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Schach [20]
4 years ago
15

The energy is water but i wanf to know how does it affect the enviorment?

Physics
1 answer:
Airida [17]4 years ago
7 0
It affects the environment because more animals and plants will grow. Down side is if there are bugs or insects that need water to live, if there's more water than more bugs. If bugs also feed on plants there could be an over population. 
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Someone help I'm stuck on question 8 and 12
Kaylis [27]
Sippen lein an hr later is theanswer to both 

4 0
4 years ago
Water flows through a cast steel pipe (k = 50 W m.K, ε = 0.8) with an outer diameter of 104mm and 2 mm wall thickness. Calculate
masha68 [24]

Answer:

The heat loss per unit length is   \frac{Q}{L}   = 2981 W/m

Explanation:

From the question we are told that

     The outer diameter of the pipe is d = 104mm = \frac{104}{1000} = 0.104 m

     The thickness is  D = 2mm = \frac{2}{1000} = 0.002m  

      The temperature  of water is  T = 90^oC = 90 + 273 = 363K  

      The outside air temperature is T_a = -10^oC = -10 +273 = 263K

        The water side heat transfer coefficient is z_1 = 300 W/ m^2 \cdot K

       The  heat transfer coefficient is  z_2 = 20 W/m^2 \cdot K

The heat lost per unit length is mathematically represented as

           \frac{Q}{L}   = \frac{2 \pi (T - Ta)}{ \frac{ln [\frac{d}{D} ]}{z_1}  +  \frac{ln [\frac{d}{D} ]}{z_2}}

Substituting values

         \frac{Q}{L}   = \frac{2 * 3.142 (363 - 263)}{ \frac{ln [\frac{0.104}{0.002} ]}{300}  +  \frac{ln [\frac{0.104}{0.002} ]}{20}}

           \frac{Q}{L}   = \frac{628}{0.2107}

           \frac{Q}{L}   = 2981 W/m

6 0
3 years ago
A concave spherical mirror has a radius of curvature of magnitude 20.0 cm. (a) Find the location of the image for object distanc
Tems11 [23]

Answer:

Answered

Explanation:

The radius of curvature of the mirror R = 20 cm

then the focal length f = R/2 = 10 cm

(a) From mirror formula

 1/f = 1/di + /1do

then the image distance

    di = fd_o / d_o - f

  = (10)(40) / 40-10

 = 30.76 cm

since the image distance is positive so the image is real

ii) when the object distance d_0=20 cm

 di = 10×20/ 20-10

= 20

Hence, the image must be  real

iii)when the object distance d_0 = 10

di = 10×10 / 10-10 =  ∞ (infinite)

the image will be formed at ∞

here also image will be real but diminished.

7 0
3 years ago
Read 2 more answers
A 2 feet piece of wire is cut into two pieces and once a piece is bent into a square and the other is bent into an equilateral t
SpyIntel [72]

Answer:

Explanation:

Let x ft be used to make square and 2-x ft be used to make equilateral triangle.

each side of square = x/4

area of square = ( x /4 )²

Each side of triangle

= (2-x) /3

Area of triangle = 1/2 (2-x)²/9 sin 60

= √3 / 36 x (2-x)²

Total area

A = ( x /4 )² +√3 / 36 (2-x)²

For maximum area

dA/dx = 0

1/16( 2x ) -√3 / 36 x2(2-x) = 0

x / 8 - √3(2-x)/ 18 = 0

x / 8 - √3/9 + √3/18 x = 0

x ( 1/8 + √3/18 ) = √3/9

x(.125 +.096 ) = .192

x = .868 ft

3 0
3 years ago
13. Suppose that the strength of the electric field about an isolated point charge has a certain value at a distance of 1 m. How
VMariaS [17]

Answer: E/4 ( one - fourth of it electric field)

Explanation:

The electric field of a point charge is given below as

E =kq/r²

E = electric field,

K = electric constant

q = magnitude of electric charge

r = distance between point charge and electric field.

It can be seen that only E and r are the only variable here and also, E is inversely proportional to r²

Which implies that

E = k/r² , k = E × r²

E1 ×(r1)² = E2 × (r2)²

Let E1 = E, r1 =1, r2 = 2 and E2 =?

Let us substitute the parameters

E × 1 = E2 × 2²

E × 1 = E2 × 4

E = E2 × 4

E2 = E/4

Which implies that the electric field at the second distance (r =4) is one fourth of the initial electric field.

5 0
3 years ago
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