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Mice21 [21]
3 years ago
14

if you knew the number of valence electrons in a nonmetal atom, how would you determine the valence of the element?

Physics
1 answer:
Vanyuwa [196]3 years ago
8 0
It would be funny because . I will not be good
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How does the density of fluid affect the magnitude of buoyancy acting on an object immersed in it
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Explanation:

more the density , more will be the buoyant force acting on it , less the density less will be the buoyant force acting on it. This is why people float in dead sea and sink in other seas

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However, sometimes levers are designed to increase the input force needed to move an object. Which of the following is most like
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Consider the following True/False statements:
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6) False

7) True

8) False

9) False

10) False

11) True

12) True

13) True

14) True

Explanation:

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3 years ago
Which situation is NOT the result of an unbalanced force acting on an object? A. an object speeds up B. an object maintains spee
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3 years ago
A tank is full of water. Find the work W required to pump the water out of the spout. (Use 9.8 m/s2 for g. Use 1000 kg/m3 as the
Sergio039 [100]

Answer:

W = 1.06 MJ

Explanation:

- We will use differential calculus to solve this problem.

- Make a differential volume of water in the tank with thickness dx. We see as we traverse up or down the differential volume of water the side length is always constant, hence, its always 8.

- As for the width of the part w we see that it varies as we move up and down the differential element. We will draw a rectangle whose base axis is x and vertical axis is y. we will find the equation of the slant line that comes out to be y = 0.5*x. And the width spans towards both of the sides its going to be 2*y = x.

- Now develop and expression of Force required:

                                             F = p*V*g

                                             F = 1000*(2*0.5*x*8*dx)*g

                                             F = 78480*x*dx

- Now, the work done is given by:

                                             W = F.s

- Where, s is the distance from top of hose to the differential volume:

                                             s = (5 - x)

- We have the work as follows:

                                            dW = 78400*x*(5-x)dx

- Now integrate the following express from 0 to 3 till the tank is empty:

                                           W = 78400*(2.5*x^2 - (1/3)*x^3)

                                           W = 78400*(2.5*3^2 - (1/3)*3^3)

                                           W = 78400*13.5 = 1058400 J

 

5 0
3 years ago
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