Answer
Given:
Wavelength = λ = 18.7 cm
= 0.187 m
Amplitude, A = 2.34 cm
Velocity, v = 0.38 m/s
A) Calculate the angular frequency.
Angular frequency,
ω = 2π f
ω = 2π x 2.03
ω = 12.75 rad/s
B) Calculate the wave number:
C)
Since the wave is traveling in the -x direction, the sign is positive between x and t
y (x, t) = A sin(k x - ω t)
y (x, t) = 2.34 sin(33.59 x - 12.75 t)
Answer:2.53*10^-10F
Explanation:
C=£o£r*A/d
Where £ represents the permittivity constant
£o= 8.85*10^-12f/m
£r=6.3
A=150mm^2=0.015m^2
d=3.3mm= 0.0033m
C=8.85*10^-12*6.3*0.015/0.0033
C=8.85*6.3*10^-12*0.015/0.0033
C=55.755*0.015^-12/0.003
C=8.36/3.3*10^-13+3
C=2.53*10^-10F
Answer:
The outcome of adding 999mm to 100m is 101m.
Explanation:
That's my belief.
Answer: SG = 2.67
The specific gravity for the sand is 2.67
Explanation:
Specific gravity is determined by the formula: density of the substance/density of water
Provided information;
Mass of sand m = 100g
The volume of sand equals the volume of water it displaces
Vs = 537.5cm^3 - 500 cm^3
Vs = 37.5cm^3
Calculating density of sand = m/Vs = 100g/37.5 cm^3
Ds = 2.67g/cm^3
Density of water Dw = 1.00 g/cm^3
Thus, the specific gravity of the sand can be expressed as
SG = Ds/Dw
SG = (2.67g/cm^3)/(1.00g/cm^3)
SG = 2.67
The specific gravity of the sand stands at 2.67