Answer:
E = k Q / [d(d+L)]
Explanation:
As the charge distribution is continuous we must use integrals to solve the problem, using the equation of the elective field
E = k ∫ dq/ r² r^
"k" is the Coulomb constant 8.9875 10 9 N / m2 C2, "r" is the distance from the load to the calculation point, "dq" is the charge element and "r^" is a unit ventor from the load element to the point.
Suppose the rod is along the x-axis, let's look for the charge density per unit length, which is constant
λ = Q / L
If we derive from the length we have
λ = dq/dx ⇒ dq = L dx
We have the variation of the cgarge per unit length, now let's calculate the magnitude of the electric field produced by this small segment of charge
dE = k dq / x²2
dE = k λ dx / x²
Let us write the integral limits, the lower is the distance from the point to the nearest end of the rod "d" and the upper is this value plus the length of the rod "del" since with these limits we have all the chosen charge consider
E = k 
We take out the constant magnitudes and perform the integral
E = k λ (-1/x)
Evaluating
E = k λ [ 1/d - 1/ (d+L)]
Using λ = Q/L
E = k Q/L [ 1/d - 1/ (d+L)]
let's use a bit of arithmetic to simplify the expression
[ 1/d - 1/ (d+L)] = L /[d(d+L)]
The final result is
E = k Q / [d(d+L)]
Answer:
The trains mass in pounds would be 40084.029 if you would round it to the hundreths
Explanation:
Answer:
The velocity after 2 seconds can be found through:
V = u +a*t
Where V is final velocity, u is initial velocity, a is acceleration and t is time.
V = 0 + 2* 2= 4 meters/second
The distance (s) can be found through:
V^2= u^2 +2*a* s
Where V is final velocity, u is initial velocity, a is acceleration.
4^2= 0^2 + 2 *2*s
16= 0 + 4s
s= 4 meters
Distance (s) can also be found through:
s= ut + 1/2 at^2
s= 0+ 1/2 *2*2^2= 1 *2*2
s= 4 meters
Explanation:
In order to make his measurements for determining the Earth-Sun distance, Aristarchus waited for the Moon's phase to be exactly half full while the Sun was still visible in the sky. For this reason, he chose the time of a half (quarter) moon.
<h3 /><h3>How did Aristarchus calculate the distance to the Sun?</h3>
It was now possible for another Greek astronomer, Aristarchus, to attempt to determine the Earth's distance from the Sun after learning the distance to the Moon. Aristarchus discovered that the Moon, the Earth, and the Sun formed a right triangle when they were all equally illuminated. Now that he was aware of the distance between the Earth and the Moon, all he needed to know to calculate the Sun's distance was the current angle between the Moon and the Sun. It was a wonderful argument that was weakened by scant evidence. Aristarchus calculated this angle to be 87 degrees using only his eyes, which was not far off from the actual number of 89.83 degrees. But when there are significant distances involved, even slight inaccuracies might suddenly become significant. His outcome was more than a thousand times off.
To know more about how Aristarchus calculate the distance to the Sun, visit:
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Answer:
the air has to be unstable as well as it needs to be moved upwards.
Explanation:
it needs to be moved upwards and also needs to have unstable air.
