Suppose that the electric field in the Earth's atmosphere is E = 8.60 101 N/C, pointing downward. Determine the electric charge
1 answer:
Answer:
q = 3.87 x 10⁵ C
Explanation:
given,
Electric field, E = 8.60 x 10¹ = 86 N/C
radius of earth, R = 6371 Km = 6.371 x 10⁶ m
Coulomb constant, K = 9 x 10⁹ N · m²/C²
Charge on the earth = ?
the electric field at the point
inserting all the values
q = 3.87 x 10⁵ C
The electric charge on the earth is equal to 3.87 x 10⁵ C
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Answer:
v ’= 21.44 m / s
Explanation:
This is a doppler effect exercise that changes the frequency of the sound due to the relative movement of the source and the observer, the expression that describes the phenomenon for body approaching s
f ’= f (v + v₀) / (v-)
where it goes is the speed of sound 343 m / s, v_{s} the speed of the source v or the speed of the observer
in this exercise both the source and the observer are moving, we will assume that both have the same speed,
v₀ = v_{s} = v ’
we substitute
f ’= f (v + v’) / (v - v ’)
f ’/ f (v-v’) = v + v ’
v (f ’/ f -1) = v’ (1 + f ’/ f)
v ’= (f’ / f-1) / (1 + f ’/ f) v
v ’= (f’-f) / (f + f’) v
let's calculate
v ’= (3400 -3000) / (3000 +3400) 343
v ’= 400/6400 343
v ’= 21.44 m / s