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Llana [10]
2 years ago
13

The acceleration of a train moving from rest and its speed reaches 36m/sec in 9 sec​

Physics
1 answer:
jenyasd209 [6]2 years ago
4 0

Answer:

4 m/s^2

Explanation:

The acceleration is defined as: Δv/Δt (the difference of the velocity over a time period in which happens that difference).

Remember that a difference is calculated by subtracting the initial value of a physical quantity from its final value.

In our case:

Δv = Vfinal - Vinitial = 36m/s - 0 m/s = 36m/s

Δt = 9s

a = Δv/Δt = 36m/s / 9s = 4m/s^2

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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

 

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