In the circuit outside of the battery the electrons have to expend all of their energy on the internal resistance of the battery which causes heating
If the applied force is in the same direction as the object's displacement, the work done on the object is:
W = Fd
W = work, F = force, d = displacement
Given values:
F = 45N
d = 12m
Plug in and solve for W:
W = 45(12)
W = 540J
Answer:
Explanation:
1 )
Here
wave length used that is λ = 580 nm
=580 x 10⁻⁹
distance between slit d = .46 mm
= .46 x 10⁻³
Angular position of first order interference maxima
= λ / d radian
= 580 x 10⁻⁹ / .46 x 10⁻³
= 0.126 x 10⁻² radian
2 )
Angular position of second order interference maxima
2 x 0.126 x 10⁻² radian
= 0.252 x 10⁻² radian
3 )
For intensity distribution the formula is
I = I₀ cos²δ/2 ( δ is phase difference of two lights.
For angular position of θ1
δ = .126 x 10⁻² radian
I = I₀ cos².126x 10⁻²/2
= I₀ X .998
For angular position of θ2
I = I₀ cos².126x2x 10⁻²/2
= I₀ cos².126x 10⁻²
It's about 6 minutes, as I seem to recall. Sort of the time for the earth to go into shadow darkness when there's an eclipse ???
Explanation:
Distance travelled = Area under the line
= ut + ½ (v-u)t
Acceleration (a) = (v-u)/t and so (v-u) = at
Therefore,
Distance travelled (s) = ut + ½ (v-u)t = ut + ½ (at)t = ut + ½ at²
Thus,proved.
