The resistance is increased when more and more bulbs are added to the circuit.
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Explanation:
Given:
m = 1.673 × 10^-27 kg
Q = q = 1.602 × 10^-19 C
r = 0.75 nm
= 0.75 × 10^-9 m
A.
Energy, U = (kQq)/r
Ut = 1/2 mv^2 + 1/2 mv^2
1.673 × 10^-27 × v^2 = (8.99 × 10^9 × (1.602 × 10^-19)^2)/0.75 × 10^-9
v = 1.356 × 10^4 m/s
B.
F = (kQq)/r^2
F = m × a
1.673 × 10^-27 × a = ((8.99 × 10^9 × (1.602 × 10-19)^2)/(0.075 × 10^-9)^2
a = 2.45 × 10^17 m/s^2.
The answer is C (the same number of valence electrons)
Answer:
The time where the avergae speed equals the instaneous speed is T/2
Explanation:
The velocity of the car is:
v(t) = v0 + at
Where v0 is the initial speed and a is the constant acceleration.
Let's find the average speed. This is given integrating the velocity from 0 to T and dividing by T:

v_ave = v0+a(T/2)
We can esaily note that when <u><em>t=T/2</em></u><u><em> </em></u>
v(T/2)=v_ave
Now we want to know where the car should be, the osition of the car is:

Where x_A is the position of point A. Therefore, the car will be at:
<u><em>x(T/2) = x_A + v_0 (T/2) + (1/8)aT^2</em></u>