Answer:
λ = 1.4 × 10^(-7) m
Explanation:
We are given;
distance of eye piece from the source;D = 1.5 m
distance between the virtual sources;d = 7.5 × 10^(-4) m
To find the wavelength, we will use the formula for fringe width;
X = λD/d
Where X is fringe width, λ is wavelength, while d and D remain as before.
Now, fringe width = eye-piece distance moved transversely/number of fringes
Eye piece distance moved transversely = 1.88 cm = 1.88 × 10^(-2) m
Thus,
Fringe width = (1.88 × 10^(-2))/10 = 1.88 × 10^(-3) m
Thus;
1.88 × 10^(-3) = λ(1.5)/(7.5 × 10^(-4))
λ = [1.88 × 10^(-3) × (7.5 × 10^(-4))]/1.5
λ = 1.4 × 10^(-7) m
Answer:

Explanation:
Since the cable touches the road at the mid point of two towers
so here we have vertex at that mid point taken to be origin
now the maximum height on the either side is given as

horizontal distance of the tower from mid point is given as

now from the equation of parabola we have



now we have

now we need to find the height at distance of 200 ft from center
so we have


Given Information:
Power = P = 100 Watts
Voltage = V = 220 Volts
Required Information:
a) Current = I = ?
b) Resistance = R = ?
Answer:
a) Current = I = 0.4545 A
b) Resistance = R = 484 Ω
Explanation:
According to the Ohm’s law, the power dissipated in the light bulb is given by

Where V is the voltage across the light bulb, I is the current flowing through the light bulb and P is the power dissipated in the light bulb.
Re-arranging the above equation for current I yields,

Therefore, 0.4545 A current is flowing through the light bulb.
According to the Ohm’s law, the voltage across the light bulb is given by

Where V is the voltage across the light bulb, I is the current flowing through the light bulb and R is the resistance of the light bulb.
Re-arranging the above equation for resistance R yields,

Therefore, the resistance of the bulb is 484 Ω
<span>Creating plasma can be dangerous because of the high amount of ENERGY needed to create it.</span>