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3 years ago

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Two cars are initially separated by 2500 m and traveling towards each other. One car travels at 4.5 m/s and the second car trave

ls at 8 m/s. How long does it take for the two cars to meet

1 answer:

mylen [45]3 years ago

7 0

Answer:

714.285s

Explanation:

use relative velocity

8-4.5 = 3.5m/s

x = 2500m

2500/3.5 = 714.285s = 700s (with sig figs)

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The pendulum consists of two slender rods AB and OC which have a mass of 3 kg/m. The thin plate has a mass of 12 kg/m2 . a) Dete

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Answer:

The answer is below

Explanation:

a) The location ӯ of the center of mass G of the pendulum is given as:

$y=\frac{0+(\pi*(0.3\ m) ^2*12kg/m^2*1.8\ m-\pi*(0.1\ m) ^2*12kg/m^2*1.8\ m)+0.75\ m*1.5\ m *3\ kg/m}{(\pi*(0.3\ m) ^2*12kg/m^2-\pi*(0.1\ m) ^2*12kg/m^2)+3\ kg/m^2*0.8\ m+3\ kg/m^2*1.5\ m} \\\\y=0.88\ m$

b) the mass moment of inertia about z axis passing the rotation center O is:

$I_G=\frac{1}{12}*3(0.8)(0.8)^2+ 3(0.8)(0.888)^2-\frac{1}{2}*(12)(\pi)(0.1)^2(0.1)^2 -(12)(\pi)(0.1)^2(1.8-\\0.888)^2+\frac{1}{2}*(12)(\pi)(0.3)^2(0.3)^2 +(12)(\pi)(0.3)^2(1.8-0.888)^2+\frac{1}{12}*3(1.5)(1.5)^2+\\3(1.5)(0.888-0.75)^2\\\\I_G=13.4\ kgm^2$

c) The mass moment of inertia about z axis passing the rotation center O is:

$I_o=\frac{1}{12}*3(0.8)(0.8)^2+ \frac{1}{3}* 3(1.5)(1.5)^2+\frac{1}{2}*(12)(\pi)(0.3)^2(0.3)^2 +(12)(\pi)(0.3)^2(1.8)^2-\\\frac{1}{2}*(12)(\pi)(0.1)^2(0.1)^2 -(12)(\pi)(0.1)^2(1.8)^2\\\\I_o=13.4\ kgm^2$

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3 years ago

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Answer:

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Explanation:

To solve this problem we must make a sketch of how will be the semicircle, for this reason we conducted an internet search, to find the scheme of the problem. This scheme is attached in the first image.

Then we make a free body diagram, with this free body diagram, we raise the forces that act on the body. Since it is a problem involving static equilibrium, the sum of forces in any direction and moments must be equal to zero.

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