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Fed [463]
3 years ago
7

A 3.04 kg particle is located on the x-axis at xm = −8 m, and a 5.61 kg particle is on the x-axis at xM = 3.56 m. Find the coord

inate of the center of mass of this two-particle system. Answer in units of m.
Physics
1 answer:
Murljashka [212]3 years ago
8 0

Answer:

center of mass = −0.50 m

Explanation:

given data

mass m1 = 3.04 kg

distance xm = -8 m

mass m2 = 5.61 kg

distance xM = 3.56 m

solution

we get here center of mass for n mass of system that is express as

center of mass = \frac{m_1x_1+m_2x_2......m_nx_n}{m_1+m_2...m_n}     ......................1

but we have only 2 particle system so we will get

center of mass = \frac{m1 \times xm+m2 \times xM}{m1+m2}      .................2

put here value and we will get

center of mass = \frac{3.04 \times (-8 )+5.61 \times 3.56}{3.04 + 5.61}

solve it we will get

center of mass = −0.50 m

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Answer:

W=173.48J

Explanation:

information we know:

Total force: F=45N

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distance: 4m

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In this case we need the formulas to calculate the components of the force (because to calculate the work we need the horizontal component of the force).

horizontal component: F_{x}=Fcos\theta

vertical component: F_{y}=Fsen\theta

but from the given information we know that F_{y}=12N

so, equation these two F_{y}=Fsen\theta and F_{y}=12N

Fsen\theta =12N

and we know the force F=45N, thus:

45sen\theta=12

now we clear for \theta

sen\theta =12/45\\\theta=sin^{-1}(12/45)\\\theta =15.466

the angle to the horizontal is 15.466°, with this information we can calculate the horizontal component of the force:

F_{x}=Fcos\theta

F_{x}=45cos(15.466)\\F_{x}=43.37N

whith this horizontal component we calculate the work to move the crate a distance of 4 m:

W=F_{x}*D\\W=(43.37N)(4m)\\W=173.48J

the work done is W=173.48J

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