The correct one is A. Technician Only.
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Answer:
The two metals expand differently.
Explanation:
The bimetallic strip has two metal strips positioned like a bridge, these strips connect the electrical circuit to the heating system. When these strips are linear or "down" they allow the electricity to move through the circuit to the heating system to turn the heat on. When the strips are "up" the disconnect the electricity flow, thus turning the heating system off, thus the room becomes cool/cold.
Answer:
because energy will be lost due to friction, sound, and heat (arguably similar to friction) and ENERGY MUST STAY THE SAME so it is IMPOSSIBLE for the ball to bounce higher than when dropped!
Answer:
1.24 C
Explanation:
We know that the magnitude of the induced emf, ε = -ΔΦ/Δt where Φ = magnetic flux and t = time. Now ΔΦ = Δ(AB) = AΔB where A = area of coil and change in magnetic flux = Now ΔB = 0 - 0.750 T = -0.750 T, since the magnetic field changes from 0.750 T to 0 T.
The are , A of the circular loop is πD²/4 where D = diameter of circular loop = 16.7 cm = 16.7 × 10⁻²m
So, ε = -ΔΦ/Δt = -AΔB/Δt= -πD²/4 × -0.750 T/Δt = 0.750πD²/4Δt.
Also, the induced emf ε = iR where i = current in the coil and R = resistance of wire = ρl/A where ρ = resistivity of copper wire =1.68 × 10⁻⁸ Ωm, l = length of wire = πD and A = cross-sectional area of wire = πd²/4 where d = diameter of wire = 2.25 mm = 2.25 × 10⁻³ m.
So, ε = iR = iρl/A = iρπD/πd²/4 = 4iρD/d²
So, 4iρD/d² = 0.750πD²/4Δt.
iΔt = 0.750πD²/4 ÷ 4iρD/d²
iΔt = 0.750πD²d²/16ρ.
So the charge Q = iΔt
= 0.750π(Dd)²/16ρ
= 0.750π(16.7 × 10⁻²m 2.25 × 10⁻³ m)²/16(1.68 × 10⁻⁸ Ωm)
= 123.76 × 10⁻² C
= 1.2376 C
≅ 1.24 C
Answer:
Heat needed = 71.19 J
Explanation:
Here heat required can be calculated by the formula
H = mL
M is the mass of water and L is the latent heat of vaporization.
Mass of water, m = 31.5 g = 0.0315 kg
Latent heat of vaporization of water = 2260 kJ/kg
Substituting
H = mL = 0.0315 x 2260 = 71.19 kJ
Heat needed = 71.19 J
