Since you are looking at a right fringe, the answer is all
the time a multiple of the wavelength for the reason that the light makes
constructive intrusion. If you are looking at the center (zeroth) fringe, the
difference is (450)(0) = 0. For the first maximum, the difference is 450 nm.
For the second, it is 450 x 2 = 900 nm. Basically, for the nth maxima, the
difference is 450n.
Answer:
This is an incomplete question. The complete question is --
An individual white LED (light-emitting diode) has an efficiency of 20% and uses 1.0 W of electric power.
How many LEDs must be combined into one light source to give a total of 3.8W of visible-light output (comparable to the light output of a 100W incandescent bulb)?
And the answer is --
19 LEDs
Explanation:
The full form of LED is Light emitting diode.
It is given that the efficiency of the LED bulb is 20 %
1 LED uses power = 1 W
So the output power of 1 LED = 0.2 W
We need to find the power required to give a 3.8 W light.
Power required for 3.8 W = Number of LEDs required = (total required power / power required for 1 LED )
= 3.8 / 0.2
= 19
Therefore, the number of LEDs required is 19.
why did my answer get deleted??
oh yeah i put a link on there- oopsies.
I wont this time!
I got 30!
<span>Density is 3.4x10^18 kg/m^3
Dime weighs 1.5x10^12 pounds
The definition of density is simply mass per volume. So let's divide the mass of the neutron star by its volume. First, we need to determine the volume. Assuming the neutron star is a sphere, the volume will be 4/3 pi r^3, so
4/3 pi 1.9x10^3
= 4/3 pi 6.859x10^3 m^3
= 2.873x10^10 m^3
Now divide the mass by the volume
9.9x10^28 kg / 2.873x10^10 m^3 = 3.44588x10^18 kg/m^3
Since we only have 2 significant digits in our data, round to 2 significant digits, giving 3.4x10^18 kg/m^3
Now to figure out how much the dime weighs, just multiply by the volume of the dime.
3.4x10^18 kg/m^3 * 2.0x10^-7 m^3 = 6.8x10^11 kg
And to convert from kg to lbs, multiply by 2.20462, so
6.8x10^11 kg * 2.20462 lb/kg = 1.5x10^12 lb</span>
