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mars1129 [50]
3 years ago
13

fins the powee of a convex lens which forms a real and inverted image of magnification -1 of an object placed at a distance from

its optical centre​
Physics
1 answer:
anzhelika [568]3 years ago
7 0

Explanation:

hope i am correct

hii

how are you

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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
La ecuación de la posición de una esferita está dada por: r(t)=(2.Cos(πt) i-3.Sen(πt) j) (m) ¿Cuál es la velocidad de la esferit
katovenus [111]

Answer:

v = (-4.44 i^ + 6.66 j^ )  m/s, a_average =( 0 i^ -2π j^) m/s²

Explanation:

The expression left corresponds to an oscillatory movement (MAS), the speed is defined by

         v = dr / dt

the function of position

         r = 2 cos πt  i^  + 3 sin πt  j^

let us note that it is a movement in two dimensions

let's perform the derivative

          v = -2π sin πt  i^  + 3π cos πt  j^

we evaluate this expression for t = 0.25 s, remember that the angle is in radians

          v = -2π sin (π 0.25) i^ + 3π cos (π 0.25) j^

           v = (-4.44 i^ + 6.66 j^ )  m/s

To calculate the mean acceleration we use the expression

           a = (v_{f} - v_{o}) / Δt

 

indicates that the time is the first 3 s

       

we look for the initial velocity t = 0 s

           v₀ = 0 i ^ + 3π j ^

         

we look for the fine velocity, t = 3 s

          v_f = - 2π sin (π 3) + 3π cos (π 3) j ^

          v_f = 0 i ^ - 3π j ^

we calculate the average acceleration

            Δt = (3 -0) = 3 s

           a_average = (0-0) / 3 i ^ + (-3π - 3π) / 3

           a_average = (0 i ^ -2π j ^ ) m/s²

6 0
4 years ago
A 50 kg mass is placed 2 meters from the fulcrum. To balance the lever, a second object is placed 4 meters
julia-pushkina [17]

Answer:

mass =25 kg

using clockwise moment = anticlockwise moment

8 0
3 years ago
I need help answering 30A and B. Help a guy out?
dmitriy555 [2]

Uh so I'm no master at this subject, but all stuffs accelerate at 9.8 m/s squared. So you multiply the 9.8 and the 0.20 it's given for reasons unknown other than that's what I see in my notes... and that gives you 1.96 m/s squared.

As for B, I have no idea. I think you may multiply the 1.96 by 4. Tell me your thoughts and maybe we can work it out together

5 0
3 years ago
What are two different examples of positive exeleration
sdas [7]

Answer:Accelerating your car to get up to speed on a freeway

An airplane accelerating to take off

Accelerating out of the starting blocks at a track meet

Explanation:

5 0
3 years ago
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