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melomori [17]
3 years ago
11

Vector a with arrow has a magnitude of 12.2 units and points due west. vector b with arrow points due north. (a) what is the mag

nitude of b with arrow if a with arrow + b with arrow has a magnitude of 17.4 units? 21.2 incorrect: your answer is incorrect. units (b) what is the direction of a with arrow + b with arrow relative to due west? ° correct: your answer is correct. (c) what is the magnitude of b with arrow if a with arrow â b with arrow has a magnitude of 17.4 units? units (d) what is the direction of a with arrow â b with arrow relative to due west? ° correct: your answer is correct.
Physics
1 answer:
Molodets [167]3 years ago
8 0
C = Sqrt (17.4^2 - 12.2^2) = 12.4
ArcTan (12.4/12.2) = 45.47 degrees above vector a,  which is 134.53 degrees measured on the xy-axis.

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A milliliter of very hot water is added to a liter of very cold water. Which of these events will occur? Assume the surrounding
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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
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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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