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klemol [59]
2 years ago
12

Find the center of mass of a 50-cm long rod, made of 25 cm of iron (density 9 g/cm3) and 25 cm of aluminum (density 3.2 g/cm3).

Express the location of the center of mass as measured from the left of the rod, assuming that the iron part of the rod is on the left and the aluminum part is on the right.
Physics
1 answer:
VashaNatasha [74]2 years ago
7 0

Answer:

The center of mass is at -1.58 cm, that is towards the left of the origin, or towards the copper end.

Explanation:

Given information: Let the length of the rod is 2l = 100cm

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You collect some more data on that horse at a later time interval, but now you are measuring thehorse’s velocity, not its positi
Monica [59]

Answer:

a)  x(t) = 10t + (2/3)*t^3

b) x*(0.1875) = 10.18 m

Explanation:

Note: The position of the horse is x = 2m. There is a typing error in the question. Otherwise, The solution to cubic equation holds a negative value of time t.

Given:

- v(t) = 10 + 2*t^2 (radar gun)

- x*(t) = 10 + 5t^2 + 3t^3  (our coordinate)

Find:

-The position x of horse as a function of time t in radar system.

-The position of the horse at x = 2m in our coordinate system

Solution:

- The position of horse according to radar gun:

                              v(t) = dx / dt = 10 + 2*t^2

- Separate variables:

                              dx = (10 + 2*t^2).dt

- Integrate over interval x = 0 @ t= 0

                             x(t) = 10t + (2/3)*t^3

- time @ x = 2 :

                              2 = 10t + (2/3)*t^3

                              0 = 10t + (2/3)*t^3 + 2

- solve for t:

                              t = 0.1875 s

- Evaluate x* at t = 0.1875 s

                              x*(0.1875) = 10 + 5(0.1875)^2 + 3(0.1875)^3

                              x*(0.1875) = 10.18 m

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V = IR
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