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Svet_ta [14]
2 years ago
13

Three small but dense objects are located in the x-y plane as shown in the figure. The objects have the following masses: mA = 3

.19 kg, mB = 2.55 kg and mC = 1.41 kg.
Determine the x and the y coordinates of the center of the mass of this system. The objects are small in size, they can be treated as point masses.


x coordinate: ?

y coordinate: ?

Physics
1 answer:
satela [25.4K]2 years ago
3 0

The x and the y coordinates of the center of the mass of this system will be 43.1 m and 3.12 m respectively.

<h3>What is the center of mass?</h3>

A location is established in relation to an object or set of objects in the center of mass. It is the system's average position across all of its components.

Given data;

There are three little objects that are densely spaced out in the x-y plane.;

The value of the masses

m₁=1.41 kg

m₂=2.55 kg

m₃=3.19 kg

\rm X_{cm} =\frac{ (W_1x_1 + W_2x_2 + W_3x_3)}{(W_1 + W_2 + W_3)} \\\\\  X_{cm} ==\frac{1.41 \times 2 + 2.55 \times 6 +3.19 \times 4}{1.41+2.55+3.19} \\\\  X_{cm} =4.31 \ m

\rm Y_{cm} =\frac{ (W_1y_1 + W_2y_2 + W_3y_3)}{(W_1 + W_2 + W_3)} \\\\\  X_{cm} ==\frac{1.41 \times 2 + 2.55 \times 4 +3.19 \times 7}{1.41+2.55+3.19} \\\\  X_{cm} =3.12 \ m

Hence, the x and the y coordinates of the center of the mass of this system will be 4.31 m and 3.12 m respectively.

To learn more about the center of mass, refer;

brainly.com/question/8662931

#SPJ1

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<em>A person, with his ear to the ground, sees a huge stone strike the concrete pavement. A moment later two sounds are heard from the impact: one travels in the air and the other in the concrete, and they are 6.4 seconds apart. How far away did the impact occur? (Sound speed in the air: 343 meters per second, sound speed in concrete: 3000 meters per second)</em>

Sound is a manifestation of mechanical waves, which needs a medium to propagate themselves. Depending on the material, sound will take more or less time to travel a given distance. From statement, we know this time difference between air and concrete (\Delta t), in seconds:

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Where:

t_{C} - Time spent by the sound in concrete, in seconds.

t_{A} - Time spent by the sound in the air, in seconds.

By suposing that sound travels the same distance and at constant speed in both materials, we have the following expression:

\Delta t = \frac{x}{v_{A}}-\frac{x}{v_{C}}

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v_{A} - Speed of the sound in the air, in meters per second.

x - Distance traveled by the sound, in meters.

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x = \frac{\Delta t}{\frac{1}{v_{A}}-\frac{1}{v_{C}}  }

x = 2478.585\,m

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