The intensity of the electric field is 30,000 N/C
Explanation:
The strength of the electric field produced by a single-point charge is given by the equation
where:
is the Coulomb's constant
q is the magnitude of the charge
r is the distance from the charge
In this problem, we have:
is the magnitude of the charge
r = 3 cm = 0.03 m is the distance at which we are calculating the field intensity
Substituting, we find:
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I don't think that 4m has anything to do with the problem.
anyway. here.
A___________________B_______C
where A is the point that the train was released.
B is where the wheel started to stick
C is where it stopped
From A to B, v=2.5m/s, it takes 2s to go A to B so t=2
AB= v*t = 2.5 * 2 = 5m
The train comes to a stop 7.7 m from the point at which it was released so AC=7.7m
then BC= AC-AB = 7.7-5 = 2.7m
now consider BC
v^2=u^2+2as
where u is initial speed, in this case is 2.5m/s
v is final speed, train stop at C so final speed=0, so v=0
a is acceleration
s is displacement, which is BC=2.7m
substitute all the number into equation, we have
0^2 = 2.5^2 + 2*a*2.7
0 = 6.25 + 5.4a
a = -6.25/5.4 = -1.157
so acceleration is -1.157m/(s^2)
The diameter of the sphere is 2cm.
<h3>How to calculate the diameter?</h3>
From the diagram, the first sphere on the ruler is at 4cn and the last sphere is at 12cm.
Therefore, the length will be:
= 12 - 4.
= 8cm
The diameter of one sphere will be:
= Length / 4
= 8/4
= 2
Therefore, the diameter of the sphere is 2cm.
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Answer:
the electric field strength of this charge is two times the strength of the other charge
Explanation:
Using the relationship between electric field and the charge, which is inversely proportionality. Let the the magnitude of the first charge be Q and the respective electric field be E. It implies that;
E1/E2 = Q2/Q1
E2 = E1 x Q1/Q2
= E x Q/ (Q/2)
= 2E
