<h2>The K.E of the charge is 1.02 x 10⁻¹⁷ J</h2>
Explanation:
When the charge of 2e is placed in between the plates .
The force applied on this charge by plates is = q E
here q is the magnitude of charge = 2 e = 2 x 1.6 x 10⁻¹⁹ C
and E is the magnitude of electric field intensity
The work done = Force x displacement
Thus W = q E x S
here S is displacement
Therefore W = 2 x 1.6 x 10⁻¹⁹ x 4 x 8
= 1.02 x 10⁻¹⁷ J
This work will be converted into the kinetic energy of charge .
Thus K.E = 1.02 x 10⁻¹⁷ J
Since toy is moving at constant speed that means that force that child is applying on toy is equal to force of friction.
Rate of speed that toy is moving is irelevant.
childs force is:
Fc = 2N
Fc = Ff (Ff -friction force)
Ff = a*Q
where Q is weight of the toy and a is friction
if we express a we get
a = F/Q = 2/8 = 0.25
Answer:

Explanation:
Given that;
I₀ = 9.55 A
f = 359 cycles/s
b = 72.2 cm
c = 32.5 cm
a = 80.2 cm
Using the formula;

where;



Replacing our values into above equation; we have:



Then the
is calculated as:



A. cleaning up after environmental disasters
To develop this problem it is necessary to apply the concepts related to Interference and the Two slit experiment.
Young's equation defines the separation between fringes this phenomenon as,

Where,
Wavelength
d = Separation between fringes
a = Slit width
D = Distance between the slits and screen
Then for the two case we have


Calculating the new separation between the fringes would be



We have then,


Therefore the correct answer is c.