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3 years ago

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As you know, the oceans cover most of the earth, and contain a huge amount of resources. Protecting these resources is one reaso

n to try and prevent the pollution of the ocean. Direct pollution into the ocean, such as trash dumping or chemical spillage, must be dealt with. Indirect pollution via the _______________ that drains into an ocean must be considered as well. A) aquifer B) bay C) pipe D) watershed

2 answers:

matrenka [14]3 years ago

8 0

A is the answer your looking for

hammer [34]3 years ago

3 0

Watershed is the correct answer

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The Large Hadron collider (LHC) is a huge piece of equipment designed and built in order to make new scientific discoveries. The

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Answer:

D

Explanation:

"The Large Hadron collider (LHC) is a <em>huge piece of equipment designed and built in order to make new scientific discoveries.</em>"

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3 years ago

4. How does the type of medium affect a sound wave?

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Answer:

The type of medium affects a sound wave as sound travels with the help of the vibration in particles.

Explanation:

As different mediums have different amount and size of particles, for example, the speed of sound is faster through solid than liquid as solids have closely packed particles whereas liquids are loosely packed.

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What type of electromagnetic radiation can be used as a medical tracer to diagnose disease by being sent through the body's circ

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2 years ago

Two neutron stars are separated by a distance of 1.0 x 1012 m. They each have a mass of 1.0 x 1028 kg and a radius of 1.0 x 103

son4ous [18]

To develop this problem it is necessary to apply the concepts related to Gravitational Potential Energy.

Gravitational potential energy can be defined as

$PE = -\frac{GMm}{R}$

As M=m, then

$PE = -\frac{Gm^2}{R}$

Where,

m = Mass

G =Gravitational Universal Constant

R = Distance /Radius

PART A) As half its initial value is u'=2u, then

$U = -\frac{2Gm^2}{R}$

$dU = -\frac{2Gm^2}{R}$

$dKE = -dU$

Therefore replacing we have that,

$\frac{1}{2}mv^2 =\frac{Gm^2}{2R}$

Re-arrange to find v,

$v= \sqrt{\frac{Gm}{R}}$

$v = \sqrt{\frac{6.67*10^{-11}*1*10^{28}}{1*10^{12}}}$

$v = 816.7m/s$

Therefore the velocity when the separation has decreased to one-half its initial value is 816m/s

PART B) With a final separation distance of 2r, we have that

$2r = 2*10^3m$

Therefore

$dU = Gm^2(\frac{1}{R}-\frac{1}{2r})$

$v = \sqrt{Gm(\frac{1}{2r}-\frac{1}{R})}$

$v = \sqrt{6.67*10^{-11}*10^{28}(\frac{1}{2*10^3}-\frac{1}{10^{12}})}$

$v = 1.83*10^7m/s$

Therefore the velocity when they are about to collide is $1.83*10^7m/s$

7 0

3 years ago

at certain times the demand for electric energy is low and electric energy is used to pump water to a reservoir 45 m above the g

Readme [11.4K]

The mass of water that must be raised is $5.25\cdot 10^7 kg$

Explanation:

Since the process is 70% efficiency, the power in output to the turbine can be written as

$P_{out} = 0.70 P_{in}$

where $P_{in}$ is the power in input.

The power in input can be written as

$P_{in} = \frac{W}{t}$

where

W is the work done in lifting the water

t = 3 h = 10,800 s is the time elapsed

The work done in lifting the water is given by

$W=mgh$

where

m is the mass of water

$g=9.8 m/s^2$ is the acceleration of gravity

h = 45 m is the height at which the water is lifted

Combining the three equations together, we get:

$P_{out} = 0.70 \frac{mgh}{t}$

Where

$P_{out} = 150 MW = 150\cdot 10^6 W$

And solving for m, we find:

$m=\frac{Pt}{0.70gh}=\frac{(1.50\cdot 10^6)(10800)}{(0.70)(9.8)(45)}=5.25\cdot 10^7 kg$

Learn more about power:

brainly.com/question/7956557

#LearnwithBrainly

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3 years ago