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miss Akunina [59]
3 years ago
11

Three masses are located in the x-y plane as follows: a mass of 6 kg is located at (0 m, 0 m), a mass of 4 kg is located at (3 m

, 0 m), and a mass of 2 kg is located at (0 m, 3 m). Where is the center of mass of the system?
A. (1 m, 2 m)
B. (2 m, 1 m)
C. (1 m, 1 m)
D. (1 m, 0.5 m)
E. (0.5 m, 1 m)
Physics
1 answer:
Natali [406]3 years ago
7 0
<h2>Answer:</h2>

D. (1m, 0.5m)

<h2>Explanation:</h2>

The center of mass (or center of gravity) of a system of particles is the point where the weight acts when the individual particles are replaced by a single particle of equivalent mass. For the three masses, the coordinates of the center of mass C(x, y) is given by;

x = (m₁x₁ + m₂x₂ + m₃x₃) / M       ----------------(i)

y = (m₁y₁ + m₂y₂ + m₃y₃) / M       ----------------(ii)

Where;

M = sum of the masses

m₁ and x₁ = mass and position of first mass in the x direction.

m₂ and x₂ = mass and position of second mass in the x direction.

m₃ and x₃ = mass and position of third mass in the x direction.

y₁ , y₂ and y₃ = positions of the first, second and third masses respectively in the y direction.

From the question;

m₁ = 6kg

m₂ = 4kg

m₃ = 2kg

x₁ = 0m

x₂ = 3m

x₃ = 0m

y₁ = 0m

y₂ = 0m

y₃ = 3m

M = m₁ + m₂ + m₃ = 6 + 4 + 2 = 12kg

Substitute these values into equations (i) and (ii) as follows;

x = ((6x0) + (4x3) + (2x0)) / 12

x = 12 / 12

x = 1 m  

y = (6x0) + (4x0) + (2x3)) / 12

y = 6 / 12

y = 0.5m

Therefore, the center of mass of the system is at (1m, 0.5m)

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We have the following data:

Speed of light: c=3(10)^{8}m/s

Circumference of the equador: r=4(10)^{4}km \frac{1000 m}{1 km}=4(10)^{7}m

The time in which light turns around equator: t=5 min \frac{60 s}{1 min}=300 s

We need to find how many tours does light do around the equator in t=300 s

Let's begin by the expression of the speed, which is a relation between the traveled distance (d) and time:

c=\frac{d}{t} (1)

Isolating d:

d=c.t (2)

d=(3(10)^{8}m/s)(300 s) (3)

d=9(10)^{10}m (4) This is the distance light travels in 5 min.

Now we need to know to how many tours is this distance equivalent, we can know this by a Rule of three:

1 tour ---- r (circumference of the equator)

? tour ---- d

Then:

?=\frac{(1 tour)(d)}{r}

?=\frac{(1 tour)(9(10)^{10}m)}{4(10)^{7}m}

?=2250 tours This is the count of tours in 5 minutes.

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