The answer is 4.0075 x 10^9
Wavelength= speed / frequency
so.....3× 10^8 / 7.26×10^14
= .413× 10^(-6)
in scientific notation= 4.13×10^(-7)
in nanometer = 413 nm
Current = (voltage) / (resistance)
= (35 volts) / (1,400 ohms)
= 25 milliamperes
Answer:
Explanation:
Using the magnification formula.
Magnification = Image distance(v)/object distance(u) = Image Height(H1)/Object Height(H2)
M = v/u = H1/H2
v/u = H1/H2...1
3) Given the radius of curvature of the concave lens R = 20cm
Focal length F = R/2
f = 20/2
f = 10cm
Object distance u = 5cm
Object height H2= 5cm
To get the image distance v, we will use the mirror formula
1/f = 1/u+1/v
1/v = 1/10-1/5
1/v = (1-2)/10
1/v =-1/10
v = -10cm
Using the magnification formula
(10)/5 = H1/5
10 = H1
H1 = 10cm
Image height of the peg is 10cm
4) If u = 15cm
1/v = 1/f-1/u
1/v = 1/10-1/15
1/v = 3-2/30
1/v = 1/30
v = 30cm
30/15 = H1/5
15H1 = 150
H1/= 10cm
5) if u = 20cm
1/v = 1/f-1/u
1/v = 1/10-1/20
1/v = 2-1/20
1/v = 1/20
v = 20cm
20/20 = H1/5
20H1 = 100
H1 = 5cm
6) If u = 30cm
1/v = 1/f-1/u
1/v = 1/10-1/30
1/v = 3-1/30
1/v = 2/30
v = 30/2 cm
v =>15cm
15/30 = Hi/5
30H1 = 75
H1 = 75/30
H1 = 2.5cm
Answer
given,
Time period= T = 1.5 s
If it's moving through equilibrium point at t₀= 0 with v = 1.0 m/s
v_max=1.00 m/s
we know,
v_ max=A ω
v = A sin (ωt)
-0.50= -1.00 sin (ωt)
sin (ωt) = 0.5



t = 0.125 s
we have time period T=1.5 it is the time to complete one oscillation
means from eq to right,then left,then eq,then left,then from right to eq
time taken for left = t/4 = 0.125/4 = 0.375 s
smallest value of time
=0.375 + 0.125
= 0.50 sec