Answer:by removing carbon dioxide from the body
Explanation:
I just got it right
Ok, so adopting that the 2nd satellite is at rest and that we're not moving anywhere near the speed of light (so no special relativity considerations), we can just add the two speed together, and say the 1st satellite is moving at 0.9m/s at the 2nd satellite. We can then set up our conservation of momentum equation, m₁v₁+m₂v₂ = m₁v₃+m₂v₄, where I'm calling v 1 and 2 the initial velocities of satellite 1 and 2 and v 3 and 4 the final velocities of satellite 1 and 2 respectively. We know, based on our chosen frame, that v₂ = 0, so that falls out to leave m₁v₁ = m₁v₃+m₂v₄, but we don't know v₃ or v₄, so we need another equation. Let's set up conversation of energy (elastic collisions conserve energy), where we only have to worry about kinetic energy (K = 1/2mv²) for each satellite before and after the collision. So we get 1/2m₁v₁²+1/2m₂v₂² = 1/2m₁v₃²+1/2m₂v₄². Now we have 2 equations and two unknown variables so let's solve with substitution. Let's solve the momentum equation for v₃, v₃ = (m₁v₁ - m₂v₄)/m₁, sub that into the energy equation, cancel the 1/2's and let's drop the v₂ terms since it's zero and we get: m₁v₁² = m₁((m₁v₁ - m₂v₄)/m₁)²+m₂v₄², then after some algebra we get v₄ = sqrt(m₁v₁/((v₁ - m₂/m₁)²+m₂)), then we plug in numbers v₄ = sqrt((4.5*10³*0.9/((0.9-(7.5/4.5))²+7.5*10³) = 0.73 m/s for the 2nd satellite after the collision. Then go back to v₃ = (m₁v₁ - m₂v₄)/m₁ and plug in numbers now that we know v₄ and we get v₃ = (4.5*10³*0.9 - 7.5*10³*0.73)/(4.5*10³) = -0.3167 m/s for the 1st satellite.
I think its inductance. If its not then I think its none of the above
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
