1. Frequency: 
The frequency of a light wave is given by:

where
is the speed of light
is the wavelength of the wave
In this problem, we have light with wavelength

Substituting into the equation, we find the frequency:

2. Period: 
The period of a wave is equal to the reciprocal of the frequency:

The frequency of this light wave is
(found in the previous exercise), so the period is:

Answer
given,
current (I) = 16 mA
circumference of the circular loop (S)= 1.90 m
Magnetic field (B)= 0.790 T
S = 2 π r
1.9 = 2 π r
r = 0.3024 m
a) magnetic moment of loop
M= I A
M=
M=
M=
b) torque exerted in the loop



Pounds
If you are talking about the unit of measurement for weight is that of force it would be Newtons.
Answer:
74.86°C
Explanation:
P₂ = Vapour pressure of water at sea level = 760 mmHg
P₁ = Pressure at base camp = 296 mmHg
T₂ = Temperature of water = 373 K
ΔH°vap for H2O = 40.7 kJ/mol = 40700 J/mol
R = Gas constant = 8.314 J/mol K
From Claussius Clapeyron equation

T₁ = 347.996 K = 74.86°C
∴Water will boil at 74.86°C