Answer:
contaminated drinking water
deaths of sea creatures that are used as a food source
limits to potential economic activities such as a fishing
Explanation:
A watershed is a large area that comprises of drainage area of all the surrounding water bodies meeting at a common affluence point before draining into sea or ocean or any other large water body. Pollution in this area can pollute the small water streams flowing through it, thereby polluting the larger water body into which it drains.
Thus, the water extracted for drinking from such area will be contaminated. Pollution in larger water body can cause death of water creature and hence pose a threat to fishing.
Answer:
0 J
Explanation:
As work is force times displacement, if no displacement occurs, no work occurs.
Answer: Shorter
Explanation: Shadow is formed when an light source is obstructed by an opaque object. The closer the source, shorter is the length of the shadow. In fact, when the source is exactly overhead, no shadow of the object is formed.
June 21 marks the Summer solstice which means the Sun passes directly overhead Tropic of cancer (23.5° N) at noon. March 21 marks the equinox which means sun passes directly overhead equator (0°).
Shadow length of an object at 42° Northern latitude will be shorter on June 21 because the Sun will be closer to this latitude as compared to March 21.
The food substance being used which turned Fehling's solution to brick red is a reducing sugar.
<h3>What is Fehling's solution?</h3>
This is an indicator which is used to test for the presence of reducing sugar or aldehydes in a solution.
Brick red precipitate of copper(I) oxide is formed when it tests positive to compounds such as glucose etc.
Read more about Fehling solution here brainly.com/question/3262179
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Complete Question
A gas gun uses high pressure gas tp accelerate projectile through the gun barrel.
If the acceleration of the projective is : a = c/s m/s2
Where c is a constant that depends on the initial gas pressure behind the projectile. The initial position of the projectile is s= 1.5m and the projectile is initially at rest. The projectile accelerates until it reaches the end of the barrel at s=3m. What is the value of the constant c such that the projectile leaves the barrel with velocity of 200m/s?
Answer:
The value of the constant is 
Explanation:
From the question we are told that
The acceleration is 
The initial position of the projectile is s= 1.5m
The final position of the projectile is 
The velocity is 
Generally 
and acceleration is 
so
=> 
integrating both sides
Now for the limit
a = 200 m/s
b = 0 m/s
c = s= 3 m
d =
= 1.5 m
So we have
![[\frac{v^2}{2} ] \left | 200} \atop {0}} \right. = c [ln s]\left | 3} \atop {1.5}} \right.](https://tex.z-dn.net/?f=%5B%5Cfrac%7Bv%5E2%7D%7B2%7D%20%5D%20%5Cleft%20%7C%20200%7D%20%5Catop%20%7B0%7D%7D%20%5Cright.%20%20%3D%20c%20%5Bln%20s%5D%5Cleft%20%7C%203%7D%20%5Catop%20%7B1.5%7D%7D%20%5Cright.)
![\frac{200^2}{2} = c ln[\frac{3}{1.5} ]](https://tex.z-dn.net/?f=%5Cfrac%7B200%5E2%7D%7B2%7D%20%20%3D%20%20c%20ln%5B%5Cfrac%7B3%7D%7B1.5%7D%20%5D)
=> 
