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gtnhenbr [62]
3 years ago
7

To be considered a source of water pollution, the source must include a chemical

Physics
2 answers:
forsale [732]3 years ago
8 0

To be considered a source of water pollution, the source must include a chemical is false.

<u>Explanation: </u>

The water pollution may be caused either by chemicals or waste materials that doesn’t contain chemicals such as domestic wastes, organic compounds, heavy metals etc. The life on Earth mostly relies on water and air, and if these two gets polluted; the results will be dangerous.

Around 70% of the planet is covered with water and the present state of its purity is a matter of a serious thought. Looking over various pollutants, only chemical wastes are not responsible to pollute the water.

Instead, there is range of elements from plastic scraps to chemical wastes that comes from either the non-point sources or the point sources. There are organic waste materials, heavy metals, volcanic eruptions, Tsunamis, earthquakes, etc. are also responsible to contaminate water and affect the amphibians as well as humans.

monitta3 years ago
8 0

To be considered a source of water pollution, the source must include a chemical.

ANSWER: false    

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Which statement below could be used as evidence to support the claim that using nuclear energy to produce electricity is better
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Answer:

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Explanation:

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3 years ago
the jet plane travels along the vertical parabolic path. when it is at point a it has speed of 200 m/s, which is increasing at t
givi [52]

Explanation:

Here is the complete question i guess. The jet plane travels along the vertical parabolic path defined by y = 0.4x². when it is at point A it has speed of 200 m/s, which is increasing at the rate .8 m/s^2. Determine the magnitude of acceleration of the plane when it is at point A.

→ The tangential component of acceleration is rate of increase in the speed of plane so,

a_{t} = v = 0.8 m/s^{2}

→ Now we have to find out the radius of curvature at point A which is 5 Km (from the figure).

dy/dx = d(0.4x²)/dx

         = 0.8x

Take the derivative again,

d²y/dx² = d(0.8x)/dx

          = 0.8

at x= 5 Km

dy/dx = 0.8(5)

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p = \frac{[1+ (\frac{dy}{dx})^{2}]^{\frac{3}{2} }   }{\frac{d^{2y} }{dx^{2} } }

now insert the values,

p = \frac{[1+(4)^{2}]^{\frac{3}{2} }  }{0.8}  = 87.62 km

→ Now the normal component of acceleration is given by

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aₙ = 0.457 m/s²

→ Now the total acceleration is,

a = [(a_{t})^{2} +(a_{n} )^{2} ]^{0.5}

a = [(0.8)^{2} + (0.457)^{2}]^{0.5}

a = 0.921 m/s²

4 0
3 years ago
Any help guys? I am stuck on two problems.
Juli2301 [7.4K]

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Answer:

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Explanation:

7 0
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a) uniform velocity

b) zero or no acceleration

c) (see picture)

EXPLANATION:

(see picture)

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