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Ronch [10]
3 years ago
12

Assume that the solar radiation incident on Earth is 1330 W/m2 (at the top of the atmosphere). Find the total power radiated by

the Sun, taking the average separation between Earth and the Sun to be 1.489 × 1011 m. Answer in units of W.
Physics
1 answer:
Temka [501]3 years ago
8 0

Answer:

The total power radiated by the sun is 3.71×10^26W

Explanation:

The Earth is spherical in shape

Therefore, Area = 4πr^2 = 4×3.143×(1.489×10^11)^2 = 2.79×10^23m^2

Power radiated by the Sun = solar radiation on Earth × Area = 1330W/m^2 × 2.79×10^23m^2 = 3.71×10^26W

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A 3-m-high, 7-m-wide rectangular gate is hinged at the top edge and is restrained by a fixed ridge. Determine the hydrostatic fo
Shalnov [3]

Answer:

The Hydrostatic force is   F  =  137.2 kN

The location of pressure center is  Z  = 1.333 \ m  

Explanation:

From the question we are told that

   The height of the gate is  h =  3 \ m

     The weight of the gate is  w =  7 \  m

      The height of the water is  h_w  =  2 \ m

       The density of water is \rho_w  =  1000 \ kg/m^3

Note used h_w for height of water and height of gate immersed by water since both have the same value

The area of the gate immersed in water  is mathematically represented as

         A =  h_w  * w

substituting values

         A =  2*  7

         A =  14  \ m^2

The hydrostatic force is mathematically represented as

          F  =  \rho_w * g * h_f * A

Where

            h_f =h-  h_w

           h_f =3 -2

           h_f = 1\ m  

So  

              F  =  1000 * 9.8 * 1 * 14

            F  =  137.2 kN

The center of pressure is mathematically represented as

        Z  =  h_f + \frac{I_g}{h_f * A}

Where I_g is the moment of inertia of the gate which mathematically represented as

            I_g =  \frac{w * h_w^2}{12}

The h_w is the height of gate immersed in water

            I_g =  \frac{7  * 2^2 }{12}

             I_g = 4.667\ kg  m^2

Thus  

        Z  = 1  + \frac{4.66}{1 * 14}

        Z  = 1.333 \ m

3 0
3 years ago
If the back of the truck is 1.3 m above the ground and the ramp is inclined at 22 ∘ , how much time do the workers have to get t
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Refer to the diagram shown below.

Assume that
(a) The piano rolls down on frictionless wheels,
(b) Wind resistance is negligible.

The distance along the ramp is
d = (1.3 m)/sin(22°) = 3.4703 m

The component of the piano's weight along the ramp is
mg sin(22°)
If the acceleration down the ramp is a, then
ma = mg sin(22°)
a = g sin(22°) = (9.8 m/s²) sin(22°) = 3.671 m/s²

The time, t, to travel down the ramp from rest is given by
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Answer: 1.375 s

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