It forces the mercury to rise, being pushed up the tube by pressing down on the dish.
I hope I could help :)
A is a good example, i suppose.
Answer:
864 mT
Explanation:
The magnetic field due to a long straight wire B = μ₀i/2πR where μ₀ = permeability of free space = 4π × 10⁻⁷ H/m, i = current in wire, and R = distance from center of wire to point of magnetic field.
The magnitude of magnetic field due to the first wire carrying current i = 2.70 A at distance R which is mid-point between the wires is B = μ₀i/2πR.
Since the other wire also carries the same current at distance R, the magnitude of the magnetic field is B = μ₀i/2πR.
The resultant magnetic field at B is B' = B + B = 2B = 2(μ₀i/2πR) = μ₀i/πR
Now R = 2.50 cm/2 = 1.25 cm = 1.25 × 10⁻² m and i = 2.70 A.
Substituting these into B' = μ₀i/πR, we have
B' = 4π × 10⁻⁷ H/m × 2.70 A/π(1.25 × 10⁻² m)
B = 10.8/1.25 × 10⁻⁵ T
B = 8.64 × 10⁻⁵ T
B = 864 × 10⁻³ T
B = 864 mT
Answer:
68.8 N 13.8°N of W
Explanation:
F₁ is 50 N 30°N of W. The terminal angle is 150°.
F₂ is 25 N 20°S of W. The terminal angle is -160°.
Graphically, you can add the vectors using head-to-tail method. Move F₂ so that the tail of the vector is at the head of F₁. The resultant vector will be from the tail of F₁ to the head of F₂.
Algebraically, find the x and y components of each vector.
F₁ₓ = 50 N cos(150°) = -43.3 N
F₁ᵧ = 50 N sin(150°) = 25 N
F₂ₓ = 25 N cos(-160°) = -23.5 N
F₂ᵧ = 25 N sin(-160°) = -8.6 N
The x and y components of the resultant vector are the sums:
Fₓ = -43.3 N + -23.5 N = -66.8 N
Fᵧ = 25 N + -8.6 N = 16.4 N
The magnitude of the resultant force is:
F = √(Fₓ² + Fᵧ²)
F = √((-66.8 N)² + (16.4 N)²)
F = 68.8 N
The direction of the resultant force is:
θ = tan⁻¹(Fᵧ / Fₓ)
θ = tan⁻¹(16.4 N / -66.8 N)
θ = 166.2°
θ = 13.8°N of W