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Gala2k [10]
3 years ago
6

The most pressing problem in the Chesapeake Bay is

Physics
2 answers:
katovenus [111]3 years ago
5 0

Answer:

pollution

Explanation:

Excess nutrients and sediment are among the leading causes of the bay's poor health. Nitrogen and phosphorus from sewage, farm and suburban runoff and air pollution can fuel the growth of algae blooms and formation of fish-stressing "dead zones" in the bay.

please give brainlest

densk [106]3 years ago
3 0

Answer:

The answer is C. Pollution

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I know how the voltage increasing/not enough/stays the same and how long the battery lasts with series and parallel connections.
kirill [66]

Answer:

It comes out the positive side of the battery and goes in to the negative side of the battery

Explanation:

There are already electrons in wires in a circuit before you add the battery. By adding the battery, you're giving the electrons the energy it needs to move along the circuit.

In a series circuit, the circuit is one continuous loop so there is only one path for the electrons to go - out of the positive side of the battery and around the circuit then goes back into the negative side of the battery.

However, with a parallel circuit, there are two or more ways the electrons can go so they take the path of least resistance. The electrons still go out the positive side of a battery but along the circuit, the electrons will go through the path of least resistance ( I tend to think of it like a net with holes in it - the lower the resistance the bigger the holes for the electrons to go through so more can fit in a set amount of time ) but the electrons still go out of the positive side and in through the negative

3 0
2 years ago
The magnetic field perpendicular to a single 16.7-cm-diameter circular loop of copper wire decreases uniformly from 0.750 T to z
sammy [17]

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

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4 years ago
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