C) Natural and Artificial sources
Answer:
The charge on the drop is
q = 1.741 x 10 ⁻²¹ C
Explanation:
Electric field due to plates
Ef = V/d
Ef = 2033 V / (2.0 * 10^-2 m )
Ef = 101650 V/m
So, we can write
Ef * q = m*g
q = m*g / E
f
The mass can be equal using the density and the volume so:
m = ρ * v
The volume can be find as:
v = 2.298 x 10 ⁻ ¹⁶ m³
q = ρ * v * g / Ef
q = 81 x 10 ³ kg/ m³ * 2.2298 x 10 ⁻ ¹⁶ m³ * 9.8 m/s² / 101650 V/m
The charge on the drop is
q = 1.741 x 10 ⁻²¹ C
The answer is electricity
The solution would be like
this for this specific problem:
<span>
The force on m is:</span>
<span>
GMm / x^2 + Gm(2m) / L^2 = 2[Gm (2m) / L^2] ->
1
The force on 2m is:</span>
<span>
GM(2m) / (L - x)^2 + Gm(2m) / L^2 = 2[Gm (2m) / L^2]
-> 2
From (1), you’ll get M = 2mx^2 / L^2 and from
(2) you get M = m(L - x)^2 / L^2
Since the Ms are the same, then
2mx^2 / L^2 = m(L - x)^2 / L^2
2x^2 = (L - x)^2
xsqrt2 = L - x
x(1 + sqrt2) = L
x = L / (sqrt2 + 1) From here, we rationalize.
x = L(sqrt2 - 1) / (sqrt2 + 1)(sqrt2 - 1)
x = L(sqrt2 - 1) / (2 - 1)
x = L(sqrt2 - 1) </span>
= 0.414L
<span>Therefore, the third particle should be located the 0.414L x
axis so that the magnitude of the gravitational force on both particle 1 and
particle 2 doubles.</span>
At speeds over 30 mph, you should maintain a following distance of at least <u>three full seconds</u> behind the vehicle ahead of you.
As a general rule and common sense at a speed of 30 mph you can leave three full seconds so that you can achieve a prudent distance between the car you are driving and the car in front in order to be able to perform some kind of maneuver if an accident or unforeseen event occurs.
To count the full three seconds you can use the technique of counting the Mississippis as follows: Mississippi one, Mississippi two, Mississippi three.
<h3>What is an accident?</h3>
An accident is an unexpected event that generally causes damage, injury or negative consequences.
Learn more about accident at: brainly.com/question/28070413
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