Answer:
386 m
Explanation:
Let's call the horizontal distance between the point of launch and the point of landing of the first package 
.
2 seconds after landing at 
, the plane has travelled a horizontal distance of 
.
At this new point, the second package is launched. Because it is launched under the same conditions as the first, its horizontal distance from its point of launch is also 
.
To determine , we determine the time of fall of the package. Being a vertical motion, the drop has an initial velocity, 
, of 0 m/s with acceleration, 
, and a distance, 
, of 160 m. We use the equation of motion:
In this time, the package, with a horizontal velocity of 50 m/s, travels the horizontal distance, , of
The distance apart between both packages is 
Answer:
no en ingleses ok porque es muy tedioso además de que no se porque estas traduciendo esto no va a a ver respuesta así que no se que harás
Explanation:
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Use the formula F = (9x10^9 Q * q) / r^2
Message me if you need help.
Answer:
C = 4,174 10³ V / m^{3/4}
, E = 7.19 10² / ∛x, E = 1.5 10³ N/C
Explanation:
For this exercise we can calculate the value of the constant and the electric field produced,
Let's start by calculating the value of the constant C
V = C 
C = V / x^{4/3}
C = 220 / (11 10⁻²)^{4/3}
C = 4,174 10³ V / m^{3/4}
To calculate the electric field we use the expression
V = E dx
E = dx / V
E = ∫ dx / C x^{4/3}
E = 1 / C x^{-1/3} / (- 1/3)
E = 1 / C (-3 / x^{1/3})
We evaluate from the lower limit x = 0 E = E₀ = 0 to the upper limit x = x, E = E
E = 3 / C (0- (-1 / x^{1/3}))
E = 3 / 4,174 10³ (1 / x^{1/3})
E = 7.19 10² / ∛x
for x = 0.110 cm
E = 7.19 10² /∛0.11
E = 1.5 10³ N/C
