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mr Goodwill [35]
3 years ago
8

When cartographers represent the three-dimensional earth in two dimensions, what is likely to occur?

Physics
2 answers:
Harlamova29_29 [7]3 years ago
7 0
Distortion is likely to occur
Sladkaya [172]3 years ago
6 0

distortion is most likely to occur.

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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
Which of the following statements best explains why a book resting on a table is in equilibrium?
timofeeve [1]
I'm pretty sure the answer is d.The weight of the book and the table's upward force on the book are equal in magnitude but opposite in direction.

3 0
3 years ago
A small drop of water is suspended motionless in air by a uniform electric field that is directed upward and has a magnitude of
Gennadij [26K]

To solve this problem we will apply the concepts related to the electric field such as the smelting of the Force and the load (In this case the force is equivalent to the weight). Later we will apply the ratio of the total charge as a function of the multiplication of the number of electrons and their individual charge.

E = \frac{mg}{q}

Here,

m = mass

g = Acceleration due to gravity

Rearranging to find the charge,

q = \frac{mg}{E}

Replacing,

q = \frac{(3.37*10^{-9})(9.8)}{11000}

q = 3.002*10^{-12}C

Since the field is acting upwards the charge on the drop should be negative to balance it in air. The equation to find the number of electrons then is

q = ne

Here,

n = Number of electrons

e = Charge of each electron

n = \frac{q}{e}

Replacing,

n = \frac{3.002*10^{-12}}{1.6*10^{-19}}

n = 2.44*10^7

Therefore the number of electrons that reside on the drop is 2.44*10^7

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Write the terms involved in s=ut+1\2​
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s = displacement; u = initial velocity; t = time of motion

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AVprozaik [17]
Current= voltage divided by resistance
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