We will determine the wavelength through the relationship given by the distance between slits, this relationship is given under the function
Here,
m = Number of order bright fringe
= Wavelength
d = Distance between slits
Both distance are the same, then
Rearranging to find the second wavelength
Therefore the wavelength of the light coming from the second monochromatic light source is 550.3nm
Answer:
213 nA
2.13 mA
851e^-t μA
Explanation:
We have a pretty straightforward question here.
Ohms Law states that the current in an electric circuit is directly proportional to the voltage and inversely proportional to the resistance in the circuit. It is mathematically written as
V = IR, since we need I, we can write that
I = V/R
a) at V = 1 mV
I = (1 * 10^-3) / 4.7 * 10^3
I = 2.13 * 10^-7 A or 213 nA
b) at V = 10 V
I = 10 / 4.7 * 10^3
I = 0.00213 A or 2.13 mA
c) at V = 4e^-t
I = 4e^-t / 4.7 * 10^3
I = 0.000851e^-t A or 851e^-t μA
Kepler's 3rd law is given as
P² = kA³
where
P = period, days
A = semimajor axis, AU
k = constant
Given:
P = 687 days
A = 1.52 AU
Therefore
k = P²/A³ = 687²/1.52³ = 1.3439 x 10⁵ days²/AU³
Answer: 1.3439 x 10⁵ (days²/AU³)
Answer:
0.96 m
Explanation:
First, convert km/h to m/s.
162.3 km/h × (1000 m/km) × (1 hr / 3600 s) = 45.08 m/s
Now find the time it takes to move 20 m horizontally.
Δx = v₀ t + ½ at²
20 m = (45.08 m/s) t + ½ (0 m/s²) t²
t = 0.4436 s
Finally, find how far the ball falls in that time.
Δy = v₀ t + ½ at²
Δy = (0 m/s) (0.4436 s) + ½ (-9.8 m/s²) (0.4436 s)²
Δy = -0.96 m
The ball will have fallen 0.96 meters.
