Using a machine can speed up work very quickly as machines are way faster than humans. Machines also do not get exhausted unlike humans so can work over a long period of time non stop.
So it doesnt get over heated ...
A cooling system<span> works by sending a liquid </span>coolant<span> through passages in the </span>engine<span>block and heads. As the </span>coolant<span> flows through these passages, it picks up heat from the </span>engine<span>. The heated fluid then makes its way through a rubber hose to the radiator in the front of the </span>car<span>.</span>
Answer:
not too sure but i think it is : force - multiplied by -mass
Answer:
5.23 C
Explanation:
The current in the wire is given by I = ε/R where ε = induced emf in the wire and R = resistance of wire.
Now, ε = -ΔΦ/Δt where ΔΦ = change in magnetic flux = AΔB and A = area of loop and ΔB = change in magnetic field intensity = B₂ - B₁
B₁ = 0.670 T and B₂ = 0 T
ΔB = B₂ - B₁ = 0 - 0.670 T = - 0.670 T
A = πD²/4 where D = diameter of circular loop = 13.2 cm = 0.132 m
A = π(0.132 m)²/4 = 0.01368 m² = `1.368 × 10⁻² m²
ε = -ΔΦ/Δt = -AΔB/Δt = -1.368 × 10⁻² m² × (-0.670 T)/Δt= 0.9166 × 10⁻² Tm²/Δt
Now, the resistance R of the circular wire R = ρl/A' where ρ = resistivity of copper wire = 1.68 x 10⁻⁸ Ω.m, 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
R = ρl/A' = 1.68 x 10⁻⁸ Ω.m × π × 0.132 m÷π(2.25 × 10⁻³ m)²/4 = 0.88704/5.0625 = 0.1752 × 10⁻² Ω = 1.752 × 10⁻³ Ω
So, I = ε/R = 0.9166 × 10⁻² Tm²/Δt1.752 × 10⁻³ Ω
IΔt = 0.9166 × 10⁻² Tm²/1.752 × 10⁻³ Ω = 0.5232 × 10 C
Since ΔQ = It = 5.232 C ≅ 5.23 C
So the charge is 5.23 C
