Response:
(b) 10 Wb
Clarification:
Given;
angle of the magnetic field, θ = 30°
initial area of the plane, A₁ = 1 m²
initial magnetic flux through the plane, Φ₁ = 5.0 Wb
The equation for magnetic flux is;
Φ = BACosθ
where;
B denotes the magnetic field strength
A represents the area of the plane
θ is the inclination angle
Φ₁ = BA₁Cosθ
5 = B(1 x cos30)
B = 5/(cos30)
B = 5.7735 T
Next, calculate the magnetic flux through a 2.0 m² section of the same plane:
Φ₂ = BA₂Cosθ
Φ₂ = 5.7735 x 2 x cos30
Φ₂ = 10 Wb
<pHence, the magnetic flux through a 2.0 m² area of the same plane is
10 Wb.Option "b"
Response:
E = ρ ( R1²) / 2 ∈o R
Clarification:
Provided information
Two cylinders are aligned parallel
Distance = d
Radial distance = R
d < (R2−R1)
To determine
Express the response using the variables ρE, R1, R2, R3, d, R, and constants
Solution
We have two parallel cylinders
therefore, area equals 2 
R × l
And we apply Gauss's Law
EA = Q(enclosed) / ∈o......1
Initially, we calculate Q(enclosed) = ρ Volume
Q(enclosed) = ρ ( R1² × l )
Thus, inserting all values into equation 1
produces
EA = Q(enclosed) / ∈o
E(2 R × l) = ρ ( R1² × l ) / ∈o
This simplifies to
E = ρ ( R1²) / 2 ∈o R
Ethylene glycol is known as the main component found in antifreeze.
The molecular formula for ethylene glycol is C₂H₆O₂.
Its molar mass is calculated as C₂H₆O₂ = (2×12) +(6×1) + (216) = 62g/mol
Given that antifreeze comprises 50% by weight, there exists 1 kg of ethylene glycol mixed with 1 kg of water.
ΔTf = Kf×m
ΔTf refers to the change in the freezing point.
= starting temperature of water - freezing temperature of the solution
= 0°C - Tf
= -Tf
Kf stands for the freezing point depression constant of water, which is 1.86°C/m
m is the molarity of the solution.
=(mass/molar mass) where mass of solvent is in kg
=1000g/62 (g/mol) /1kg
=16.13m
Substituting the value into the equation gives us
-Tf = 1.86 × 16.13 = 30
thus Tf = -30°C
The string does not experience any force of tension, as it balances two forces acting in the same direction. Hence, the tension is zero.
Explanation:
If tension existed in the string, it would mean that two equal but opposite forces are exerting pull in contrary directions.
When a force of f newtons is applied from the right and another force of f newtons from the left, the resulting action occurs through one force. Because there is action on the same string in opposing directions, the tension in the string can only be equal to the magnitude of the string itself.
Therefore, the string indeed has no tension since it is dealing with two forces acting in the same direction. Thus, the tension is zero.
