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lyudmila [28]
3 years ago
12

What might be collected to monitor the amount of excess nitrogen in the local water system

Physics
1 answer:
Debora [2.8K]3 years ago
4 0

Answer:

If excess nitrogen is found in the crop fields, the drainage water can introduce it into ... have aquifers that can supply a lot of freshwater very near the land surface.

You might be interested in
At a certain altitude above the Earth's surface, the electric field has a magnitude of 135 V/m. How much energy is stored in 1.0
Paul [167]

Answer:

Explanation:

Given that,

Electric field E=135V/m

Energy stored in 1m³of air=?

The energy stored in an electric field is given as

u = ½ εo E²

Where

U is the energy stored

εo is permissivity and it value is 8.85×10^-12C²/N..m²

And E is the electric field

Then,

U=½×8.85×10^-12×135²

U=8.06×10^-8J/m³

Then, the energy stored in 1m³ of air is 8.06×10^-8 J/m³

4 0
3 years ago
What is the kinetic energy of a 2,000-kg boat moving at 5.0 m/s?
vlada-n [284]

Answer:

25kJ

Explanation:

K = 1/2 mv^2 = 1/2 x 2000kg x (5.0m/s)^2

= 25 x 10^3 J = 25kJ

6 0
3 years ago
To meet a U.S. Postal Service requirement, employees' footwear must have a coefficient of static friction of 0.5 or more on a sp
liq [111]

Answer:

Minimum time interval (t2)=0.90 SECONDS

Explanation:

  • coefficient of friction for employees footwear = 0.5
  • coefficient of friction for typical athletic shoe = 0.810
  • frictional force = coefficient of friction X acceleration due to gravity X mass of body
  • Acceleration due to gravity is a constant = 9.81 m/s
  • Let frictional force for employee footwear = FF1
  • Let frictional force for athletic footwear =FF2

                 FF1 = O.5 X 9.81 X mass of body

                         = 4.905 x mass of body

                  FF2 = 0.810 X 9.81 X mass of body

                          = 7.9461 x mass of body

The body started from rest there by making the initial velocity zero ( u = 0)

From d= ut + 1/2 a x t^{2}

  •      d = \frac{1}{2} x a x t^{2}  .....................................i  

            where d= distance and it is given as 3.25m

  •          F =ma  ...................................ii

making acceleration subject of the formula from equation ii

  •              a =\frac{F}{m}

         Making t subject of formula from equation (i)

  • t=\sqrt{\frac{2d}{(f/m} }

    where

  • \frac{FF1}{Mass of body} = 4.905
  • \frac{FF2}{Mass of body} =7.9461

  Let

  •            t1 = minimum time taken for frictional force for employee foot wear
  •                                 t1 = \sqrt{\frac{6.5}{4.905} } =1.15 seconds

  •                                  t2 = \sqrt{\frac{6.5}{7.9461} } = 0.90 seconds

 

THANK YOU

5 0
3 years ago
Earth’s orbital speed is approximately _____. 1600 kilometers per hour 110 kilometers per hour 43,000 kilometers per hour 107,00
zhannawk [14.2K]

Answer:

107,000 kilometers per hour.

Explanation:

5 0
4 years ago
Read 2 more answers
A rectangular field is of length 42 cm and breadth 25 m. Find the area of the field in SI unit. EXPLAIN STEP BY STEP​
lorasvet [3.4K]

Answer:

the area of the rectangular field is 10.5 m²

Explanation:

Given;

length of the rectangular field, L = 42 cm = 0.42 m

breadth of the rectangular field, b = 25 m

The area of the rectangular field is calculated as follows;

Area = Length x breadth

Area = 0.42 m x 25 m

Area = 10.5 m²

Therefore, the area of the rectangular field is 10.5 m²

3 0
3 years ago
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