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2 years ago

The most pressing problem in the Chesapeake Bay is

Physics

2 answers:

katovenus [111]2 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]2 years ago

3 0

Answer:

The answer is C. Pollution

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Timed! I would really appreciate some help! thank you!

GenaCL600 [577]

Answer:

x = 5[km]

Explanation:

We must convert the time from minutes to hours.

$t=30[min]*\frac{1h}{60min}= 0.5[h]\\$

We know that speed is defined as the relationship between space and time.

$v=x/t$

where:

x = space [m]

t = time = 0.5 [h]

v = velocity [m/s]

Now replacing:

$x = 10[\frac{km}{h} ]*0.5[h]\\x=5[km]$

4 0

2 years ago

Air (14.5 lb) undergoes a polytropic process in a closed system from p1 = 80 lbf/in2, υ1 = 4 ft3/lb to a final state where p2 =

Yanka [14]

The energy transfer in terms of work has the equation:

W = mΔ(PV)

To be consistent with units, let's convert them first as follows:

P₁ = 80 lbf/in² * (1 ft/12 in)² = 5/9 lbf/ft²
P₂ = 20 lbf/in² * (1 ft/12 in)² = 5/36 lbf/ft²
V₁ = 4 ft³/lbm
V₂ = 11 ft³/lbm

W = m(P₂V₂ - P₁V₁)
W = (14.5 lbm)[(5/36 lbf/ft²)(4 ft³/lbm) - (5/9 lbf/ft²)(11 lbm/ft³)]
W = -80.556 ft·lbf

In 1 Btu, there is 779 ft·lbf. Thus, work in Btu is:
W = -80.556 ft·lbf(1 Btu/779 ft·lbf)
<em>W = -0.1034 BTU</em>

4 0

2 years ago

What do you get when you subtract the force of air resistance from the force<br> of gravity?

atroni [7]

Answer:

Net force

Explanation:

3 0

2 years ago

Read 2 more answers

What is the chemical name for a sand

vovikov84 [41]

It is known as silicon dioxide or silica!

Hope this helps!

7 0

2 years ago

A typical helium-neon laser found in supermarket checkout scanners emits 633-nm-wavelength light in a 1.0-mm-diameter beam with

Fofino [41]

Answer:

Eo = 9.796 x 10^2 N/C

Bo = 3.266 x 10^-6 T

Explanation:

Given

Wavelength λ = 633 nm

Diameter of the beam D = 1.0 mm

Power P = 1.0 mW

Solution

Radius of the beam r = D/2 = 0.5 mm = 0.0005 m

Area of cross section

$A = \pi r^{2} \\A = 3.15 \times 0.0005^{2}\\A = 7.58 \times 10^{-7} m^{2}\\$

Intensity

$I = \frac{P}{A} \\I = \frac{0.001}{7.85\times 10^{-7}} \\I = 1273.885 {W}/{m^{2} }$

Amplitude of Electric Field

$E_{o} = \sqrt{\frac{2I}{ \epsilon_{o}c } } \\E_{o} = \sqrt{\frac{2 \times 1273.88}{ 8.85 \times 10^{-12} \times 3 \times 10^{8} } }\\E_{o} = 9.796 \times 10^{2}N/C$

Amplitude of Magnetic Field

$B_{o} = \sqrt{\frac{2 \mu_{o}I}{c } } \\B_{o} = \sqrt{\frac{2 \times 4 \times \pi \times 10^{-7} \times 1273.88}{ 3 \times 10^{8} } }\\B_{o} = 3.266 \times 10^{-6} T$

5 0

3 years ago