Answer:

Explanation:
Given that
J(r) = Br
We know that area of small element
dA = 2 π dr
I = J A
dI = J dA
Now by putting the values
dI = B r . 2 π dr
dI= 2π Br² dr
Now by integrating above equation


Given that
B= 2.35 x 10⁵ A/m³
r₁ = 2 mm
r₂ = 2+ 0.0115 mm
r₂ = 2.0115 mm

By putting the values


Answer:
1320336992.2512 m²
1320.33 kilometers or 509.79 miles
Explanation:
Energy transferred by the sun

Energy required by the United States is
(assumed)
Power

Area

Area of the solar collector would be 1320336992.2512 m²
Converting to km²


Converting to mi²


Each side of the square would be 1320.33 kilometers or 509.79 miles
Answer:
a. y(x,t)= 2.05 mm cos[( 6.98 rad/m)x + (744 rad/s).
b. third harmonic
c. to calculate frequency , we compare with general wave equation
y(x,t)=Acos(kx+ωt)
from ωt=742t
ω=742
ω=2*pi*f
742/2*pi
f=118.09Hz
Explanation:
A fellow student of mathematical bent tells you that the wave function of a traveling wave on a thin rope is y(x,t)=2.30mmcos[(6.98rad/m)x+(742rad/s)t]. Being more practical-minded, you measure the rope to have a length of 1.35 m and a mass of 3.38 grams. Assume that the ends of the rope are held fixed and that there is both this traveling wave and the reflected wave traveling in the opposite direction.
A) What is the wavefunction y(x,t) for the standing wave that is produced?
B) In which harmonic is the standing wave oscillating?
C) What is the frequency of the fundamental oscillation?
a. y(x,t)= 2.05 mm cos[( 6.98 rad/m)x + (744 rad/s).
b. lambda=2L/n
when comparing the wave equation with the general wave equation , we get the wavelength to be
2*pi*x/lambda=6.98x
lambda=0.9m
we use the equation
lambda=2L/n
n=number of harmonics
L=length of string
0.9=2(1.35)/n
n=2.7/0.9
n=3
third harmonic
c. to calculate frequency , we compare with general wave equation
y(x,t)=Acos(kx+ωt)
from ωt=742t
ω=742
ω=2*pi*f
742/2*pi
f=118.09Hz
Answer:
the yield of product is YP=46.835 % and the concentration of solids is
Cs = 27.33%
Explanation:
Assuming that all the solids and fats remains in the milk after the evaporation, then the mass of product mP will be
Mass of fat in 100 kg of milk = 100 kg* 0.037 = mP* 0.079
mP = 100 kg* 0.037/0.079 = 46.835 kg
then the yield YP of the product is
YP= mP / 100 kg = 46.835 kg / 100 kg = 46.835 %
YP= 46.835 %
the concentration of solids Cs is
Mass of solids in 100 kg of milk = 100 kg* 0.128 = 46.835 kg * Cs
Cs = 100 kg* 0.128 / 46.835 kg = 0.2733 = 27.33%
Cs = 27.33%
Answer:
Final mass=0.89kg
Final pressure=5.6bar
Explanation:
To find mass,m=v/v1
But v1=vf + x(vg-vf)
Vf= 0.001093m^3/kg
Vg= 0.3748m^3/kg
V1= 0.001093+0.5(0.3748-0.001093)
V1= 0.225m^3/kg
M= 0.20/0.225 =0.89kg
Final pressure will be:
V/V1= P/P1
Cross multiply
VP1=V1P
P1= 0.225×5/0.2
P1=:5.6 bar