference (v), charge(Q) and the potential energy(U) is expressed as

U=qV

from the question, we asked to determine potential energy given the charge and the potential difference.

Hence if we substitute values into the equation, we arrive at

U=qV\\U=3.0*10^{9}*1e\\U=3.0*10^{9}eV

Hence the magnitude of the change in the electric potential energy of an electron that moves between the ground and the cloud is 3.0*10^{9}eV

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Given an integer k, a set C of n cities c1, . . . , cn, and the distances between these cities dij = d(ci , cj ), for 1 ⤠i &lt
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Answer:

See explaination

Explanation:

2-Approximation Algorithm

Step 1: Choose any one city from the given set of cities C arbitrarily and put it in to a set H which is initially empty.

Step 2: For every city c in set C that is currently not present in set H compute min_distc = Minimum[ d(c, c1), d(c, c2), d(c, c3), ..... . . . . d(c, ci) ]

where c1, c2, ... ci are the cities in set H

and d(x, y) is the euclidean distance between city x and city y

Step 3: H = H ∪ {cx} where cx is the city have maximum value of min_dist over all possible cities c, computed in Step-2.

Step 4: Step-2 and Step-3 are iterated for k-1 times so that k cities are included int set H.

The set H is the required set of cities.

Example

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C = {0, 1, 2, 3}

d(0,1) = 10, d(0,2) = 7, d(0,3) = 6, d(1,2) = 8, d(1,3) = 5, d(2,3) = 12

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min_dist2 = min{dist(0,2)} = min{7} = 7

min_dist3 = min{dist(0,3)} = min{6} = 6

Step-3: Max{10, 7, 6} = 10

Step-4: cx = 1

Step-5: H = H ∪ cx = {0} \cup {1} = {0, 1}

Step-6: Cities c \not\in H are {2, 3}

min_dist2 = min{dist(0,2), dist(1,2)} = min{7, 8} = 7

min_dist3 = min{dist(0,3), dist(1,3)} = min{6, 5} = 5

Step-7: Max{7, 5} = 7

Step-8: cx = 2

Step-9: H = H \cup cx = {0, 1} \cup {2} = {0, 1, 2}

Result: The set H is {0, 1, 2}.

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